Question about primes and divisibility abstract algebra/number theory

[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!

 

[Hurkyl]

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.

 

[qntty]

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.

 

[AxiomOfChoice]

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

 

[qntty]

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.

 

[HallsofIvy]

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.

 

[qntty]

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