Determine the GCD of all number that can be writen under the form of $p^8-1$, with $p$ prime and $p>5$

Question

First of all they are all even... So it is at least $2$. But is it bigger than that? How do I find out?

Answer 1

HINT:

As $(p,2)=1$ using Carmichael Function,

for $n\ge3,\lambda(2^n)=2^{n-2}$ we have $2^{n-2}=8\iff n=5$

Again, for any prime $q,\lambda(q^m)=q^{m-1}(q-1)$

So, $q-1$ must divide $8\implies q=3,5$

Observe that $(p,3)=(p,5)=1$

For $q=3,\lambda(3^{a+1})=3^a(3-1)$ which must divide $8\implies a=0$

Can you handle $q=5?$

Answer 2

We see $p^8-1 =(p^4+1)(p^4-1)=(p^4+1)(p^2+1)(p+1)(p-1)$. Each of these is divisible by $2$ so the product is divisible by $16$. Also, one of $p+1$,$p-1$ is divisible by $3$ and $p^4-1$ is divisible by $5$ by Fermat's little theorem so multiplying these divisors, we have that the original expression is divisible by $240$. I'm not sure whether we can find a larger divisor.

EDIT: In the comments above @DanielFischer points out that we can get an extra factor of $2$. It's likely the answer is $480$.