What is the greatest integer that divides $p^4-1$ for every prime number $p$ greater than $5$? What is the greatest integer that divides $p^4-1$ for every prime number $p > 5$?  This was on a practice math GRE so it's probably really easy.

Let $n$ be the number. Then the only prime factors of $n$ are 2,3,5, because any  prime number $p>5$ doesn't divide $p^4-1$ for all other $p>5$.
So $n=2^a3^b5^c$, and using the fact that $p^2-1 \cong 1 mod 24$, $p^2+1$ is even and $p^4 \cong 1 \mod 5$, the answer is easy to figure..
To finish it:
If $p=5 mod 8$ (13 works) then $p+1, p^2+1$ are $2 \mod 4$ and $P-1 = 4 mod 8$, thus $a \leq 4$.
If $p= 4 mod 9$ (again 13 works) $p^2-1= 6 \mod 9$ and $p^2+1 =8 \mod 9$ thus $b \leq 1$.
If $p= 6 \mod 25$, then $p^2-1= 10 \mod 25$ and  $p^2-1= 10 \mod 25$ so $c \leq 1$. I think that actually $p=7$ works nicer in this case....
