# Question about primes and divisibility abstract algebra/number theory

• AxiomOfChoice

#### AxiomOfChoice

Can someone please tell me how to go about answering a question like this? I've been racking my brain for a long time and still don't have a clue...I guess because my background in algebra/number theory really isn't that strong.

"What is the greatest integer that divides $p^4 - 1$ for every prime number $p$ greater than 5?"

Thanks!

There seems an obvious first thing to try:
Compute the greatest integer that divides $p^4 - 1$ for every prime number p in the range 5 < p < N​

where N is whatever number you like. I'd probably start with 10 and then increase it a few times until I had an idea what was going on.

What is the greatest integer that divides $p^4 - 1$ for every prime number $p$ greater than 5?"

Wouldn't it be $$p^4-1$$? Maybe I'm not understanding the question.

Sorry; this is a multiple choice question off of an old Math Subject GRE exam. There are five answer choices:

(A) 12
(B) 30
(C) 48
(D) 120
(E) 240

This is what I have so far.

$$p^4-1= (p+1)(p-1)(p^2+1)$$

p is odd so $$p = 1 \text{ or } 3$$ (mod 4) so there are three 2's in (p+1) and (p-1) plus another in $$(p^2+1)$$ so $$16|p^4-1$$. Furthermore, 3 does not divide p (since p>5) so (p-1) or (p+1) does and so $$3|p^4-1$$. Now it's between 240 and 48.

Wouldn't it be $$p^4-1$$? Maybe I'm not understanding the question.

Yes, you are. The question is about a single number that divides $p^4- 1$ for all primes p> 5. It cannot depend on p.

Alright I found the the last factor.

1^2 = 1 mod 5
2^2 = 4
3^2 = 4
4^2 = 1

So $$p^2$$ = 1 or 4 mod 5

$$(p^2)^2 = 1$$ mod 5

$$p^4-1 = 0$$ mod 5