CoT Reasoning Models – Which One Reigns Supreme in 2025?

So I have this problem: I need to compute the greatest common divisor (GCD) of the set of numbers of the form n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) where n>1n > 1n>1. At first glance, this seems a bit tricky, but let's break it down step by step.

First, I need to understand what the set looks like. For each integer n>1n > 1n>1, I can plug it into the expression n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) and get a number. So, for example, if n=2n = 2n=2, I'd get 299(260−1)2^{99}(2^{60} - 1)299(260−1), for n=3n = 3n=3, it's 399(360−1)3^{99}(3^{60} - 1)399(360−1), and so on. The goal is to find the GCD of all such numbers for every integer n>1n > 1n>1.

Now, finding the GCD of an infinite set might sound impossible, but probably there's a pattern or a property that can help simplify this. One thing I notice is that the expression involves n99n^{99}n99 and n60−1n^{60} - 1n60−1. Since n60−1n^{60} - 1n60−1 can be factored, maybe that will help.

Wait, actually, I recall that n60−1n^{60} - 1n60−1 can be factored using the difference of powers. Specifically, n60−1=(n30−1)(n30+1)n^{60} - 1 = (n^{30} - 1)(n^{30} + 1)n60−1=(n30−1)(n30+1), and each of those can be factored further. But I'm not sure if that's immediately helpful.

Alternatively, perhaps I can think about the properties of GCD. Since I'm dealing with a set of numbers, maybe I can find a common divisor and then show that it's the greatest one.

Let me try to compute n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) for a couple of small nnn to see if I can spot a pattern. For n=2n = 2n=2, as I said, it's 299(260−1)2^{99}(2^{60} - 1)299(260−1). For n=3n = 3n=3, it's 399(360−1)3^{99}(3^{60} - 1)399(360−1), and so on. But calculating these directly is impractical because the numbers are huge.

Maybe instead of specific values, I can consider the expression in a more general way. Since nnn is an integer greater than 1, perhaps I can think about the prime factors involved.

Wait, another idea: since the GCD is the largest number that divides all elements in the set, maybe I can find a number that divides n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) for all n>1n > 1n>1, and then show that no larger number does so.

But what could such a number be? Let's consider if there's a constant that divides n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) for all n>1n > 1n>1.

Hmm, actually, since n99n^{99}n99 is a factor, but it depends on nnn, it's not a constant. Wait, perhaps I can factor out something from n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1), but again, since nnn varies, that might not help.

Let me think differently. Maybe I can use properties of GCD with polynomials or something, but since nnn is an integer, perhaps number theory is more appropriate.

Wait a minute, I recall that in number theory, there are results about the GCD of values of a polynomial at integer points. For example, if you have a polynomial f(x)f(x)f(x), and you consider f(n)f(n)f(n) for integers nnn, sometimes you can find the GCD of all such f(n)f(n)f(n).

In this case, my expression is n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1), which is a polynomial in nnn, specifically of degree 99+60=15999 + 60 = 15999+60=159. So, perhaps there's a way to find the GCD of all f(n)f(n)f(n) where f(x)=x99(x60−1)f(x) = x^{99}(x^{60} - 1)f(x)=x99(x60−1) for integers x>1x > 1x>1.

I think there's a theorem that says that for a polynomial f(x)f(x)f(x) with integer coefficients, the GCD of all f(n)f(n)f(n) for integers nnn is equal to the GCD of the values of fff at certain points, or something like that. Wait, more precisely, I think that if f(x)f(x)f(x) is a polynomial with integer coefficients, then the GCD of f(n)f(n)f(n) for all integers nnn is equal to the GCD of the content of fff and f(k)f(k)f(k) for some kkk, but I'm not sure.

Actually, upon second thought, since f(x)f(x)f(x) has integer coefficients, the GCD of all f(n)f(n)f(n) should divide any specific f(a)f(a)f(a) for integer aaa, but that might not directly help.

Wait, perhaps I can consider the polynomial f(x)=x99(x60−1)f(x) = x^{99}(x^{60} - 1)f(x)=x99(x60−1), and think about its roots or something, but since we're dealing with integers, maybe not.

Another idea: maybe I can use the Euclidean algorithm in some way. For example, if I take two different values, say f(a)f(a)f(a) and f(b)f(b)f(b) for distinct integers a,b>1a, b > 1a,b>1, and compute gcd⁡(f(a),f(b))\gcd(f(a), f(b))gcd(f(a),f(b)), and see if there's a pattern or if it stabilizes.

Let me try that. Suppose I take n=2n = 2n=2 and n=3n = 3n=3, so f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1) and f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1). Now, since 2 and 3 are coprime, maybe I can find some common factors.

Wait, actually, since 2992^{99}299 and 3993^{99}399 are coprime, the common factors would have to come from (260−1)(2^{60} - 1)(260−1) and (360−1)(3^{60} - 1)(360−1). So, perhaps gcd⁡(f(2),f(3))\gcd(f(2), f(3))gcd(f(2),f(3)) is related to gcd⁡(260−1,360−1)\gcd(2^{60} - 1, 3^{60} - 1)gcd(260−1,360−1).

In fact, since 2992^{99}299 divides f(2)f(2)f(2) but not f(3)f(3)f(3), and similarly 3993^{99}399 divides f(3)f(3)f(3) but not f(2)f(2)f(2), the GCD should be gcd⁡(260−1,360−1)\gcd(2^{60} - 1, 3^{60} - 1)gcd(260−1,360−1), because the other parts don't share common factors.

Wait, is that accurate? Let's see: actually, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1). Since 2 and 3 are distinct primes, 2992^{99}299 and 3993^{99}399 are coprime, and also 2992^{99}299 doesn't divide 360−13^{60} - 1360−1 because 360−1≡−1(mod2)3^{60} - 1 \equiv -1 \pmod{2}360−1≡−1(mod2), and similarly 3993^{99}399 doesn't divide 260−12^{60} - 1260−1. So yes, the GCD should be gcd⁡(260−1,360−1)\gcd(2^{60} - 1, 3^{60} - 1)gcd(260−1,360−1).

Now, this is interesting. Maybe I can generalize this. Perhaps for any two distinct integers a,b>1a, b > 1a,b>1, gcd⁡(f(a),f(b))=gcd⁡(a60−1,b60−1)\gcd(f(a), f(b)) = \gcd(a^{60} - 1, b^{60} - 1)gcd(f(a),f(b))=gcd(a60−1,b60−1).

Wait, in the example above, it's gcd⁡(260−1,360−1)\gcd(2^{60} - 1, 3^{60} - 1)gcd(260−1,360−1), so yes, that seems to hold.

Moreover, perhaps the GCD of the entire set is related to the GCD of (n60−1)(n^{60} - 1)(n60−1) for all n>1n > 1n>1, but wait, that doesn't make sense because (n60−1)(n^{60} - 1)(n60−1) varies with nnn, and their GCD would be the GCD of all (n60−1)(n^{60} - 1)(n60−1), which might be 1 or something small.

Actually, let's think about gcd⁡(n60−1,m60−1)\gcd(n^{60} - 1, m^{60} - 1)gcd(n60−1,m60−1) for different n,mn, mn,m. I recall there's a property that gcd⁡(ak−1,bk−1)=(gcd⁡(a,b))k−1\gcd(a^k - 1, b^k - 1) = ( \gcd(a,b) )^k - 1gcd(ak−1,bk−1)=(gcd(a,b))k−1, but wait, that's not quite right. Actually, for positive integers a,b,ka, b, ka,b,k, it's known that gcd⁡(ak−1,bk−1)=(gcd⁡(a,b))k−1\gcd(a^k - 1, b^k - 1) = (\gcd(a,b))^k - 1gcd(ak−1,bk−1)=(gcd(a,b))k−1, but again, that's not accurate.

Wait, more precisely, if d=gcd⁡(a,b)d = \gcd(a,b)d=gcd(a,b), then gcd⁡(ak−1,bk−1)=dk−1\gcd(a^k - 1, b^k - 1) = d^k - 1gcd(ak−1,bk−1)=dk−1, but actually, that's also not true. For example, take a=2,b=4,k=1a = 2, b = 4, k = 1a=2,b=4,k=1, then gcd⁡(2−1,4−1)=gcd⁡(1,3)=1\gcd(2 - 1, 4 - 1) = \gcd(1,3) = 1gcd(2−1,4−1)=gcd(1,3)=1, and d=2d = 2d=2, 21−1=12^1 - 1 = 121−1=1, so it holds, but for k=2k = 2k=2, gcd⁡(22−1,42−1)=gcd⁡(3,15)=3\gcd(2^2 - 1, 4^2 - 1) = \gcd(3,15) = 3gcd(22−1,42−1)=gcd(3,15)=3, and 22−1=32^2 - 1 = 322−1=3, so yes, it seems gcd⁡(ak−1,bk−1)=(gcd⁡(a,b))k−1\gcd(a^k - 1, b^k - 1) = (\gcd(a,b))^k - 1gcd(ak−1,bk−1)=(gcd(a,b))k−1.

Wait, but in my earlier example, for a=2,b=3a = 2, b = 3a=2,b=3, since gcd⁡(2,3)=1\gcd(2,3) = 1gcd(2,3)=1, then gcd⁡(260−1,360−1)=160−1=0\gcd(2^{60} - 1, 3^{60} - 1) = 1^{60} - 1 = 0gcd(260−1,360−1)=160−1=0, but that can't be right because 260−12^{60} - 1260−1 and 360−13^{60} - 1360−1 are both positive integers, so their GCD is at least 1, but actually, since 2 and 3 are coprime, perhaps there's a better way.

Wait, actually, I think the correct statement is that for integers a,b>1a, b > 1a,b>1, gcd⁡(a−1,b−1)\gcd(a - 1, b - 1)gcd(a−1,b−1) divides gcd⁡(ak−1,bk−1)\gcd(a^k - 1, b^k - 1)gcd(ak−1,bk−1), but it's not necessarily equal. For instance, gcd⁡(2−1,3−1)=gcd⁡(1,2)=1\gcd(2 - 1, 3 - 1) = \gcd(1,2) = 1gcd(2−1,3−1)=gcd(1,2)=1, and gcd⁡(260−1,360−1)\gcd(2^{60} - 1, 3^{60} - 1)gcd(260−1,360−1) could be larger.

In fact, I know that ak−1=(a−1)(ak−1+ak−2+⋯+a+1)a^k - 1 = (a - 1)(a^{k-1} + a^{k-2} + \cdots + a + 1)ak−1=(a−1)(ak−1+ak−2+⋯+a+1), so a−1a - 1a−1 divides ak−1a^k - 1ak−1, and similarly b−1b - 1b−1 divides bk−1b^k - 1bk−1. But for different aaa and bbb, the GCD of ak−1a^k - 1ak−1 and bk−1b^k - 1bk−1 can be more complicated.

Alternatively, perhaps I can use the fact that ak−1a^k - 1ak−1 and bk−1b^k - 1bk−1 share common factors related to the divisors of kkk. For example, if k=60k = 60k=60, then a60−1a^{60} - 1a60−1 has factors like ad−1a^d - 1ad−1 for divisors ddd of 60.

Wait, actually, there's a standard result in number theory: for positive integers a,b,ka, b, ka,b,k, gcd⁡(ak−1,bk−1)=agcd⁡(k,log⁡ba)−1\gcd(a^k - 1, b^k - 1) = a^{\gcd(k, \log_b a)} - 1gcd(ak−1,bk−1)=agcd(k,logba)−1, but that doesn't make sense because log⁡ba\log_b alogba might not be integer.

Wait, more accurately, if aaa and bbb are powers of the same base, say a=cm,b=cna = c^m, b = c^na=cm,b=cn, then gcd⁡(ak−1,bk−1)=cgcd⁡(mk,nk)−1\gcd(a^k - 1, b^k - 1) = c^{\gcd(mk, nk)} - 1gcd(ak−1,bk−1)=cgcd(mk,nk)−1, but that's not helpful here.

Actually, for general a,ba, ba,b, I think the GCD of ak−1a^k - 1ak−1 and bk−1b^k - 1bk−1 is related to the GCD of a−1,b−1a - 1, b - 1a−1,b−1, but in a different way. For example, if a≡r(modm)a \equiv r \pmod{m}a≡r(modm) and b≡s(modm)b \equiv s \pmod{m}b≡s(modm), then ak≡rk(modm)a^k \equiv r^k \pmod{m}ak≡rk(modm), so ak−1≡rk−1(modm)a^k - 1 \equiv r^k - 1 \pmod{m}ak−1≡rk−1(modm), and similarly for bk−1b^k - 1bk−1. But I'm not sure.

Wait, perhaps a better approach is to consider specific primes or something. Let's try to find a candidate for the GCD of the set.

Let's denote d=gcd⁡({n99(n60−1)∣n>1})d = \gcd(\{n^{99}(n^{60} - 1) | n > 1\})d=gcd({n99(n60−1)∣n>1}). Then ddd divides n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) for every integer n>1n > 1n>1.

In particular, for each prime ppp, since p>1p > 1p>1, ddd divides p99(p60−1)p^{99}(p^{60} - 1)p99(p60−1). But actually, since ppp is prime, perhaps I can use that.

Wait, another idea: maybe I can factor n60−1n^{60} - 1n60−1 using its standard factorization. For example, n60−1=(n30−1)(n30+1)=(n15−1)(n15+1)(n30+1)n^{60} - 1 = (n^{30} - 1)(n^{30} + 1) = (n^{15} - 1)(n^{15} + 1)(n^{30} + 1)n60−1=(n30−1)(n30+1)=(n15−1)(n15+1)(n30+1), and so on, but perhaps more usefully, I can write it in terms of cyclotomic polynomials or something, but that might be overkill.

Alternatively, let's recall that for any integer mmm, if I take n=m+1n = m + 1n=m+1, then n60−1n^{60} - 1n60−1 can be related to mmm, but that might not help directly.

Wait, here's a thought: perhaps I can use the Euclidean algorithm on the polynomial level. For example, since nnn is an integer, maybe I can consider the GCD of f(n)f(n)f(n) for all n>1n > 1n>1, and relate it to the GCD of f(a)f(a)f(a) and f(b)f(b)f(b) for some a,ba, ba,b.

Actually, there's a standard technique for finding the GCD of values of a polynomial at integer points. Specifically, for a polynomial f(x)f(x)f(x) with integer coefficients, the GCD of f(n)f(n)f(n) for all integers nnn is equal to f(c)f(c)f(c), where ccc is chosen such that f(c)≠0f(c) \neq 0f(c)=0, and then adjusted for the content, but actually, it's more precise to say that if f(x)f(x)f(x) is not identically zero, then the GCD of {f(n)∣n∈Z}\{f(n) | n \in \mathbb{Z}\}{f(n)∣n∈Z} divides f(k)f(k)f(k) for any integer kkk, but since f(x)f(x)f(x) can be zero for some xxx, perhaps it's better to consider n>1n > 1n>1, as in the problem.

Wait, in this problem, since n>1n > 1n>1, and f(n)=n99(n60−1)f(n) = n^{99}(n^{60} - 1)f(n)=n99(n60−1), note that for n>1n > 1n>1, n99≠0n^{99} \neq 0n99=0, and n60−1≠0n^{60} - 1 \neq 0n60−1=0 because n60>1n^{60} > 1n60>1, so f(n)≠0f(n) \neq 0f(n)=0, and in fact, since n>1n > 1n>1, n60−1>0n^{60} - 1 > 0n60−1>0, so f(n)>0f(n) > 0f(n)>0.

So, perhaps I can use the fact that for a polynomial f(x)f(x)f(x) with integer coefficients, the GCD of f(n)f(n)f(n) for all integers nnn such that f(n)≠0f(n) \neq 0f(n)=0 is related to the fixed divisors or something, but I'm not recalling the exact theorem.

Alternatively, maybe I can use the property that if I take two values, say f(a)f(a)f(a) and f(a+1)f(a+1)f(a+1), and compute their GCD, and see if there's a pattern.

Let's try that. For example, take n=2n = 2n=2, so f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and n=3n = 3n=3, f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and as I thought earlier, gcd⁡(f(2),f(3))=gcd⁡(260−1,360−1)\gcd(f(2), f(3)) = \gcd(2^{60} - 1, 3^{60} - 1)gcd(f(2),f(3))=gcd(260−1,360−1).

But perhaps I can find a way to compute gcd⁡(260−1,360−1)\gcd(2^{60} - 1, 3^{60} - 1)gcd(260−1,360−1). Actually, there's a formula for gcd⁡(am−1,bm−1)\gcd(a^m - 1, b^m - 1)gcd(am−1,bm−1) when a,b>1a, b > 1a,b>1 and mmm is a positive integer. Specifically, it's known that gcd⁡(am−1,bm−1)=agcd⁡(m,log⁡ba)−1\gcd(a^m - 1, b^m - 1) = a^{\gcd(m, \log_b a)} - 1gcd(am−1,bm−1)=agcd(m,logba)−1, but again, since log⁡ba\log_b alogba may not be integer, that's not directly applicable.

Wait, more accurately, for integers a,b>1a, b > 1a,b>1, gcd⁡(am−1,bm−1)=(agcd⁡(m,k)−1)\gcd(a^m - 1, b^m - 1) = (a^{\gcd(m, k)} - 1)gcd(am−1,bm−1)=(agcd(m,k)−1), where kkk is such that b=akb = a^kb=ak, but if aaa and bbb are not powers of the same base, it's more complicated.

Actually, I think the correct generalization is that gcd⁡(am−1,bm−1)=gcd⁡(agcd⁡(m,n)−1,bgcd⁡(m,n)−1)\gcd(a^m - 1, b^m - 1) = \gcd(a^{\gcd(m, n)} - 1, b^{\gcd(m, n)} - 1)gcd(am−1,bm−1)=gcd(agcd(m,n)−1,bgcd(m,n)−1) or something, but I'm getting confused.

Wait, perhaps it's better to recall that for any integers a,b>1a, b > 1a,b>1 and positive integer mmm, gcd⁡(am−1,bm−1)\gcd(a^m - 1, b^m - 1)gcd(am−1,bm−1) divides (ab)gcd⁡(m)−1 (ab)^{\gcd(m)} - 1 (ab)gcd(m)−1, but that's not precise.

Alternatively, let's consider specific cases. For example, since 60 is a multiple of many numbers, perhaps I can use the fact that am−1a^m - 1am−1 has a certain structure.

Wait, here's a key observation: for any integer n>1n > 1n>1, since n60−1=(n30−1)(n30+1)n^{60} - 1 = (n^{30} - 1)(n^{30} + 1)n60−1=(n30−1)(n30+1), and further, n30−1=(n15−1)(n15+1)n^{30} - 1 = (n^{15} - 1)(n^{15} + 1)n30−1=(n15−1)(n15+1), etc., but maybe more importantly, I can think about the factorization over the integers or over the rationals.

Actually, since we're dealing with GCD, perhaps I can consider the prime factors that divide n60−1n^{60} - 1n60−1 for all n>1n > 1n>1.

But wait, for different nnn, n60−1n^{60} - 1n60−1 will have different prime factors. For example, for n=2n = 2n=2, 260−12^{60} - 1260−1 has certain prime factors, for n=3n = 3n=3, 360−13^{60} - 1360−1 has different ones, and so on. So, the common prime factors would be those that divide n60−1n^{60} - 1n60−1 for all n>1n > 1n>1.

But is there any prime that divides n60−1n^{60} - 1n60−1 for all n>1n > 1n>1? Let's see, for instance, take n=2n = 2n=2, then 260−12^{60} - 1260−1 is divisible by, say, 3, since 22≡1(mod3)2^2 \equiv 1 \pmod{3}22≡1(mod3), but actually 60=2⋅3060 = 2 \cdot 3060=2⋅30, so 260≡130≡1(mod3)2^{60} \equiv 1^{30} \equiv 1 \pmod{3}260≡130≡1(mod3), wait, 2≡−1(mod3)2 \equiv -1 \pmod{3}2≡−1(mod3), so 260≡(−1)60≡1(mod3)2^{60} \equiv (-1)^{60} \equiv 1 \pmod{3}260≡(−1)60≡1(mod3), thus 260−1≡0(mod3)2^{60} - 1 \equiv 0 \pmod{3}260−1≡0(mod3), so 3 divides 260−12^{60} - 1260−1.

But for n=3n = 3n=3, 360−1≡0(mod3)3^{60} - 1 \equiv 0 \pmod{3}360−1≡0(mod3), since 3≡0(mod3)3 \equiv 0 \pmod{3}3≡0(mod3), so 360≡060≡0(mod3)3^{60} \equiv 0^{60} \equiv 0 \pmod{3}360≡060≡0(mod3), and 0−1≡−1(mod3)0 - 1 \equiv -1 \pmod{3}0−1≡−1(mod3), wait, actually, 360=3⋅359≡0(mod3)3^{60} = 3 \cdot 3^{59} \equiv 0 \pmod{3}360=3⋅359≡0(mod3), so 360−1≡−1(mod3)3^{60} - 1 \equiv -1 \pmod{3}360−1≡−1(mod3), thus 3 does not divide 360−13^{60} - 1360−1, because −1≢0(mod3)-1 \not\equiv 0 \pmod{3}−1≡0(mod3).

Wait, really? Let's check for small mmm. For example, take m=1m = 1m=1, 31−1=23^1 - 1 = 231−1=2, which is not divisible by 3. For m=2m = 2m=2, 32−1=83^2 - 1 = 832−1=8, not divisible by 3. For m=3m = 3m=3, 33−1=263^3 - 1 = 2633−1=26, 26 / 3 = 8.666, not integer. Wait, but actually, since 3≡0(mod3)3 \equiv 0 \pmod{3}3≡0(mod3), 3k≡0(mod3)3^k \equiv 0 \pmod{3}3k≡0(mod3) for k≥1k \geq 1k≥1, so 3k−1≡−1(mod3)3^k - 1 \equiv -1 \pmod{3}3k−1≡−1(mod3), indeed not divisible by 3.

So, for n=3n = 3n=3, 360−13^{60} - 1360−1 is not divisible by 3, while for n=2n = 2n=2, 260−12^{60} - 1260−1 is divisible by 3. Therefore, there is no prime that divides n60−1n^{60} - 1n60−1 for all n>1n > 1n>1, because for each prime ppp, I can choose n=pn = pn=p, and then p60−1≡060−1≡−1(modp)p^{60} - 1 \equiv 0^{60} - 1 \equiv -1 \pmod{p}p60−1≡060−1≡−1(modp), so ppp does not divide p60−1p^{60} - 1p60−1.

Wait, more precisely, since p60≡0(modp)p^{60} \equiv 0 \pmod{p}p60≡0(modp), so p60−1≡−1(modp)p^{60} - 1 \equiv -1 \pmod{p}p60−1≡−1(modp), thus ppp does not divide p60−1p^{60} - 1p60−1. Therefore, for each prime ppp, there exists n=p>1n = p > 1n=p>1 such that ppp does not divide n60−1n^{60} - 1n60−1, and since n99n^{99}n99 is a power of ppp, but p99(p60−1)p^{99}(p^{60} - 1)p99(p60−1) is still not divisible by ppp, wait no, actually p99p^{99}p99 is divisible by ppp, but since p60−1≢0(modp)p^{60} - 1 \not\equiv 0 \pmod{p}p60−1≡0(modp), so p99(p60−1)≡0⋅(−1)=0(modp)p^{99}(p^{60} - 1) \equiv 0 \cdot (-1) = 0 \pmod{p}p99(p60−1)≡0⋅(−1)=0(modp), wait, actually, since p99≡0(modp)p^{99} \equiv 0 \pmod{p}p99≡0(modp), and p60−1≡−1(modp)p^{60} - 1 \equiv -1 \pmod{p}p60−1≡−1(modp), so p99(p60−1)≡0⋅(−1)=0(modp)p^{99}(p^{60} - 1) \equiv 0 \cdot (-1) = 0 \pmod{p}p99(p60−1)≡0⋅(−1)=0(modp), so yes, ppp does divide p99(p60−1)p^{99}(p^{60} - 1)p99(p60−1), but that doesn't help because for other nnn, it might not.

Wait, actually, since for each n>1n > 1n>1, n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) is an integer, and for different nnn, the prime factors can vary, but perhaps there's a way to find a common divisor.

Let's try to think of it differently. Maybe I can consider the expression n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) and see if there's a constant factor that divides it for all n>1n > 1n>1.

For example, is there a number ddd such that ddd divides n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) for all integers n>1n > 1n>1? Well, since n99n^{99}n99 is present, but it depends on nnn, perhaps not directly.

Wait, here's another idea: perhaps I can use the lifting-the-exponent lemma or something similar, but that might be for specific primes.

Alternatively, maybe I can consider the polynomial f(x)=x99(x60−1)f(x) = x^{99}(x^{60} - 1)f(x)=x99(x60−1), and think about its constant term or something, but since f(0)=0f(0) = 0f(0)=0, that doesn't help.

Wait, actually, let's consider what happens when n=k+1n = k + 1n=k+1 for some integer kkk. But that might not be straightforward.

Let's try to find a specific number that divides n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) for all n>1n > 1n>1. For instance, perhaps n60−1n^{60} - 1n60−1 has some common factors.

Wait, earlier I saw that for different nnn, n60−1n^{60} - 1n60−1 doesn't share all prime factors, but maybe there's a specific number that divides all n60−1n^{60} - 1n60−1 for n>1n > 1n>1.

But as I thought, since for n=pn = pn=p, a prime, p60−1p^{60} - 1p60−1 is not divisible by ppp, so there's no universal prime divisor.

Wait, perhaps the GCD is 1. Is that possible? Well, let's check if 1 is the only common divisor. But actually, since for large nnn, n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) can be very large, but that doesn't prevent them from sharing a common divisor greater than 1.

For example, let's see if 2 divides all n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1). Well, for n=2n = 2n=2, 299(260−1)2^{99}(2^{60} - 1)299(260−1), since 260−12^{60} - 1260−1 is odd, so 299⋅(odd)=even2^{99} \cdot (\text{odd}) = \text{even}299⋅(odd)=even, so yes, 2 divides it. But for n=3n = 3n=3, 399(360−1)3^{99}(3^{60} - 1)399(360−1), now 333 is odd, and 360≡160≡1(mod2)3^{60} \equiv 1^{60} \equiv 1 \pmod{2}360≡160≡1(mod2), so 360−1≡0(mod2)3^{60} - 1 \equiv 0 \pmod{2}360−1≡0(mod2), thus 399⋅(even)=even3^{99} \cdot ( \text{even} ) = \text{even}399⋅(even)=even, since 3993^{99}399 is odd, but odd times even is even, so yes, 2 divides 399(360−1)3^{99}(3^{60} - 1)399(360−1). Similarly, for any n>1n > 1n>1, if nnn is even, then n99n^{99}n99 is even, so f(n)f(n)f(n) is even; if nnn is odd, then n60n^{60}n60 is odd, so n60−1n^{60} - 1n60−1 is even, and n99n^{99}n99 is odd, so odd times even is even. So indeed, 2 divides n99(n60−1)n^{99}(n^{60} - 1)n99(n60−1) for all n>1n > 1n>1.

So, at least 2 is a common divisor. Is there a larger common divisor? For example, does 4 divide all f(n)f(n)f(n)? Let's check for n=2n = 2n=2, 299(260−1)2^{99}(2^{60} - 1)299(260−1), since 260−1≡(−1)60−1≡1−1=0(mod4)2^{60} - 1 \equiv (-1)^{60} - 1 \equiv 1 - 1 = 0 \pmod{4}260−1≡(−1)60−1≡1−1=0(mod4), wait, actually, for n=2n = 2n=2, since 2 is congruent to 2 mod 4, and 22≡0(mod4)2^2 \equiv 0 \pmod{4}22≡0(mod4), so for k≥2k \geq 2k≥2, 2k≡0(mod4)2^k \equiv 0 \pmod{4}2k≡0(mod4), but 260≡0(mod4)2^{60} \equiv 0 \pmod{4}260≡0(mod4), so 260−1≡−1(mod4)2^{60} - 1 \equiv -1 \pmod{4}260−1≡−1(mod4), thus 299(260−1)≡0⋅(−1)=0(mod4)2^{99}(2^{60} - 1) \equiv 0 \cdot (-1) = 0 \pmod{4}299(260−1)≡0⋅(−1)=0(mod4), but wait, actually, since 2992^{99}299 is divisible by 22=42^2 = 422=4, and 260−1≡−1(mod4)2^{60} - 1 \equiv -1 \pmod{4}260−1≡−1(mod4), but since 2992^{99}299 is divisible by higher powers of 2, in fact, 299(260−1)2^{99}(2^{60} - 1)299(260−1) is divisible by 2992^{99}299, so yes, 4 divides it.

But for n=3n = 3n=3, let's check 399(360−1)(mod4)3^{99}(3^{60} - 1) \pmod{4}399(360−1)(mod4). First, note that 3 ≡ -1 \pmod{4}, so 360≡(−1)60≡1(mod4)3^{60} \equiv (-1)^{60} \equiv 1 \pmod{4}360≡(−1)60≡1(mod4), thus 360−1≡0(mod4)3^{60} - 1 \equiv 0 \pmod{4}360−1≡0(mod4), so 360−13^{60} - 1360−1 is divisible by 4. Also, 399≡(−1)99≡−1(mod4)3^{99} \equiv (-1)^{99} \equiv -1 \pmod{4}399≡(−1)99≡−1(mod4), so 399(360−1)≡(−1)⋅0≡0(mod4)3^{99}(3^{60} - 1) \equiv (-1) \cdot 0 \equiv 0 \pmod{4}399(360−1)≡(−1)⋅0≡0(mod4), but to check if it's divisible by 4, since 360−1≡0(mod4)3^{60} - 1 \equiv 0 \pmod{4}360−1≡0(mod4), and 399≡−1(mod4)3^{99} \equiv -1 \pmod{4}399≡−1(mod4), but actually, since 360−1=(330−1)(330+1)3^{60} - 1 = (3^{30} - 1)(3^{30} + 1)360−1=(330−1)(330+1), and each factor is even, but let's see further. Actually, since 3 ≡ -1 \pmod{4}, and 32≡1(mod4)3^2 \equiv 1 \pmod{4}32≡1(mod4), so for even exponents, 32k≡1(mod4)3^{2k} \equiv 1 \pmod{4}32k≡1(mod4), thus 360=(32)30≡130≡1(mod4)3^{60} = (3^2)^{30} \equiv 1^{30} \equiv 1 \pmod{4}360=(32)30≡130≡1(mod4), so 360−1≡0(mod4)3^{60} - 1 \equiv 0 \pmod{4}360−1≡0(mod4), but in fact, since 32≡1(mod4)3^2 \equiv 1 \pmod{4}32≡1(mod4), and the order is 2, but for divisibility by higher powers, perhaps not.

Wait, actually, let's check for small exponents. For example, 32−1=83^2 - 1 = 832−1=8, which is divisible by 4, 34−1=803^4 - 1 = 8034−1=80, also divisible by 4, but 31−1=23^1 - 1 = 231−1=2, not divisible by 4, but since 60 is even, perhaps 360−13^{60} - 1360−1 is divisible by 4. In fact, since 32≡1(mod8)3^2 \equiv 1 \pmod{8}32≡1(mod8), wait, 3^2 = 9 ≡ 1 \pmod{8}, so for exponents that are multiples of 2, 32k≡1(mod8)3^{2k} \equiv 1 \pmod{8}32k≡1(mod8), but actually, for k≥1k \geq 1k≥1, 32k=(32)k≡1k≡1(mod8)3^{2k} = (3^2)^k \equiv 1^k \equiv 1 \pmod{8}32k=(32)k≡1k≡1(mod8), so 32k−1≡0(mod8)3^{2k} - 1 \equiv 0 \pmod{8}32k−1≡0(mod8), wait, 1 - 1 = 0, but 0 mod 8 is 0, so yes, 32k−1≡0(mod8)3^{2k} - 1 \equiv 0 \pmod{8}32k−1≡0(mod8), but since 8 divides 0, actually, since 32k≡1(mod8)3^{2k} \equiv 1 \pmod{8}32k≡1(mod8), so 32k−1≡0(mod8)3^{2k} - 1 \equiv 0 \pmod{8}32k−1≡0(mod8), and in fact, since 32−1=83^2 - 1 = 832−1=8, 34−1=80=8⋅103^4 - 1 = 80 = 8 \cdot 1034−1=80=8⋅10, etc., but for k=1k = 1k=1, 32−1=8=8⋅13^2 - 1 = 8 = 8 \cdot 132−1=8=8⋅1, for k=2k = 2k=2, 34−1=80=8⋅103^4 - 1 = 80 = 8 \cdot 1034−1=80=8⋅10, but 10 is not divisible by 2, so 34−1=803^{4} - 1 = 8034−1=80 is divisible by 16? 80 / 16 = 5, yes, so 34−1=16⋅53^4 - 1 = 16 \cdot 534−1=16⋅5, and 36−1=7283^6 - 1 = 72836−1=728, 728 / 16 = 45.5, not integer, wait, 728 / 8 = 91, but 91 is odd, so 728 = 8 * 91, not divisible by 16. So for m=2km = 2km=2k, 3m−13^m - 13m−1 is divisible by 8, but not necessarily by 16.

Similarly, for n=3n = 3n=3, since 360−13^{60} - 1360−1 is divisible by 8, and 3993^{99}399 is odd, so 399(360−1)3^{99}(3^{60} - 1)399(360−1) is divisible by 8. But for n=4n = 4n=4, let's check if 16 divides f(n)f(n)f(n).

Wait, perhaps instead, let's consider if there's a larger common divisor. For example, does 3 divide all f(n)f(n)f(n)? Let's check for n=2n = 2n=2, 299(260−1)2^{99}(2^{60} - 1)299(260−1), as I calculated earlier, since 2≡2(mod3)2 \equiv 2 \pmod{3}2≡2(mod3), and 21≡2(mod3)2^1 \equiv 2 \pmod{3}21≡2(mod3), 22≡1(mod3)2^2 \equiv 1 \pmod{3}22≡1(mod3), 23≡2(mod3)2^3 \equiv 2 \pmod{3}23≡2(mod3), so 260=(22)30≡130≡1(mod3)2^{60} = (2^2)^{30} \equiv 1^{30} \equiv 1 \pmod{3}260=(22)30≡130≡1(mod3), thus 260−1≡0(mod3)2^{60} - 1 \equiv 0 \pmod{3}260−1≡0(mod3), so 3 divides 260−12^{60} - 1260−1, and since 2992^{99}299 is not divisible by 3, but since 260−12^{60} - 1260−1 is divisible by 3, so f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1) is divisible by 3.

For n=3n = 3n=3, 399(360−1)3^{99}(3^{60} - 1)399(360−1), clearly divisible by 3 since 3993^{99}399 is.

For n=4n = 4n=4, 499(460−1)4^{99}(4^{60} - 1)499(460−1), since 4 ≡ 1 \pmod{3}, so 460≡160≡1(mod3)4^{60} \equiv 1^{60} \equiv 1 \pmod{3}460≡160≡1(mod3), thus 460−1≡0(mod3)4^{60} - 1 \equiv 0 \pmod{3}460−1≡0(mod3), so 3 divides 460−14^{60} - 1460−1, and 499≡199≡1(mod3)4^{99} \equiv 1^{99} \equiv 1 \pmod{3}499≡199≡1(mod3), so f(4)f(4)f(4) is divisible by 3.

In fact, for any n>1n > 1n>1, if n≡0(mod3)n \equiv 0 \pmod{3}n≡0(mod3), then n99n^{99}n99 is divisible by 3, so f(n)f(n)f(n) is divisible by 3. If n≢0(mod3)n \not\equiv 0 \pmod{3}n≡0(mod3), then since the multiplicative order of nnn mod 3 is 1 or 2, and since 60 is even, n60≡(n2)30≡130≡1(mod3)n^{60} \equiv (n^2)^{30} \equiv 1^{30} \equiv 1 \pmod{3}n60≡(n2)30≡130≡1(mod3), because n2≡1(mod3)n^2 \equiv 1 \pmod{3}n2≡1(mod3) for n≢0(mod3)n \not\equiv 0 \pmod{3}n≡0(mod3), since n≡1n \equiv 1n≡1 or 2≡−1(mod3)2 \equiv -1 \pmod{3}2≡−1(mod3), and (−1)2=1(-1)^2 = 1(−1)2=1, so yes, n60≡1(mod3)n^{60} \equiv 1 \pmod{3}n60≡1(mod3), thus n60−1≡0(mod3)n^{60} - 1 \equiv 0 \pmod{3}n60−1≡0(mod3), so 3 divides n60−1n^{60} - 1n60−1, and hence 3 divides f(n)f(n)f(n).

Therefore, 3 divides f(n)f(n)f(n) for all n>1n > 1n>1.

Similarly, perhaps I can check for other primes. For example, does 5 divide all f(n)f(n)f(n)? Let's see for n=2n = 2n=2, 299(260−1)2^{99}(2^{60} - 1)299(260−1), need to check if 5 divides 260−12^{60} - 1260−1. Since the order of 2 mod 5 is 4, and 60 is divisible by 4, so 24≡1(mod5)2^4 \equiv 1 \pmod{5}24≡1(mod5), thus 260=(24)15≡115≡1(mod5)2^{60} = (2^4)^{15} \equiv 1^{15} \equiv 1 \pmod{5}260=(24)15≡115≡1(mod5), so 260−1≡0(mod5)2^{60} - 1 \equiv 0 \pmod{5}260−1≡0(mod5), thus 5 divides 260−12^{60} - 1260−1, and since 2992^{99}299 is not divisible by 5, but f(2)f(2)f(2) is divisible by 5.

For n=5n = 5n=5, 599(560−1)5^{99}(5^{60} - 1)599(560−1), clearly divisible by 5.

For n=3n = 3n=3, let's check 360−1(mod5)3^{60} - 1 \pmod{5}360−1(mod5). Since 3 mod 5 has order 4, and 60 ≡ 0 mod 4, so 34≡1(mod5)3^4 \equiv 1 \pmod{5}34≡1(mod5), thus 360=(34)15≡115≡1(mod5)3^{60} = (3^4)^{15} \equiv 1^{15} \equiv 1 \pmod{5}360=(34)15≡115≡1(mod5), so 360−1≡0(mod5)3^{60} - 1 \equiv 0 \pmod{5}360−1≡0(mod5), thus 5 divides 360−13^{60} - 1360−1, and hence 5 divides f(3)f(3)f(3).

Similarly, for any n>1n > 1n>1, if n≡0(mod5)n \equiv 0 \pmod{5}n≡0(mod5), then n99n^{99}n99 is divisible by 5, so f(n)f(n)f(n) is. If n≢0(mod5)n \not\equiv 0 \pmod{5}n≡0(mod5), then since the order of nnn mod 5 divides 4, and 60 is divisible by 4, so n60≡1(mod5)n^{60} \equiv 1 \pmod{5}n60≡1(mod5), thus n60−1≡0(mod5)n^{60} - 1 \equiv 0 \pmod{5}n60−1≡0(mod5), so 5 divides n60−1n^{60} - 1n60−1, and hence 5 divides f(n)f(n)f(n).

Therefore, 5 also divides f(n)f(n)f(n) for all n>1n > 1n>1.

Wait a minute, this seems promising. So far, I've shown that 2, 3, and 5 divide f(n)f(n)f(n) for all n>1n > 1n>1. Maybe there's a pattern here.

Let's generalize this. For a prime ppp, does ppp divide f(n)f(n)f(n) for all n>1n > 1n>1? Well, from above, for p=2,3,5p = 2,3,5p=2,3,5, it seems yes, but let's check for a larger prime, say p=7p = 7p=7.

For n=2n = 2n=2, check if 7 divides 260−12^{60} - 1260−1. The order of 2 mod 7 is 3, since 2^1 = 2, 2^2 = 4, 2^3 = 8 ≡ 1 \pmod{7}, so order is 3. Since 3 does not divide 60? Wait, 3 does divide 60, since 60 = 3*20, so actually, since 3 divides 60, 260≡120≡1(mod7)2^{60} \equiv 1^{20} \equiv 1 \pmod{7}260≡120≡1(mod7), so 260−1≡0(mod7)2^{60} - 1 \equiv 0 \pmod{7}260−1≡0(mod7), thus 7 divides 260−12^{60} - 1260−1, and hence 7 divides f(2)f(2)f(2).

For n=7n = 7n=7, 799(760−1)7^{99}(7^{60} - 1)799(760−1), clearly divisible by 7.

For n=3n = 3n=3, check 360−1(mod7)3^{60} - 1 \pmod{7}360−1(mod7). The order of 3 mod 7 is 6, since 3^1 = 3, 3^2 = 9 ≡ 2, 3^3 ≡ 6 ≡ -1, 3^6 ≡ 1 \pmod{7}, so order is 6. Since 6 divides 60, 360≡1(mod7)3^{60} \equiv 1 \pmod{7}360≡1(mod7), so 360−1≡0(mod7)3^{60} - 1 \equiv 0 \pmod{7}360−1≡0(mod7), thus 7 divides f(3)f(3)f(3).

Similarly, for any n>1n > 1n>1, if n≡0(mod7)n \equiv 0 \pmod{7}n≡0(mod7), then f(n)f(n)f(n) is divisible by 7. If n≢0(mod7)n \not\equiv 0 \pmod{7}n≡0(mod7), then the order of nnn mod 7 divides 6, and since 6 divides 60, n60≡1(mod7)n^{60} \equiv 1 \pmod{7}n60≡1(mod7), so n60−1≡0(mod7)n^{60} - 1 \equiv 0 \pmod{7}n60−1≡0(mod7), thus 7 divides f(n)f(n)f(n).

Wait, so this seems to hold for p=7p = 7p=7 as well. In fact, for any prime ppp, since 60 is a multiple of ϕ(p)=p−1\phi(p) = p - 1ϕ(p)=p−1, where ϕ\phiϕ is Euler's totient function, and for n≢0(modp)n \not\equiv 0 \pmod{p}n≡0(modp), nϕ(p)≡1(modp)n^{\phi(p)} \equiv 1 \pmod{p}nϕ(p)≡1(modp), so np−1≡1(modp)n^{p-1} \equiv 1 \pmod{p}np−1≡1(modp), and since 60 is a multiple of p−1p - 1p−1 for p=2,3,5,7,p = 2,3,5,7,p=2,3,5,7,, but actually, for larger ppp, say p=61p = 61p=61, since 61 - 1 = 60, and 60 divides 60, so yes, n60≡1(mod61)n^{60} \equiv 1 \pmod{61}n60≡1(mod61) for n≢0(mod61)n \not\equiv 0 \pmod{61}n≡0(mod61), thus n60−1≡0(mod61)n^{60} - 1 \equiv 0 \pmod{61}n60−1≡0(mod61), so 61 divides n60−1n^{60} - 1n60−1 for n≢0(mod61)n \not\equiv 0 \pmod{61}n≡0(mod61), and for n≡0(mod61)n \equiv 0 \pmod{61}n≡0(mod61), n99n^{99}n99 is divisible by 61, so in either case, 61 divides f(n)f(n)f(n).

Wait, so this suggests that every prime ppp divides f(n)f(n)f(n) for all n>1n > 1n>1. But that can't be right, because if every prime divides f(n)f(n)f(n), then f(n)f(n)f(n) would have to be divisible by all primes, which is impossible unless f(n)=0f(n) = 0f(n)=0, but f(n)≠0f(n) \neq 0f(n)=0.

Wait, what's wrong here? Oh, wait, for a fixed nnn, f(n)f(n)f(n) is a specific number, and it can't be divisible by all primes, but in the argument above, for each ppp, there is some nnn such that ppp does not divide f(n)f(n)f(n), but the point is that for each ppp, ppp divides f(n)f(n)f(n) for all n>1n > 1n>1, but that's not true, because for example, take p=3p = 3p=3, and n=3n = 3n=3, then f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and since 360−1≢0(mod3)3^{60} - 1 \not\equiv 0 \pmod{3}360−1≡0(mod3), but wait, actually, 360≡0(mod3)3^{60} \equiv 0 \pmod{3}360≡0(mod3), so 360−1≡−1(mod3)3^{60} - 1 \equiv -1 \pmod{3}360−1≡−1(mod3), and 399≡0(mod3)3^{99} \equiv 0 \pmod{3}399≡0(mod3), so f(3)=399(360−1)≡0⋅(−1)=0(mod3)f(3) = 3^{99}(3^{60} - 1) \equiv 0 \cdot (-1) = 0 \pmod{3}f(3)=399(360−1)≡0⋅(−1)=0(mod3), but since 3993^{99}399 is divisible by 32=93^2 = 932=9, say, but actually, for p=3p = 3p=3, when n=3n = 3n=3, f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and 360−13^{60} - 1360−1 is not divisible by 3, so the overall f(3)f(3)f(3) is divisible by 3993^{99}399, but not by higher powers of 3 necessarily.

Wait, but in the earlier logic, I showed that for each ppp, ppp divides f(n)f(n)f(n) for all n>1n > 1n>1, but actually, that's not accurate. Wait, let's clarify: for a fixed ppp, for each n>1n > 1n>1, if n≡0(modp)n \equiv 0 \pmod{p}n≡0(modp), then ppp divides n99n^{99}n99, so ppp divides f(n)f(n)f(n); if n≢0(modp)n \not\equiv 0 \pmod{p}n≡0(modp), then since ppp is prime and p−1∣60p - 1 | 60p−1∣60, because 60 is a multiple of many numbers, but actually, for p>61p > 61p>61, say p=67p = 67p=67, since 67 - 1 = 66, and 66 does not divide 60, since 60 / 66 < 1, so for p=67p = 67p=67, it's not necessarily true that n60≡1(mod67)n^{60} \equiv 1 \pmod{67}n60≡1(mod67) for n≢0(mod67)n \not\equiv 0 \pmod{67}n≡0(mod67), because the order might be larger than 60.

For example, take p=67p = 67p=67, and n=2n = 2n=2, then does 67 divide 260−12^{60} - 1260−1? Well, the order of 2 mod 67 would be the smallest kkk such that 2k≡1(mod67)2^k \equiv 1 \pmod{67}2k≡1(mod67), and since 67 is prime, the order divides 66. Since 66 = 2 * 3 * 11, and 60 = 2^2 * 3 * 5, so 11 does not divide 60, so possibly, if 2 has order 66 mod 67, then 260≢1(mod67)2^{60} \not\equiv 1 \pmod{67}260≡1(mod67), because 60 does not divide 66. Wait, actually, since 60 does not divide 66, but 66 does not divide 60 either, wait, more precisely, if the order is 66, then since 66 does not divide 60, 260≢1(mod67)2^{60} \not\equiv 1 \pmod{67}260≡1(mod67), but actually, to check if 260≡1(mod67)2^{60} \equiv 1 \pmod{67}260≡1(mod67), we can see if 60 is a multiple of the order, but since the order divides 66, and 66 and 60 have gcd 6, since 66 = 611, 60 = 610, so the possible orders are divisors of 66, and since 60 is not a multiple of 66, but rather, for some elements, the order could be 11 or 22, etc.

Wait, actually, let's compute directly. Perhaps I can find if there's an nnn such that n60≡1(mod67)n^{60} \equiv 1 \pmod{67}n60≡1(mod67) for n≢0(mod67)n \not\equiv 0 \pmod{67}n≡0(mod67), but since 67 is prime, the multiplicative group is cyclic of order 66, and since 60 does not divide 66, there are elements whose order does not divide 60, so for those nnn, n60≢1(mod67)n^{60} \not\equiv 1 \pmod{67}n60≡1(mod67), thus n60−1≢0(mod67)n^{60} - 1 \not\equiv 0 \pmod{67}n60−1≡0(mod67), so 67 does not divide n60−1n^{60} - 1n60−1, and since n99n^{99}n99 may not be divisible by 67, so for such nnn, f(n)f(n)f(n) may not be divisible by 67.

For example, if I take nnn such that nnn has order 66 mod 67, then n60≢1(mod67)n^{60} \not\equiv 1 \pmod{67}n60≡1(mod67), since 66 does not divide 60, so n60−1≢0(mod67)n^{60} - 1 \not\equiv 0 \pmod{67}n60−1≡0(mod67), and since 67 is prime, n99≡n99mod  66=n33(mod67)n^{99} \equiv n^{99 \mod 66} = n^{33} \pmod{67}n99≡n99mod66=n33(mod67), and since n≠0n \neq 0n=0, n33≠0(mod67)n^{33} \neq 0 \pmod{67}n33=0(mod67), so f(n)=n99(n60−1)≢0(mod67)f(n) = n^{99}(n^{60} - 1) \not\equiv 0 \pmod{67}f(n)=n99(n60−1)≡0(mod67), thus 67 does not divide f(n)f(n)f(n) for such nnn.

Therefore, for primes p>60p > 60p>60, say, or more precisely, for primes ppp such that p−1p - 1p−1 does not divide 60, wait, actually, in the above, for p=67p = 67p=67, since 66 = p - 1, and 66 does not divide 60, there are nnn such that n60≢1(mod67)n^{60} \not\equiv 1 \pmod{67}n60≡1(mod67), so 67 does not divide f(n)f(n)f(n) for those nnn.

So, to summarize, for a prime ppp, if p−1p - 1p−1 divides 60, then for all n>1n > 1n>1, ppp divides f(n)f(n)f(n), because if n≡0(modp)n \equiv 0 \pmod{p}n≡0(modp), then ppp divides n99n^{99}n99, if n≢0(modp)n \not\equiv 0 \pmod{p}n≡0(modp), then since p−1∣60p - 1 | 60p−1∣60, so n60≡1(modp)n^{60} \equiv 1 \pmod{p}n60≡1(modp), thus ppp divides n60−1n^{60} - 1n60−1, so ppp divides f(n)f(n)f(n).

On the other hand, if p−1p - 1p−1 does not divide 60, then there exists n>1n > 1n>1 such that ppp does not divide f(n)f(n)f(n).

So, the primes ppp such that p−1∣60p - 1 | 60p−1∣60 are the primes that divide f(n)f(n)f(n) for all n>1n > 1n>1.

Let's find such primes. First, note that 60 = 2^2 * 3 * 5, so the positive divisors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60.

Therefore, the possible values for p−1p - 1p−1 are these divisors, so p=2,3,4,5,6,7,11,13,16,21,31,61p = 2,3,4,5,6,7,11,13,16,21,31,61p=2,3,4,5,6,7,11,13,16,21,31,61, but among these, the primes are p=2,3,5,7,11,13,31,61p = 2,3,5,7,11,13,31,61p=2,3,5,7,11,13,31,61, since 4 is not prime, 6 not prime, etc.

Wait, let's list them: for d∣60d | 60d∣60, p=d+1p = d + 1p=d+1 is prime if d+1d + 1d+1 is prime. So for d=1d = 1d=1, p=2p = 2p=2, prime; d=2d = 2d=2, p=3p = 3p=3, prime; d=3d = 3d=3, p=4p = 4p=4, not prime; d=4d = 4d=4, p=5p = 5p=5, prime; d=5d = 5d=5, p=6p = 6p=6, not prime; d=6d = 6d=6, p=7p = 7p=7, prime; d=10d = 10d=10, p=11p = 11p=11, prime; d=12d = 12d=12, p=13p = 13p=13, prime; d=15d = 15d=15, p=16p = 16p=16, not prime; d=20d = 20d=20, p=21p = 21p=21, not prime; d=30d = 30d=30, p=31p = 31p=31, prime; d=60d = 60d=60, p=61p = 61p=61, prime.

So the primes ppp such that p−1∣60p - 1 | 60p−1∣60 are p=2,3,5,7,11,13,31,61p = 2,3,5,7,11,13,31,61p=2,3,5,7,11,13,31,61.

Therefore, for these primes, ppp divides f(n)f(n)f(n) for all n>1n > 1n>1, while for other primes, there exists some n>1n > 1n>1 such that ppp does not divide f(n)f(n)f(n).

Now, since the GCD ddd of the set {f(n)∣n>1}\{f(n) | n > 1\}{f(n)∣n>1} must be divisible only by the primes that divide all f(n)f(n)f(n), so ddd can only be divisible by the primes in {2,3,5,7,11,13,31,61}\{2,3,5,7,11,13,31,61\}{2,3,5,7,11,13,31,61}.

Moreover, since for each of these primes ppp, ppp divides f(n)f(n)f(n) for all n>1n > 1n>1, in fact, perhaps ddd is the product of these primes raised to some powers.

But wait, more precisely, since f(n)f(n)f(n) is divisible by each of these ppp, and since there are infinitely many nnn, perhaps ddd is 1, but that can't be, because as we saw, 2 divides all f(n)f(n)f(n), so d≥2d \geq 2d≥2.

Actually, to find the exact ddd, perhaps I can compute gcd⁡(f(2),f(3),f(4),… )\gcd(f(2), f(3), f(4), \dots)gcd(f(2),f(3),f(4),…), but since it's an infinite set, maybe I can find a way to compute it using properties of the polynomial.

Wait, here's another idea: perhaps I can consider the number f(1)f(1)f(1), but wait, n>1n > 1n>1, so n=1n = 1n=1 is not included, but maybe I can use the fact that for polynomials, the GCD of f(n)f(n)f(n) for all nnn can be related to the constant term or something, but since the set is n>1n > 1n>1, perhaps not directly.

Alternatively, let's recall that for two values, gcd⁡(f(a),f(b))=gcd⁡(a60−1,b60−1)\gcd(f(a), f(b)) = \gcd(a^{60} - 1, b^{60} - 1)gcd(f(a),f(b))=gcd(a60−1,b60−1) for distinct a,b>1a,b > 1a,b>1, as I thought earlier, but actually, more precisely, since f(a)=a99(a60−1)f(a) = a^{99}(a^{60} - 1)f(a)=a99(a60−1), and f(b)=b99(b60−1)f(b) = b^{99}(b^{60} - 1)f(b)=b99(b60−1), and since a99a^{99}a99 and b99b^{99}b99 are coprime if aaa and bbb are distinct, but actually, if aaa and bbb share common factors, it might be different, but for simplicity, let's take a=2,b=3a = 2, b = 3a=2,b=3, then gcd⁡(f(2),f(3))=gcd⁡(260−1,360−1)\gcd(f(2), f(3)) = \gcd(2^{60} - 1, 3^{60} - 1)gcd(f(2),f(3))=gcd(260−1,360−1), since 2992^{99}299 and 3993^{99}399 are coprime.

Now, similarly, perhaps I can take multiple values. For example, let's consider gcd⁡(f(2),f(3),f(4),… )\gcd(f(2), f(3), f(4), \dots)gcd(f(2),f(3),f(4),…), but since it's infinite, maybe I can find a way to express it.

Wait, actually, there's a theorem that says that for a polynomial f(x)f(x)f(x) with integer coefficients, the GCD of f(n)f(n)f(n) for all integers nnn such that f(n)≠0f(n) \neq 0f(n)=0 is equal to the GCD of the values of fff at points where xxx is greater than some bound or something, but perhaps more accurately, in this case, since f(n)≠0f(n) \neq 0f(n)=0 for n>1n > 1n>1, maybe I can use that.

But let's think about what the common divisor is. From earlier, I know that the primes dividing ddd are among 2,3,5,7,11,13,31,612,3,5,7,11,13,31,612,3,5,7,11,13,31,61, so perhaps ddd is the product of these primes to some powers.

In fact, maybe I can compute gcd⁡(260−1,360−1,460−1,… )\gcd(2^{60} - 1, 3^{60} - 1, 4^{60} - 1, \dots)gcd(260−1,360−1,460−1,…), but since f(n)=n99(n60−1)f(n) = n^{99}(n^{60} - 1)f(n)=n99(n60−1), and the n99n^{99}n99 varies, but in the GCD, since n99n^{99}n99 is coprime to other terms, perhaps the GCD is indeed related to the GCD of (n60−1)(n^{60} - 1)(n60−1) for all n>1n > 1n>1.

Wait, let's be more precise. Actually, since for each n>1n > 1n>1, f(n)=n99(n60−1)f(n) = n^{99}(n^{60} - 1)f(n)=n99(n60−1), and since n99n^{99}n99 is an integer that depends on nnn, but when taking GCD over all n>1n > 1n>1, the common factors should be those that appear in every f(n)f(n)f(n).

But perhaps a better way is to note that since for any two distinct a,b>1a, b > 1a,b>1, gcd⁡(f(a),f(b))\gcd(f(a), f(b))gcd(f(a),f(b)) divides ddd, and in particular, if I can find the GCD over all pairs, but that might not directly give ddd.

Wait, alternatively, let's consider that for any prime ppp not in {2,3,5,7,11,13,31,61}\{2,3,5,7,11,13,31,61\}{2,3,5,7,11,13,31,61}, there exists some n>1n > 1n>1 such that ppp does not divide f(n)f(n)f(n), so ppp does not divide ddd.

Moreover, for the primes in {2,3,5,7,11,13,31,61}\{2,3,5,7,11,13,31,61\}{2,3,5,7,11,13,31,61}, since each ppp divides f(n)f(n)f(n) for all n>1n > 1n>1, in fact, perhaps ddd is the product of these primes to the minimal exponent that appears in f(n)f(n)f(n).

But actually, since f(n)f(n)f(n) can be very large, the minimal exponent might be zero, but wait, for example, for p=2p = 2p=2, since 2 divides f(n)f(n)f(n) for all n>1n > 1n>1, but the exponent of 2 in f(n)f(n)f(n) can be arbitrarily large as nnn increases, so the minimal exponent is 1, meaning that 2^1 divides all f(n)f(n)f(n), but 2^2 may not divide all f(n)f(n)f(n).

For instance, let's check for n=3n = 3n=3, since 333 is odd, 3993^{99}399 is odd, and 360−13^{60} - 1360−1 is even, but let's see the exponent of 2 in 360−13^{60} - 1360−1. Earlier, I saw that 360−13^{60} - 1360−1 is divisible by 8, since 32−1=83^2 - 1 = 832−1=8, and for larger even exponents, it may be divisible by higher powers, but for n=3n = 3n=3, f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and since 3993^{99}399 is odd, the exponent of 2 in f(3)f(3)f(3) is the same as in 360−13^{60} - 1360−1. Let's compute the exponent of 2 in 360−13^{60} - 1360−1.

Actually, there's a formula for the exponent of 2 in ak−1a^k - 1ak−1, where aaa is odd. Specifically, if aaa is odd, then the exponent of 2 in ak−1a^k - 1ak−1 can be determined based on kkk. For example, if kkk is even, then since ak−1=(ak/2−1)(ak/2+1)a^k - 1 = (a^{k/2} - 1)(a^{k/2} + 1)ak−1=(ak/2−1)(ak/2+1), and both ak/2−1a^{k/2} - 1ak/2−1 and ak/2+1a^{k/2} + 1ak/2+1 are even, since aaa is odd, ak/2a^{k/2}ak/2 is odd, so odd - 1 = even, odd + 1 = even, so at least two factors of 2. In fact, for a=3a = 3a=3, let's see for small kkk: for k=1k = 1k=1, 31−1=23^1 - 1 = 231−1=2, so exponent 1; for k=2k = 2k=2, 32−1=8=233^2 - 1 = 8 = 2^332−1=8=23, exponent 3; for k=4k = 4k=4, 34−1=80=24∗53^4 - 1 = 80 = 2^4 * 534−1=80=24∗5, exponent 4; for k=8k = 8k=8, 38−1=6560=25∗2053^8 - 1 = 6560 = 2^5 * 20538−1=6560=25∗205, wait, 6560 / 16 = 410, 410 / 2 = 205, so 2^5, exponent 5. Actually, there's a pattern: for k=2mk = 2^mk=2m, the exponent of 2 in 32m−13^{2^m} - 132m−1 is m+2m + 2m+2, since for m=0m = 0m=0, 31−1=2=213^1 - 1 = 2 = 2^131−1=2=21, but wait, for m=1m = 1m=1, k=21=2k = 2^1 = 2k=21=2, 32−1=8=23=21+23^2 - 1 = 8 = 2^3 = 2^{1+2}32−1=8=23=21+2; for m=2m = 2m=2, k=4k = 4k=4, 34−1=80=24∗5=22+23^4 - 1 = 80 = 2^4 * 5 = 2^{2+2}34−1=80=24∗5=22+2; for m=3m = 3m=3, k=8k = 8k=8, 38−1=6560=25∗205=23+23^8 - 1 = 6560 = 2^5 * 205 = 2^{3+2}38−1=6560=25∗205=23+2, wait, 5 = 3 + 2, yes, so in general, for a=3a = 3a=3, the exponent of 2 in 32m−13^{2^m} - 132m−1 is m+2m + 2m+2.

But for k=60k = 60k=60, since 60 = 2^2 * 3 * 5, perhaps I can use the lifting-the-exponent lemma or something to find the exact exponent.

Actually, there's a general formula for the exponent of 2 in ak−1a^k - 1ak−1 when aaa is odd. Specifically, if aaa is odd, then the exponent of 2 in ak−1a^k - 1ak−1 is equal to the exponent of 2 in a−1a - 1a−1 plus the number of times 2 divides kkk, but wait, more precisely, it's related to the order.

Wait, actually, for a≡1(mod2)a \equiv 1 \pmod{2}a≡1(mod2), since aaa is odd, and for kkk even, as above, it's at least 2, but for exact computation, perhaps it's better to note that since 3≡−1(mod4)3 \equiv -1 \pmod{4}3≡−1(mod4), wait, 3 ≡ 3 ≡ -1 \pmod{4}, no, 3 ≡ 3 ≡ 1 \pmod{4}, since 3 - 1 = 2, not divisible by 4.

Wait, actually, for a≡1(mod4)a \equiv 1 \pmod{4}a≡1(mod4), then the exponent of 2 in ak−1a^k - 1ak−1 can be determined. Specifically, if a≡1(mod8)a \equiv 1 \pmod{8}a≡1(mod8), then for kkk even, ak−1≡0(mod8)a^k - 1 \equiv 0 \pmod{8}ak−1≡0(mod8), but for higher powers, it depends.

Alternatively, perhaps for a=3a = 3a=3, since 3≡3(mod8)3 \equiv 3 \pmod{8}3≡3(mod8), and 32=9≡1(mod8)3^2 = 9 \equiv 1 \pmod{8}32=9≡1(mod8), so for even kkk, 3k≡1(mod8)3^k \equiv 1 \pmod{8}3k≡1(mod8), so 3k−1≡0(mod8)3^k - 1 \equiv 0 \pmod{8}3k−1≡0(mod8), but not necessarily higher. For example, 32−1=83^2 - 1 = 832−1=8, which is 8 * 1, so divisible by 8 but not by 16; 34−1=80=16∗53^4 - 1 = 80 = 16 * 534−1=80=16∗5, divisible by 16 but not by 32; 36−1=728=8∗913^6 - 1 = 728 = 8 * 9136−1=728=8∗91, since 91 = 7*13, not divisible by 16, so for k=6k = 6k=6, 36−1=728=8∗913^6 - 1 = 728 = 8 * 9136−1=728=8∗91, so exponent of 2 is 3.

Wait, so for k=60k = 60k=60, since 60 = 4 * 15, and 34≡1(mod16)3^4 \equiv 1 \pmod{16}34≡1(mod16), since 34=81=5∗16+13^4 = 81 = 516 + 134=81=5∗16+1, yes, 81−1=80=5∗1681 - 1 = 80 = 51681−1=80=5∗16, so 34−1=80=16∗53^4 - 1 = 80 = 16534−1=80=16∗5, and for higher powers, since 34≡1(mod16)3^4 \equiv 1 \pmod{16}34≡1(mod16), so 34m≡1m≡1(mod16)3^{4m} \equiv 1^m \equiv 1 \pmod{16}34m≡1m≡1(mod16), thus 34m−1≡0(mod16)3^{4m} - 1 \equiv 0 \pmod{16}34m−1≡0(mod16), but since 16 divides 34−13^4 - 134−1, and for m>1m > 1m>1, 34m−1=(34)m−1=(1+16t)m−1≡1+m∗16t−1=16mt(mod162)3^{4m} - 1 = (3^4)^m - 1 = (1 + 16t)^m - 1 \equiv 1 + m16t - 1 = 16mt \pmod{16^2}34m−1=(34)m−1=(1+16t)m−1≡1+m∗16t−1=16mt(mod162), so actually, for m≥1m \geq 1m≥1, 34m−1≡0(mod16)3^{4m} - 1 \equiv 0 \pmod{16}34m−1≡0(mod16), but not necessarily divisible by higher powers unless mmm is large.

In particular, for k=60=4∗15k = 60 = 4 * 15k=60=4∗15, since 34≡1(mod16)3^4 \equiv 1 \pmod{16}34≡1(mod16), and 360=(34)15≡115≡1(mod16)3^{60} = (3^4)^{15} \equiv 1^{15} \equiv 1 \pmod{16}360=(34)15≡115≡1(mod16), so 360−1≡0(mod16)3^{60} - 1 \equiv 0 \pmod{16}360−1≡0(mod16), but let's check if it's divisible by 32. Since 34≡1(mod16)3^4 \equiv 1 \pmod{16}34≡1(mod16), but to check modulo 32, let's compute 34=81≡81−2∗32=81−64=17(mod32)3^4 = 81 \equiv 81 - 232 = 81 - 64 = 17 \pmod{32}34=81≡81−2∗32=81−64=17(mod32), wait, actually 81 - 64 = 17, yes, so 34≡17(mod32)3^4 \equiv 17 \pmod{32}34≡17(mod32), and 172=289≡289−9∗32=289−288=1(mod32)17^2 = 289 \equiv 289 - 932 = 289 - 288 = 1 \pmod{32}172=289≡289−9∗32=289−288=1(mod32), so indeed 38=(34)2≡172≡1(mod32)3^8 = (3^4)^2 \equiv 17^2 \equiv 1 \pmod{32}38=(34)2≡172≡1(mod32), thus for m≥1m \geq 1m≥1, 38m≡1(mod32)3^{8m} \equiv 1 \pmod{32}38m≡1(mod32), so since 60 = 8*7 + 4, but actually, since 8 does not divide 60, perhaps 360≢1(mod32)3^{60} \not\equiv 1 \pmod{32}360≡1(mod32), but let's compute 360=(38)7∗34≡17∗17≡17(mod32)3^{60} = (3^8)^7 * 3^4 \equiv 1^7 * 17 \equiv 17 \pmod{32}360=(38)7∗34≡17∗17≡17(mod32), so 360≡17(mod32)3^{60} \equiv 17 \pmod{32}360≡17(mod32), thus 360−1≡17−1=16(mod32)3^{60} - 1 \equiv 17 - 1 = 16 \pmod{32}360−1≡17−1=16(mod32), so 360−1≡16(mod32)3^{60} - 1 \equiv 16 \pmod{32}360−1≡16(mod32), which means 360−1=32∗q+163^{60} - 1 = 32 * q + 16360−1=32∗q+16 for some qqq, so it's divisible by 16 but not by 32.

Therefore, for n=3n = 3n=3, f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and since 360−13^{60} - 1360−1 is divisible by 16 but not by 32, and 3993^{99}399 is odd, so the exponent of 2 in f(3)f(3)f(3) is 4, since 16 = 2^4.

Similarly, for other nnn, the exponent of 2 in f(n)f(n)f(n) may be higher, but since for n=3n = 3n=3, it's only 4, so the common divisor ddd can have 2 raised to at most the 4th power.

Wait, actually, wait, for n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and 260−12^{60} - 1260−1 is odd, so the exponent of 2 in f(2)f(2)f(2) is 99, which is larger than 4, but since we're taking GCD over all n>1n > 1n>1, the minimal exponent is what matters. So for n=3n = 3n=3, the exponent is 4, for other nnn, it might be higher, but since 2^4 divides f(n)f(n)f(n) for all n>1n > 1n>1, and for some nnn, like n=3n = 3n=3, 2^5 does not divide f(n)f(n)f(n), because if f(3)f(3)f(3) were divisible by 32 = 2^5, but since 360−1≡16(mod32)3^{60} - 1 \equiv 16 \pmod{32}360−1≡16(mod32), and 16 is not 0 mod 32, so 360−13^{60} - 1360−1 is not divisible by 32, and since 3993^{99}399 is odd, f(3)f(3)f(3) is not divisible by 32.

Therefore, the highest power of 2 that divides all f(n)f(n)f(n) is 2^4 = 16.

Wait, is that correct? Wait, actually, let's confirm for another nnn. For example, take n=4n = 4n=4, since 4 is even, f(4)=499(460−1)f(4) = 4^{99}(4^{60} - 1)f(4)=499(460−1), and 4 = 2^2, so 499=21984^{99} = 2^{198}499=2198, and 460−1=(22)60−1=2120−14^{60} - 1 = (2^2)^{60} - 1 = 2^{120} - 1460−1=(22)60−1=2120−1, and since 2^{120} is divisible by 2^{120}, but 2^{120} - 1 is odd, so f(4)=2198⋅(odd)f(4) = 2^{198} \cdot (\text{odd})f(4)=2198⋅(odd), so the exponent of 2 is 198, which is larger than 4.

But for n=5n = 5n=5, let's say, 599(560−1)5^{99}(5^{60} - 1)599(560−1), since 5 is odd, similar to 3, we need to find the exponent of 2 in 560−15^{60} - 1560−1. Similarly, since 5 ≡ 1 \pmod{4}, actually 5 ≡ 1 \pmod{8}, since 5 - 1 = 4, not divisible by 8, but 5^2 = 25 ≡ 1 \pmod{8}, so for even exponents, 52k≡1(mod8)5^{2k} \equiv 1 \pmod{8}52k≡1(mod8), so 52k−1≡0(mod8)5^{2k} - 1 \equiv 0 \pmod{8}52k−1≡0(mod8), but for higher powers, since 5 ≡ 5 \pmod{16}, 5^2 = 25 ≡ 9 \pmod{16}, 5^4 = (5^2)^2 ≡ 9^2 = 81 ≡ 1 \pmod{16}, so 54≡1(mod16)5^4 \equiv 1 \pmod{16}54≡1(mod16), thus for k≥1k \geq 1k≥1, 54k≡1(mod16)5^{4k} \equiv 1 \pmod{16}54k≡1(mod16), so 560=54∗15≡1(mod16)5^{60} = 5^{415} \equiv 1 \pmod{16}560=54∗15≡1(mod16), thus 560−1≡0(mod16)5^{60} - 1 \equiv 0 \pmod{16}560−1≡0(mod16), but to check modulo 32, since 5^4 = 625 ≡ 625 - 1932 = 625 - 608 = 17 \pmod{32}, wait, 625 / 32 = 19.53125, 3219 = 608, 625 - 608 = 17, so 5^4 ≡ 17 \pmod{32}, and then 5^8 ≡ 17^2 = 289 ≡ 289 - 932 = 289 - 288 = 1 \pmod{32}, so actually 5^8 ≡ 1 \pmod{32}, thus for m≥1m \geq 1m≥1, 58m≡1(mod32)5^{8m} \equiv 1 \pmod{32}58m≡1(mod32), and since 60 = 87 + 4, but 8 does not divide 60, so 560=58∗7∗54≡1∗54≡17(mod32)5^{60} = 5^{87} * 5^4 ≡ 1 * 5^4 ≡ 17 \pmod{32}560=58∗7∗54≡1∗54≡17(mod32), so 560−1≡17−1=16(mod32)5^{60} - 1 ≡ 17 - 1 = 16 \pmod{32}560−1≡17−1=16(mod32), thus 560−1≡16(mod32)5^{60} - 1 \equiv 16 \pmod{32}560−1≡16(mod32), not 0, so again, divisible by 16 but not by 32.

Therefore, for n=5n = 5n=5, f(5)=599(560−1)f(5) = 5^{99}(5^{60} - 1)f(5)=599(560−1), and since 5995^{99}599 is odd, the exponent of 2 is 4, same as for n=3n = 3n=3.

So, it seems that for odd nnn, the exponent of 2 in f(n)f(n)f(n) is determined by the exponent in n60−1n^{60} - 1n60−1, which for many nnn is 4, since n60−1≡0(mod16)n^{60} - 1 \equiv 0 \pmod{16}n60−1≡0(mod16) but not necessarily higher.

Wait, actually, to confirm, let's check for n=9n = 9n=9, say. For n=9n = 9n=9, since 9 is odd, f(9)=999(960−1)f(9) = 9^{99}(9^{60} - 1)f(9)=999(960−1), and 960=(32)60=31209^{60} = (3^2)^{60} = 3^{120}960=(32)60=3120, so 960−1=3120−19^{60} - 1 = 3^{120} - 1960−1=3120−1, and since 3^120 - 1, from earlier, for a=3a = 3a=3, 360−1≡16(mod32)3^{60} - 1 \equiv 16 \pmod{32}360−1≡16(mod32), but wait, actually, since 60 = 4*15, and 34≡17(mod32)3^4 \equiv 17 \pmod{32}34≡17(mod32), but perhaps for larger exponents, let's see if there's a pattern.

Alternatively, perhaps I can use the fact that since there are infinitely many nnn where the exponent of 2 in f(n)f(n)f(n) is 4, for example, for n=3kn = 3^kn=3k, but actually, for large kkk, f(3k)f(3^k)f(3k) might have higher exponents, wait, let's see for n=3n = 3n=3, as above, exponent is 4, and for n=32=9n = 3^2 = 9n=32=9, let's compute 960−1=(32)60−1=3120−19^{60} - 1 = (3^2)^{60} - 1 = 3^{120} - 1960−1=(32)60−1=3120−1, and since 120 = 815, and from above, 38≡1(mod32)3^8 \equiv 1 \pmod{32}38≡1(mod32), wait, earlier I had 34≡17(mod32)3^4 \equiv 17 \pmod{32}34≡17(mod32), 38≡172=289≡289−9∗32=289−288=1(mod32)3^8 \equiv 17^2 = 289 \equiv 289 - 932 = 289 - 288 = 1 \pmod{32}38≡172=289≡289−9∗32=289−288=1(mod32), yes, so 38≡1(mod32)3^8 \equiv 1 \pmod{32}38≡1(mod32), thus 3120=(38)15≡115≡1(mod32)3^{120} = (3^8)^{15} \equiv 1^{15} \equiv 1 \pmod{32}3120=(38)15≡115≡1(mod32), so 3120−1≡0(mod32)3^{120} - 1 \equiv 0 \pmod{32}3120−1≡0(mod32), but let's check modulo 64. Since 38≡1(mod32)3^8 \equiv 1 \pmod{32}38≡1(mod32), but for modulo 64, let's compute 38=65613^8 = 656138=6561, and 6561 / 64 = 102.515625, 64102 = 6528, 6561 - 6528 = 33, so 38≡33(mod64)3^8 \equiv 33 \pmod{64}38≡33(mod64), and 33 = 32 + 1 = 2^5 + 1, so 38≡1(mod32)3^8 \equiv 1 \pmod{32}38≡1(mod32), but 38≡33≡1(mod32)3^8 \equiv 33 \equiv 1 \pmod{32}38≡33≡1(mod32), yes, but for higher, actually, since 38≡33(mod64)3^8 \equiv 33 \pmod{64}38≡33(mod64), and 332=1089≡1089−17∗64=1089−1088=1(mod64)33^2 = 1089 \equiv 1089 - 1764 = 1089 - 1088 = 1 \pmod{64}332=1089≡1089−17∗64=1089−1088=1(mod64), so 316≡1(mod64)3^16 \equiv 1 \pmod{64}316≡1(mod64), thus for m≥1m \geq 1m≥1, 316m≡1(mod64)3^{16m} \equiv 1 \pmod{64}316m≡1(mod64), and since 120 = 167 + 8, but 16 does not divide 120, so 3120=316∗7∗38≡1∗38≡33(mod64)3^{120} = 3^{167} * 3^8 \equiv 1 * 3^8 \equiv 33 \pmod{64}3120=316∗7∗38≡1∗38≡33(mod64), thus 3120−1≡33−1=32(mod64)3^{120} - 1 \equiv 33 - 1 = 32 \pmod{64}3120−1≡33−1=32(mod64), so 3120−1≡32(mod64)3^{120} - 1 \equiv 32 \pmod{64}3120−1≡32(mod64), which means it's divisible by 32 but not by 64, since 32 ≡ 32 \pmod{64}), not 0.

Therefore, for n=9n = 9n=9, f(9)=999(960−1)=999(3120−1)f(9) = 9^{99}(9^{60} - 1) = 9^{99}(3^{120} - 1)f(9)=999(960−1)=999(3120−1), and since 3120−1≡32(mod64)3^{120} - 1 \equiv 32 \pmod{64}3120−1≡32(mod64), so the exponent of 2 in 3120−13^{120} - 13120−1 is 5, since 32 = 2^5, and 9^{99} is odd, so f(9)f(9)f(9) has exponent of 2 as 5.

Similarly, for other nnn, the exponent can be higher, but since for n=3n = 3n=3, it's 4, and for n=5n = 5n=5, it's also 4, etc., so the minimal exponent is 4, meaning that 2^4 divides all f(n)f(n)f(n), but 2^5 does not divide all f(n)f(n)f(n), since for n=3n = 3n=3, f(3)f(3)f(3) is only divisible by 2^4.

Therefore, in the GCD ddd, the exponent of 2 is 4.

Now, let's check for other primes. For example, for p=3p = 3p=3, what's the minimal exponent in f(n)f(n)f(n) for all n>1n > 1n>1? Well, for n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and since 2 ≡ 2 \pmod{3}, 260≡260mod  ϕ(3)=260mod  2=20=1(mod3)2^{60} \equiv 2^{60 \mod \phi(3)} = 2^{60 \mod 2} = 2^0 = 1 \pmod{3}260≡260modϕ(3)=260mod2=20=1(mod3), wait, since ϕ(3)=2\phi(3) = 2ϕ(3)=2, and 60 is even, so 260≡20≡1(mod3)2^{60} \equiv 2^0 \equiv 1 \pmod{3}260≡20≡1(mod3), thus 260−1≡0(mod3)2^{60} - 1 \equiv 0 \pmod{3}260−1≡0(mod3), so 3 divides 260−12^{60} - 1260−1, and in fact, since 2^2 ≡ 1 \pmod{3}, and the order is 2, so for even kkk, 2k≡1(mod3)2^k \equiv 1 \pmod{3}2k≡1(mod3), thus 2k−1≡0(mod3)2^k - 1 \equiv 0 \pmod{3}2k−1≡0(mod3), but the exponent is 1, since 3^1 divides 2k−12^k - 12k−1 for even kkk, but not higher, because, for example, 22−1=32^2 - 1 = 322−1=3, divisible by 3 but not by 9.

So for n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and since 260−1=3∗m2^{60} - 1 = 3 * m260−1=3∗m for some mmm not divisible by 3, and 2992^{99}299 is not divisible by 3, so f(2)f(2)f(2) is divisible by 3^1 but not by 3^2.

Similarly, for other nnn, if n≡0(mod3)n \equiv 0 \pmod{3}n≡0(mod3), then f(n)=n99(n60−1)f(n) = n^{99}(n^{60} - 1)f(n)=n99(n60−1), and since n≡0(mod3)n \equiv 0 \pmod{3}n≡0(mod3), n99≡0(mod3)n^{99} \equiv 0 \pmod{3}n99≡0(mod3), and n60−1≡−1(mod3)n^{60} - 1 \equiv -1 \pmod{3}n60−1≡−1(mod3), so f(n)≡0∗(−1)=0(mod3)f(n) \equiv 0 * (-1) = 0 \pmod{3}f(n)≡0∗(−1)=0(mod3), but the exponent depends on nnn. For example, for n=3n = 3n=3, f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and 360−1≡−1(mod3)3^{60} - 1 \equiv -1 \pmod{3}360−1≡−1(mod3), so the exponent of 3 in f(3)f(3)f(3) is 99, since 360−13^{60} - 1360−1 is not divisible by 3.

Wait, actually, since 360≡0(mod3)3^{60} \equiv 0 \pmod{3}360≡0(mod3), 360−1≡−1(mod3)3^{60} - 1 \equiv -1 \pmod{3}360−1≡−1(mod3), so 3 does not divide 360−13^{60} - 1360−1, thus the exponent of 3 in f(3)f(3)f(3) is exactly 99.

But since for n=2n = 2n=2, the exponent is 1, and for other nnn, it could be higher, but the minimal exponent is 1, so in the GCD ddd, the exponent of 3 is 1.

Similarly, for p=5p = 5p=5, let's check the exponent in f(n)f(n)f(n). For example, for n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and since 2^4 ≡ 1 \pmod{5}, and 4|60, so 260≡1(mod5)2^{60} \equiv 1 \pmod{5}260≡1(mod5), thus 260−1≡0(mod5)2^{60} - 1 \equiv 0 \pmod{5}260−1≡0(mod5), and in fact, since the order is 4, and 4|60, so 260−1=(24−1)∗(something)=15∗(something)2^{60} - 1 = (2^4 - 1) * (\text{something}) = 15 * (\text{something})260−1=(24−1)∗(something)=15∗(something), and 15 = 3*5, so 5 divides 260−12^{60} - 1260−1, but to find the exact exponent, note that since 2^4 ≡ 1 \pmod{5}, and actually, for higher powers, since 5 is prime, we can use the lifting-the-exponent lemma, which states that for odd prime ppp, and integers a,ba, ba,b such that ppp does not divide aaa or bbb, then the exponent of ppp in an±bna^n \pm b^nan±bn can be determined.

In this case, for 260−1=260−1602^{60} - 1 = 2^{60} - 1^{60}260−1=260−160, and since 5 does not divide 2 or 1, and since 60 is divisible by the order of 2 mod 5, which is 4, so the exponent of 5 in 260−12^{60} - 1260−1 is 1, because the lemma says that if nnn is even, and p∣(a+b)p | (a + b)p∣(a+b), but here a=2,b=1a = 2, b = 1a=2,b=1, so a+b=3a + b = 3a+b=3, and 5 does not divide 3, so the exponent is 1.

Wait, more precisely, the lifting-the-exponent lemma for an−bna^n - b^nan−bn when ppp is an odd prime, and ppp does not divide aaa or bbb, and p∣(a−b)p | (a - b)p∣(a−b), but here a−b=2−1=1a - b = 2 - 1 = 1a−b=2−1=1, and 5 does not divide 1, so the exponent is 0, but wait, actually, since 260≡1(mod5)2^{60} \equiv 1 \pmod{5}260≡1(mod5), so 5 divides 260−12^{60} - 1260−1, but the exponent is 1, meaning 51∣(260−1)5^1 | (2^{60} - 1)51∣(260−1), but 525^252 may or may not divide, depending on further conditions.

In fact, for p=5p = 5p=5, since 24≡1(mod5)2^4 \equiv 1 \pmod{5}24≡1(mod5), but to check modulo 25, let's see 24=16≡16−3∗5=16−15=1(mod5)2^4 = 16 \equiv 16 - 35 = 16 - 15 = 1 \pmod{5}24=16≡16−3∗5=16−15=1(mod5), but actually 16≡16(mod25)16 \equiv 16 \pmod{25}16≡16(mod25), wait, 16 < 25, so 24=16≡16(mod25)2^4 = 16 \equiv 16 \pmod{25}24=16≡16(mod25), and since 16 ≠ 1 mod 25, so the order of 2 mod 25 is greater than 4, actually, since 2^5 = 32 ≡ 7, 2^10 ≡ 7^2 = 49 ≡ 24 ≡ -1, 2^20 ≡ (-1)^2 ≡ 1 \pmod{25}, so the order is 20. Therefore, since 20 does not divide 60, wait, 60 = 320, so actually 20|60, yes, so 260=(220)3≡13≡1(mod25)2^{60} = (2^{20})^3 \equiv 1^3 \equiv 1 \pmod{25}260=(220)3≡13≡1(mod25), thus 260−1≡0(mod25)2^{60} - 1 \equiv 0 \pmod{25}260−1≡0(mod25), so 25 divides 260−12^{60} - 1260−1, meaning the exponent of 5 in 260−12^{60} - 1260−1 is at least 2.

Wait, so for n=2n = 2n=2, since 260−12^{60} - 1260−1 is divisible by 25, and 2992^{99}299 is not divisible by 5, so f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1) is divisible by 25.

Similarly, for other nnn, say n=3n = 3n=3, f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and let's check the exponent of 5 in 360−13^{60} - 1360−1. Since 3 mod 5 has order 4, and 4|60, so 360≡1(mod5)3^{60} \equiv 1 \pmod{5}360≡1(mod5), and actually, since the order is 4, and for modulo 25, let's check if 34≡1(mod25)3^4 \equiv 1 \pmod{25}34≡1(mod25), but 3^4 = 81 ≡ 6 \pmod{25}), since 81 - 325 = 81 - 75 = 6, and 6 ≠ 1, so the order is larger, in fact, since 3^5 = 243 ≡ 243 - 925 = 243 - 225 = 18, 3^10 ≡ 18^2 = 324 ≡ 324 - 12*25 = 324 - 300 = 24 ≡ -1 \pmod{25}, so 3^20 ≡ 1 \pmod{25}, thus the order is 20. Since 20|60, 360≡1(mod25)3^{60} \equiv 1 \pmod{25}360≡1(mod25), so 360−1≡0(mod25)3^{60} - 1 \equiv 0 \pmod{25}360−1≡0(mod25), thus 25 divides 360−13^{60} - 1360−1, and since 3993^{99}399 is not divisible by 5, so f(3)f(3)f(3) is divisible by 25.

Similarly, for n=4n = 4n=4, f(4)=499(460−1)f(4) = 4^{99}(4^{60} - 1)f(4)=499(460−1), and 4 ≡ 4 \pmod{5}, 4^2 = 16 ≡ 1 \pmod{5}, so order is 2, and since 2|60, 460≡1(mod5)4^{60} \equiv 1 \pmod{5}460≡1(mod5), and for modulo 25, since 4^2 = 16 ≡ 16 \pmod{25}, and 16^2 = 256 ≡ 256 - 1025 = 256 - 250 = 6, wait, actually, let's compute 4^4 = (4^2)^2 ≡ 16^2 = 256 ≡ 256 - 1025 = 256 - 250 = 6 \pmod{25}, 4^5 ≡ 4*6 = 24 ≡ -1 \pmod{25}, so 4^10 ≡ 1 \pmod{25}, thus the order is 10. Since 10|60, 460≡1(mod25)4^{60} \equiv 1 \pmod{25}460≡1(mod25), so 460−1≡0(mod25)4^{60} - 1 \equiv 0 \pmod{25}460−1≡0(mod25), and since 4 is not divisible by 5, 4994^{99}499 is not divisible by 5, so f(4)f(4)f(4) is divisible by 25.

In fact, perhaps for all n>1n > 1n>1, since for n≢0(mod5)n \not\equiv 0 \pmod{5}n≡0(mod5), the order of nnn mod 5 is a divisor of 4, and since 4|60, n60≡1(mod5)n^{60} \equiv 1 \pmod{5}n60≡1(mod5), and actually, since 5 is prime, and the group is cyclic, for n≢0(mod5)n \not\equiv 0 \pmod{5}n≡0(mod5), n4≡1(mod5)n^{4} \equiv 1 \pmod{5}n4≡1(mod5), and for higher powers, depending on the order mod 25, but since 20 is the maximal order mod 25, and 20|60, because 60 = 3*20, so for any n≢0(mod5)n \not\equiv 0 \pmod{5}n≡0(mod5), n20≡1(mod25)n^{20} \equiv 1 \pmod{25}n20≡1(mod25), thus n60=(n20)3≡13≡1(mod25)n^{60} = (n^{20})^3 \equiv 1^3 \equiv 1 \pmod{25}n60=(n20)3≡13≡1(mod25), so n60−1≡0(mod25)n^{60} - 1 \equiv 0 \pmod{25}n60−1≡0(mod25), therefore 25 divides n60−1n^{60} - 1n60−1, and for n≡0(mod5)n \equiv 0 \pmod{5}n≡0(mod5), n99n^{99}n99 is divisible by 5^{99}, so in either case, f(n)f(n)f(n) is divisible by 25.

Moreover, to check if it's divisible by higher powers, say 5^3 = 125, let's see for some nnn, for example, take n=2n = 2n=2, since 260−12^{60} - 1260−1, and since 2 has order 20 mod 25, but to check mod 125, perhaps we can use the lifting-the-exponent lemma again. Actually, for an−1a^n - 1an−1, when p∣(a−1)p | (a - 1)p∣(a−1), but here a=2,p=5a = 2, p = 5a=2,p=5, and 2 - 1 = 1, 5 does not divide 1, so the exponent is determined by the first kkk such that ak≡1(modp)a^k \equiv 1 \pmod{p}ak≡1(modp), but since the order is 4 mod 5, and for mod 25, the order is 20, and since 20|60, so 260≡1(mod25)2^{60} \equiv 1 \pmod{25}260≡1(mod25), but for mod 125, let's see if 260≡1(mod125)2^{60} \equiv 1 \pmod{125}260≡1(mod125). Actually, the order of 2 mod 125 can be found; since 125 = 5^3, and the group is cyclic of order ϕ(125)=100\phi(125) = 100ϕ(125)=100, so the order divides 100, and since 2^4 = 16 ≡ 16 \pmod{125}, 2^5 = 32, 2^10 = 1024 ≡ 1024 - 8125 = 1024 - 1000 = 24 ≡ -101, wait, perhaps better to compute sequentially. Actually, it's known that 2 is a generator mod 5^k for k≥1k \geq 1k≥1, but wait, for 5^2 = 25, 2 has order 20, but 20 < ϕ(25)=20\phi(25) = 20ϕ(25)=20, wait, 20 = 20, so order is 20, but for 125, since 2^20 ≡ 1 \pmod{25}), but perhaps not mod 125. Actually, for p=5, since 2 is a primitive root mod 5, and 2^{p-1} = 2^4 = 16 ≡ 16 - 35 = 1 \pmod{5}), wait, 16 ≡ 1 \pmod{5}, but for higher, the condition for 2 to be a primitive root mod 5^k is that 2 is a primitive root mod 5 and 2^2 ≠ 1 mod 5^2, but 2^2 = 4 ≠ 1 mod 25, so yes, 2 is a primitive root mod 5^k for all k, meaning that the order of 2 mod 125 is 100. Therefore, since 100 does not divide 60, 260≢1(mod125)2^{60} \not\equiv 1 \pmod{125}260≡1(mod125), so 260−1≢0(mod125)2^{60} - 1 \not\equiv 0 \pmod{125}260−1≡0(mod125), thus 125 does not divide 260−12^{60} - 1260−1, and since 2992^{99}299 is not divisible by 5, so f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1) is not divisible by 125.

Therefore, the highest power of 5 that divides all f(n)f(n)f(n) is 5^2 = 25.

Similarly, for other primes in the list, I need to find the minimal exponent.

In general, for each prime ppp in {2,3,5,7,11,13,31,61}\{2,3,5,7,11,13,31,61\}{2,3,5,7,11,13,31,61}, I can find the minimal exponent epe_pep such that pepp^{e_p}pep divides f(n)f(n)f(n) for all n>1n > 1n>1, and then d=∏pepd = \prod p^{e_p}d=∏pep.

For p=2p = 2p=2, as above, e2=4e_2 = 4e2=4, since 2^4 divides all f(n)f(n)f(n), but 2^5 does not divide all f(n)f(n)f(n).

For p=3p = 3p=3, from earlier, for n=2n = 2n=2, f(2)f(2)f(2) is divisible by 3^1 but not by 3^2, since 260−1=3∗m2^{60} - 1 = 3 * m260−1=3∗m with mmm not divisible by 3, so e3=1e_3 = 1e3=1.

For p=5p = 5p=5, as above, e5=2e_5 = 2e5=2, since 5^2 divides all f(n)f(n)f(n), but 5^3 does not divide all f(n)f(n)f(n).

Similarly, for p=7p = 7p=7, let's find the minimal exponent. For example, take n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and since 2^3 ≡ 1 \pmod{7}, and 3|60, so 260≡1(mod7)2^{60} \equiv 1 \pmod{7}260≡1(mod7), thus 7 divides 260−12^{60} - 1260−1, and actually, since the order is 3, and 3|60, so the exponent of 7 in 260−12^{60} - 1260−1 can be found using the lifting-the-exponent lemma. Since 7 is prime, and 2^3 ≡ 1 \pmod{7}, and 3|60, in fact, for ak−1a^k - 1ak−1 where a≡1(mod7)a \equiv 1 \pmod{7}a≡1(mod7), but wait, more precisely, since 2^3 ≡ 1 \pmod{7}, and 60 = 320, so 260=(23)20≡120≡1(mod7)2^{60} = (2^3)^{20} \equiv 1^{20} ≡ 1 \pmod{7}260=(23)20≡120≡1(mod7), but to find the exponent, we can check mod 49, since 7^2 = 49. Let's see if 23≡1(mod49)2^3 \equiv 1 \pmod{49}23≡1(mod49), but 2^3 = 8 ≠ 1 mod 49, so the order is larger. Actually, since 7 is prime, and 2 is not a multiple of 7, the order divides ϕ(49)=42\phi(49) = 42ϕ(49)=42, and since 2^3 = 8 ≡ 8, 2^6 ≡ 64 ≡ 64 - 49 = 15, 2^7 ≡ 215 = 30, 2^14 ≡ 30^2 = 900 ≡ 900 - 1849 = 900 - 882 = 18, wait, perhaps better to find that the multiplicative order of 2 mod 49 is 21, since 2^21 = (2^7)^3 = 128^3, but perhaps compute step by step. Actually, it's known that 2 has order 42 mod 49, but since 2^3 ≡ 8, and 8^7 = (2^3)^7 = 2^21, etc., wait, alternatively, since 49 = 7^2, and the group is cyclic, but for simplicity, let's assume that since 2 is a generator mod 7, and check if 2^6 ≡ 1 mod 49, but 2^6 = 64 ≡ 15, 15 ≠ 1, 2^7 = 128 ≡ 128 - 249 = 128 - 98 = 30, 2^14 ≡ 30^2 = 900 ≡ 900 - 1849 = 900 - 882 = 18, 2^21 ≡ 2^7 * 2^14 ≡ 3018 = 540 ≡ 540 - 1149 = 540 - 539 = 1 \pmod{49}, so yes, 2^21 ≡ 1 \pmod{49}, thus the order of 2 mod 49 is 21. Since 21 does not divide 60, because 60 = 212 + 18, wait, 212 = 42 < 60, but 21 does not divide 60, since 60 / 21 ≈ 2.857, not integer, so 260=221∗2+18=(221)2∗218≡12∗218≡218(mod49)2^{60} = 2^{212 + 18} = (2^{21})^2 * 2^{18} ≡ 1^2 * 2^{18} ≡ 2^{18} \pmod{49}260=221∗2+18=(221)2∗218≡12∗218≡218(mod49), and since 2^18 is not 1 mod 49, so 260≢1(mod49)2^{60} \not\equiv 1 \pmod{49}260≡1(mod49), thus 260−1≢0(mod49)2^{60} - 1 \not\equiv 0 \pmod{49}260−1≡0(mod49), meaning that 49 does not divide 260−12^{60} - 1260−1, so the exponent of 7 in 260−12^{60} - 1260−1 is 1, since 7 divides 260−12^{60} - 1260−1 but 7^2 does not.

Therefore, for n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and since 260−12^{60} - 1260−1 is divisible by 7 but not by 7^2, and 2992^{99}299 is not divisible by 7, so f(2)f(2)f(2) is divisible by 7^1 but not by 7^2.

Hence, the minimal exponent for p=7p = 7p=7 is 1.

Similarly, for other primes, I can check the exponent.

Wait, actually, let's try to find a general pattern. For a prime ppp in {2,3,5,7,11,13,31,61}\{2,3,5,7,11,13,31,61\}{2,3,5,7,11,13,31,61}, since p−1∣60p - 1 | 60p−1∣60, and 60 = (p - 1) * k for some kkk, but actually, since p−1∣60p - 1 | 60p−1∣60, and for n≢0(modp)n \not\equiv 0 \pmod{p}n≡0(modp), np−1≡1(modp)n^{p-1} \equiv 1 \pmod{p}np−1≡1(modp), so n60≡1(modp)n^{60} \equiv 1 \pmod{p}n60≡1(modp), thus p∣(n60−1)p | (n^{60} - 1)p∣(n60−1), and for n≡0(modp)n \equiv 0 \pmod{p}n≡0(modp), p∣n99p | n^{99}p∣n99, so in either case, p∣f(n)p | f(n)p∣f(n).

Now, to find the exact exponent, it depends on the specific ppp. For example, for p=5p = 5p=5, since 5 - 1 = 4 | 60, and in fact, for n≢0(mod5)n \not\equiv 0 \pmod{5}n≡0(mod5), n60≡1(mod52)n^{60} \equiv 1 \pmod{5^2}n60≡1(mod52), as we saw earlier, because the order divides 20, and 20|60, so yes, n60≡1(mod25)n^{60} \equiv 1 \pmod{25}n60≡1(mod25), thus 25 | (n^{60} - 1), so for n≢0(mod5)n \not\equiv 0 \pmod{5}n≡0(mod5), 52∣(n60−1)5^2 | (n^{60} - 1)52∣(n60−1), and for n≡0(mod5)n \equiv 0 \pmod{5}n≡0(mod5), 5∣n995 | n^{99}5∣n99, but actually, since n=5mn = 5mn=5m, n99=(5m)99=599m99n^{99} = (5m)^{99} = 5^{99} m^{99}n99=(5m)99=599m99, so 599∣n995^{99} | n^{99}599∣n99, so in this case, f(n)=n99(n60−1)f(n) = n^{99}(n^{60} - 1)f(n)=n99(n60−1) is divisible by 5995^{99}599, but since we're looking for the minimal exponent, and for n≢0(mod5)n \not\equiv 0 \pmod{5}n≡0(mod5), 52∣(n60−1)5^2 | (n^{60} - 1)52∣(n60−1), but perhaps not higher.

Similarly, for p=2p = 2p=2, earlier we saw that for odd nnn, 24∣(n60−1)2^4 | (n^{60} - 1)24∣(n60−1), but not necessarily higher, so the minimal exponent is 4.

For p=3p = 3p=3, for n≢0(mod3)n \not\equiv 0 \pmod{3}n≡0(mod3), since 3 - 1 = 2 | 60, and the order of nnn mod 3 is 1 or 2, so n2≡1(mod3)n^{2} \equiv 1 \pmod{3}n2≡1(mod3), thus n60≡1(mod3)n^{60} \equiv 1 \pmod{3}n60≡1(mod3), so 3 | (n^{60} - 1), but for higher powers, say mod 9, for example, take n=2n = 2n=2, 2 ≡ 2 \pmod{9}, 2^2 = 4, 2^4 = 16 ≡ 7, 2^8 ≡ 7^2 = 49 ≡ 4, wait, actually, 2^3 = 8 ≡ -1 \pmod{9}, 2^6 ≡ (-1)^2 ≡ 1 \pmod{9}, so the order is 6. Since 6|60, 260≡1(mod9)2^{60} \equiv 1 \pmod{9}260≡1(mod9), thus 9 | (2^{60} - 1), so for n=2n = 2n=2, since 2 ≡ 2 \pmod{9}, and 260≡1(mod9)2^{60} \equiv 1 \pmod{9}260≡1(mod9), wait, actually, since 2^6 ≡ 1 \pmod{9}, and 60 = 610, so yes, 260≡1(mod9)2^{60} \equiv 1 \pmod{9}260≡1(mod9), thus 9 | (2^{60} - 1). Similarly, for other n≢0(mod3)n \not\equiv 0 \pmod{3}n≡0(mod3), since the multiplicative group mod 9 is cyclic of order 6, and 6|60, so n60≡1(mod9)n^{60} \equiv 1 \pmod{9}n60≡1(mod9) for n≢0(mod3)n \not\equiv 0 \pmod{3}n≡0(mod3), thus 9 | (n^{60} - 1). Therefore, for n≢0(mod3)n \not\equiv 0 \pmod{3}n≡0(mod3), 32∣(n60−1)3^2 | (n^{60} - 1)32∣(n60−1), and for n≡0(mod3)n \equiv 0 \pmod{3}n≡0(mod3), 3∣n993 | n^{99}3∣n99, so in general, 32∣f(n)3^2 | f(n)32∣f(n) for all n>1n > 1n>1, wait, no, for n≡0(mod3)n \equiv 0 \pmod{3}n≡0(mod3), since n99n^{99}n99 is divisible by 3, but n60−1≡−1(mod3)n^{60} - 1 \equiv -1 \pmod{3}n60−1≡−1(mod3), so f(n)=n99(n60−1)f(n) = n^{99}(n^{60} - 1)f(n)=n99(n60−1) is divisible by 3, but not necessarily by 9, because if n=3n = 3n=3, f(3)=399(360−1)f(3) = 3^{99}(3^{60} - 1)f(3)=399(360−1), and 360−1≡−1(mod9)3^{60} - 1 \equiv -1 \pmod{9}360−1≡−1(mod9), since for example 31−1=23^1 - 1 = 231−1=2, 2 ≡ 2 \pmod{9}, not 0, so 360−1≢0(mod9)3^{60} - 1 \not\equiv 0 \pmod{9}360−1≡0(mod9), thus 9 does not divide 360−13^{60} - 1360−1, so the exponent of 3 in f(3)f(3)f(3) is exactly 99, but for the GCD, since for n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and since 2^{60} - 1) is divisible by 9, as above, but wait, is it divisible by higher powers? Actually, for p=3p = 3p=3, to find the exponent in 260−12^{60} - 1260−1, since 2^6 ≡ 1 \pmod{9}, and the order is 6, so for k=60=6∗10k = 60 = 610k=60=6∗10, 260−1=(26)10−1=(1+9m)10−1≡1+10∗9m−1=90m(mod81)2^{60} - 1 = (2^6)^{10} - 1 = (1 + 9m)^{10} - 1 \equiv 1 + 109m - 1 = 90m \pmod{81}260−1=(26)10−1=(1+9m)10−1≡1+10∗9m−1=90m(mod81), so since 90 = 910, and 81 = 9*9, so 260−1≡0(mod9)2^{60} - 1 \equiv 0 \pmod{9}260−1≡0(mod9), but not necessarily 0 mod 27, depending on mmm, but in general, the exponent is at least 2, but for some nnn, it might be exactly 2.

Wait, actually, let's check for n=4n = 4n=4, since 4 ≡ 1 \pmod{3}, 4^2 = 16 ≡ 1 \pmod{3}, actually, 4 ≡ 1 \pmod{9}, since 4 - 1 = 3, not divisible by 9, but 4^2 = 16 ≡ 7 ≡ -2 \pmod{9}, wait, 16 - 9 = 7, 7 - 9 = -2, but perhaps better, 4 ≡ 4 \pmod{9}, 4^2 = 16 ≡ 7, 4^3 ≡ 47 = 28 ≡ 1, since 28 - 39 = 28 - 27 = 1, so 4^3 ≡ 1 \pmod{9}, thus the order of 4 mod 9 is 3, and since 3|60, 460≡1(mod9)4^{60} \equiv 1 \pmod{9}460≡1(mod9), so 460−1≡0(mod9)4^{60} - 1 \equiv 0 \pmod{9}460−1≡0(mod9), but to check if it's divisible by 27, let's see mod 27. For mod 27, let's compute 4^3 = 64 ≡ 64 - 227 = 64 - 54 = 10, wait, 64 - 27 = 37, 37 - 27 = 10, so 4^3 ≡ 10 \pmod{27}, and 10^2 = 100 ≡ 100 - 327 = 100 - 81 = 19, so 4^6 ≡ 10^2 ≡ 19 \pmod{27}, 4^9 ≡ 4^6 * 4^3 ≡ 1910 = 190 ≡ 190 - 727 = 190 - 189 = 1 \pmod{27}, so the order of 4 mod 27 is 9, since 4^9 ≡ 1 \pmod{27}, and 9|60, since 60 = 96 + 6, wait, 96 = 54 < 60, but 9 does not divide 60, since 60 / 9 = 6.666..., so 460=(49)6∗46≡16∗46≡46(mod27)4^{60} = (4^9)^6 * 4^6 ≡ 1^6 * 4^6 ≡ 4^6 \pmod{27}460=(49)6∗46≡16∗46≡46(mod27), and from above, 4^6 ≡ 19 \pmod{27}, so 460−1≡19−1=18(mod27)4^{60} - 1 ≡ 19 - 1 = 18 \pmod{27}460−1≡19−1=18(mod27), thus 460−1≡18(mod27)4^{60} - 1 \equiv 18 \pmod{27}460−1≡18(mod27), which is divisible by 9 but not by 27, since 18 ≡ 18 \pmod{27}), not 0.

Therefore, for n=4n = 4n=4, f(4)=499(460−1)f(4) = 4^{99}(4^{60} - 1)f(4)=499(460−1), and since 4^{60} - 1 ≡ 18 \pmod{27}), so the exponent of 3 in 460−14^{60} - 1460−1 is 1, since 18 is divisible by 3 but not by 9, wait, actually, 18 = 2*9, so divisible by 9 = 3^2, but wait, 18 ≡ 0 \pmod{9}), since 9|18, but 27 does not divide 18, so exponent is 2, since 9| (4^{60} - 1) but 27 does not.

Wait, earlier I said 460−1≡18(mod27)4^{60} - 1 ≡ 18 \pmod{27}460−1≡18(mod27), and 18 = 29 = 23^2, so yes, divisible by 3^2 but not by 3^3.

Therefore, for n=4n = 4n=4, since 4 is not divisible by 3, 4994^{99}499 is not divisible by 3, so f(4)=499(460−1)f(4) = 4^{99}(4^{60} - 1)f(4)=499(460−1) is divisible by 3^2 but not by 3^3.

So, for p=3p = 3p=3, the minimal exponent is 2, because for some nnn, like n=4n = 4n=4, f(n)f(n)f(n) is divisible by 3^2 but not by 3^3.

Wait, earlier for n=2n = 2n=2, since 260−12^{60} - 1260−1 is divisible by 9, as 22−1=32^2 - 1 = 322−1=3, but for larger kkk, since 2^6 ≡ 1 \pmod{9}, so 260=(26)10≡110≡1(mod9)2^{60} = (2^6)^{10} ≡ 1^{10} ≡ 1 \pmod{9}260=(26)10≡110≡1(mod9), but to check mod 27, let's compute 2^6 = 64 ≡ 64 - 227 = 64 - 54 = 10 \pmod{27}, and 10^2 = 100 ≡ 100 - 327 = 100 - 81 = 19, so 2^12 ≡ 19 \pmod{27}, 2^18 ≡ 2^6 * 2^12 ≡ 1019 = 190 ≡ 190 - 727 = 190 - 189 = 1 \pmod{27}, so 2^18 ≡ 1 \pmod{27}, thus the order of 2 mod 27 is 18, since 18|60, 260=(218)3∗26≡13∗26≡64≡10(mod27)2^{60} = (2^18)^3 * 2^6 ≡ 1^3 * 2^6 ≡ 64 ≡ 10 \pmod{27}260=(218)3∗26≡13∗26≡64≡10(mod27), wait, actually, since 60 = 183 + 6, but since 2^18 ≡ 1, so 260=218∗3∗26≡1∗26≡64≡10(mod27)2^{60} = 2^{183} * 2^6 ≡ 1 * 2^6 ≡ 64 ≡ 10 \pmod{27}260=218∗3∗26≡1∗26≡64≡10(mod27), so 260−1≡10−1=9(mod27)2^{60} - 1 ≡ 10 - 1 = 9 \pmod{27}260−1≡10−1=9(mod27), thus 260−1≡9(mod27)2^{60} - 1 ≡ 9 \pmod{27}260−1≡9(mod27), which means it's divisible by 9 but not by 27, since 9 ≡ 9 \pmod{27}), not 0.

Therefore, for n=2n = 2n=2, f(2)=299(260−1)f(2) = 2^{99}(2^{60} - 1)f(2)=299(260−1), and 260−1≡9(mod27)2^{60} - 1 ≡ 9 \pmod{27}260−1≡9(mod27), so exponent of 3 is 2, since 9 = 3^2.

Similarly, for other nnn, the exponent may be higher, but since for n=2n = 2n=2, it's 2, and for n=4n = 4n=4, it's also 2, etc., so the minimal exponent for p=3p = 3p=3 is 2.

Wait, earlier I said for n=4n = 4n=4, it's divisible by 3^2, yes.

Therefore, for p=3p = 3p=3, e3=2e_3 = 2e3=2.

Similarly, for p=5p = 5p=5, as previously, e5=2e_5 = 2e5=2, since for some nnn, like n=2n = 2n=2, f(2)f(2)f(2) is divisible by 5^2 but not by 5^3.

For p=7p = 7p=7, from earlier, for n=2n = 2n=2, f(2)f(2)f(2) is divisible by 7^1 but not by 7^2, so e7=1e_7 = 1e7=1.

For p=11p = 11p=11, let's check. First, note that 11 - 1 = 10 | 60, since 60 = 6*10, yes. So for n≢0(mod11)n \not\equiv 0 \pmod{11}n≡0(mod11), n10≡1(mod11)n^{10} \equiv 1 \pmod{11}n10≡1(mod11), thus n60≡1(mod11)n^{60} \equiv 1 \pmod{11}n60≡1(mod11), so 11 | (n^{60} - 1). Actually, since 11 is prime, and the group is cyclic, for n≢0(mod11)n \not\equiv 0 \pmod{11}n≡0(mod11), n10≡1(mod11)n^{10} \equiv 1 \pmod{11}n10≡1(mod11), but for mod 121 = 11^2, the order could be larger. In fact, since 11^2 = 121, and ϕ(121)=110\phi(121) = 110ϕ(121)=110, the maximal order is 110, and since 10|60, but 110 does not divide 60, so for some nnn, n60≢1(mod121)n^{60} \not\equiv 1 \pmod{121}n60≡1(mod121), thus for those nnn, 121 does not divide n60−1n^{60} - 1n60−1, so the exponent of 11 in n60−1n^{60} - 1n60−1 is 1 for those nnn, and since n99n^{99}n99 may not be divisible by 11, so for such nnn, f(n)f(n)f(n) is divisible by 11^1 but not by 11^2. Therefore, e11=1e_{11} = 1e11=1.

Similarly, for p=13p = 13p=13, since 13 - 1 = 12 | 60, and 12|60, so for n≢0(mod13)n \not\equiv 0 \pmod{13}n≡0(mod13), n12≡1(mod13)n^{12} \equiv 1 \pmod{13}n12≡1(mod13), thus n60≡1(mod13)n^{60} \equiv 1 \pmod{13}n60≡1(mod13), and for mod 169 = 13^2, the maximal order is ϕ(169)=156\phi(169) = 156ϕ(169)=156, and since 12|60 but 156 does not divide 60, so there are nnn such that n60≢1(mod169)n^{60} \not\equiv 1 \pmod{169}n60≡1(mod169), thus 169 does not divide n60−1n^{60} - 1n60−1, so e13=1e_{13} = 1e13=1.

For p=31p = 31p=31, since 31 - 1 = 30 | 60, and 30|60, so similarly, for n≢0(mod31)n \not\equiv 0 \pmod{31}n≡0(mod31), n30≡1(mod31)n^{30} \equiv 1 \pmod{31}n30≡1(mod31), thus n60≡1(mod31)n^{60} \equiv 1 \pmod{31}n60≡1(mod31), and for mod 961 = 31^2, the maximal order is ϕ(961)=930\phi(961) = 930ϕ(961)=930, and 30|60 but 930 does not divide 60, so there are nnn such that n60≢1(mod961)n^{60} \not\equiv 1 \pmod{961}n60≡1(mod961), thus e31=1e_{31} = 1e31=1.

Finally, for p=61p = 61p=61, since 61 - 1 = 60 | 60, so for n≢0(mod61)n \not\equiv 0 \pmod{61}n≡0(mod61), n60≡1(mod61)n^{60} \equiv 1 \pmod{61}n60≡1(mod61), and actually, since 60|60, in fact, n60≡1(mod61)n^{60} \equiv 1 \pmod{61}n60≡1(mod61), but for mod 3721 = 61^2, the maximal order is ϕ(3721)=3660\phi(3721) = 3660ϕ(3721)=3660, and since 60|60, but 3660 does not divide 60, wait, actually, for elements of order dividing 60, but since the group is cyclic, there are elements with order up to 3660, so for those nnn, n60≢1(mod3721)n^{60} \not\equiv 1 \pmod{3721}n60≡1(mod3721), thus 61^2 does not divide n60−1n^{60} - 1n60−1, so e61=1e_{61} = 1e61=1.

Therefore, for all primes ppp in {7,11,13,31,61}\{7,11,13,31,61\}{7,11,13,31,61}, the minimal exponent ep=1e_p = 1ep=1.

For p=2p = 2p=2, e2=4e_2 = 4e2=4, for p=3p = 3p=3, e3=2e_3 = 2e3=2, for p=5p = 5p=5, e5=2e_5 = 2e5=2, and for the others, ep=1e_p = 1ep=1.

Wait, but earlier for p=5p = 5p=5, I said e5=2e_5 = 2e5=2, but let's confirm for p=5p = 5p=5, is there a higher exponent? Wait, no, since for some nnn, like n=2n = 2n=2, f(2)f(2)f(2) is divisible by 5^2 but not by 5^3, as previously established.

Similarly, for p=3p = 3p=3, e3=2e_3 = 2e3=2, since for n=2n = 2n=2, f(2)f(2)f(2) is divisible by 3^2 but not by 3^3.

For p=2p = 2p=2, e2=4e_2 = 4e2=4, as determined earlier.

Therefore, the GCD ddd of the set {n99(n60−1)∣n>1}\{n^{99}(n^{60} - 1) | n > 1\}{n99(n60−1)∣n>1} is 24⋅32⋅52⋅7⋅11⋅13⋅31⋅612^4 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 31 \cdot 6124⋅32⋅52⋅7⋅11⋅13⋅31⋅61.

So, to compute the numerical value, let's calculate each part: 24=162^4 = 1624=16, 32=93^2 = 932=9, 52=255^2 = 2552=25, and then the product of the primes 7, 11, 13, 31, 61.

First, compute 7⋅11=777 \cdot 11 = 777⋅11=77, 77⋅13=100177 \cdot 13 = 100177⋅13=1001, 1001⋅31=310311001 \cdot 31 = 310311001⋅31=31031, 31031⋅61=31031⋅60+31031⋅1=1861860+31031=189289131031 \cdot 61 = 31031 \cdot 60 + 31031 \cdot 1 = 1861860 + 31031 = 189289131031⋅61=31031⋅60+31031⋅1=1861860+31031=1892891.

Then, 16⋅9=14416 \cdot 9 = 14416⋅9=144, 144⋅25=3600144 \cdot 25 = 3600144⋅25=3600, and 3600⋅18928913600 \cdot 18928913600⋅1892891.

But perhaps it's better to leave it in factored form, since the problem likely expects the answer in terms of the product.

Alternatively, since the problem asks to compute the GCD, and given the context, probably the answer is expected to be the product of these primes to their respective exponents.

Therefore, the greatest common divisor of the set {n99(n60−1)∣n>1}\{n^{99}(n^{60} - 1) \mid n > 1\}{n99(n60−1)∣n>1} is 24⋅32⋅52⋅7⋅11⋅13⋅31⋅612^4 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 31 \cdot 6124⋅32⋅52⋅7⋅11⋅13⋅31⋅61.

To confirm, let's note that for each prime ppp not in this list, there exists n>1n > 1n>1 such that ppp does not divide f(n)f(n)f(n), so ddd cannot be divisible by any other primes.

Moreover, for the primes in the list, the exponents are chosen to be the minimal such that pepp^{e_p}pep divides f(n)f(n)f(n) for all n>1n > 1n>1.

Hence, the final answer is 24⋅32⋅52⋅7⋅11⋅13⋅31⋅612^4 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 31 \cdot 6124⋅32⋅52⋅7⋅11⋅13⋅31⋅61.

So, to box the answer:

24⋅32⋅52⋅7⋅11⋅13⋅31⋅61 \boxed{2^{4} \cdot 3^{2} \cdot 5^{2} \cdot 7 \cdot 11 \cdot 13 \cdot 31 \cdot 61} 24⋅32⋅52⋅7⋅11⋅13⋅31⋅61

Final Answer

24⋅32⋅52⋅7⋅11⋅13⋅31⋅61 \boxed{2^{4} \cdot 3^{2} \cdot 5^{2} \cdot 7 \cdot 11 \cdot 13 \cdot 31 \cdot 61} 24⋅32⋅52⋅7⋅11⋅13⋅31⋅61