TITLE: The largest integer that divides $p^4-1$, in which p is a prime greater than $5$
QUESTION [5 upvotes]: The question is to find the largest integer that divides all $p^4-1$, where p is a prime greater than 5. Being asked this question, I just assume this number exists. Set $p = 7$, then $p^4-1=2400$. I don't have any background in number theory and not sure what to do next. Thank you for your help!

REPLY [2 votes]: Let $n$ be the largest integer that divides $p^4-1$ for all prime $p\geq 7.$ 
We have $11^4-1=14640$ and $7^4-1=2400.$ The $gcd$ of $14640$ and $2400$ is $240.$ So  $$n\leq 240.$$ If $p$ is odd then modulo $16$ we have $p^4\in \{(\pm 1)^4, (\pm 3)^4,(\pm 5)^4,(\pm 7)^4\}=\{1^2,9^2, 25^2, 49^2\}=\{1^2,9^2,9^2,1^2\}=$
 $=\{1,81,81,1\}=\{1\}.$  
If $p$ is not divisible by $3$ then modulo $3$ we have $p^4\in \{(\pm 1)^4\}=\{1\}.$
If $p$ is not divisible by $5$ then modulo $5$ we have $p^4\in \{(\pm 1)^4,(\pm 2)^4\}=\{1,16\}=\{1\}.$
So for any integer $p$ that is not divisible by $2,3,$ or $5$  we have $p^4\equiv 1 \pmod {16}$ and $p^4  \equiv 1 \pmod 3$ and $p^4 \equiv 1 \pmod 5;$ and  since $16,3,$ and 5 are pair-wise co-prime, therefore  $p^4\equiv 1 \pmod {16\cdot 3\cdot 5}=240,$ so $$n\geq 240.$$