# Greatest integer divides p^4 -1

1. Dec 20, 2003

### yxgao

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

It is 240. Why?

Thanks!!

2. Dec 20, 2003

### Hurkyl

Staff Emeritus
Try factoring $p^4-1$.

3. Dec 20, 2003

### yxgao

I already figured this out. (p^4-1) = (p^2+1)(p-1)(p+1).
i.) Each term is divisible by 2 since p is odd. Also, either p-1 or p+1 is divisible by 4.
So divisible by 16.
ii.) either p mod 3 = 1 or p mod 3 = 2. If the first case, p-1 = 0 mod 3, second case, p^2 -1 = 0 mod 3.
So divisible by 3.
iii.) either p mod 5 = 1, p mod 5 = 2, p mod 5 = 3, or p mod 5 = 4.
If p mod 5 = 1, p-1 = 0 mod 5. If p mod 4 = 1, p+1 = 0 mod 5, and if p mod 5 = 2 or p mod 3 = 1, then p^2+1 = 0 mod 5.
So divisible by 5.

Hence, divisible by 16*3*5 = 240.

4. Dec 20, 2003

### Hurkyl

Staff Emeritus
Er, my mistake, I misread the problem.

Well, you've verified that $240 | p^4 - 1$ for any prime number greater than 5, correct? (In fact, $240 | n^4 - 1$ if $(240, n) = 1$)

The easiest way to proceed from here is, I think, to start looking at some explicit examples, and finish the proof from a small number of those. For instance, if $m | p^4 - 1$ for all primes $p > 5$, then $m | 7^4 - 1 = 2400$.