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.
 A: $\rm\:p\nmid p^4-1\:$ so $\rm\:2,3,5\:$ are the only possible prime divisors. $240$ is the maximal product of such since $11^4-1 = 122\cdot 120 = 2^4\cdot3\cdot5\cdot 61\:.$ Now, as I mentioned on sci.math on Sep 1, 2003, for every n coprime to $\:2,3,5\:,\:$ hence certainly for every prime > $5$, we have that $\rm\:n^4 \equiv 1\ (mod\ 240)\:$ since $240$ has Carmichael lambda  function $\rm\:\lambda(240) = \lambda(2^4\cdot3\cdot 5) = lcm(2^{4-2},\phi(3),\phi(5)) = 4\:.$
See also my my post on Carmichael's generalizations of little Fermat and Euler's phi function, and many applications of such in prior posts. It's well-worth learning these generalizations since they are ubiquitous in number theory.
A: I suppose this is more of a brute force method. By FlT, you know $p^4\equiv 1\pmod{5}$, so $5$ is a factor. Also, factoring, you get 
$$p^4-1=(p-1)(p+1)(p^2+1),$$
and since $p$ is odd, each factor is divisible by $2$. Moreover, one of the factors $p-1$ or $p+1$ is divisible by $4$, so $4\cdot2\cdot 2=16$ is another factor. Finally, of the three consecutive integers, $p-1$, $p$, and $p+1$, $3$ is a factor of one. Since $p$ is prime and larger than $5$, $3\nmid p$, so $3|p^4-1$. Hence $16\cdot 3\cdot 5=240$ is a factor of $p^4-1$ for any prime $p\gt 5$. 
It still remains to see that $240$ is the greatest such divisor. This can be confirmed by looking at $7^4-1=2400$ and $11^4-1=14640$.
A: 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....
A: Kind of heuristic but after factoring $p^4-1$, consider the prime factorisation of $240$:
$$240 = 2^4 \cdot 3 \cdot 5$$
The greatest prime here is $5$. Now, observe that our assumption is $p > 5$.
