Prove that p^3 - 1 is a composite number given P > 2

Question

In order to prove a number composite I have to prove that p^3 - 1 = ab With a and b not being 1 and itself. Its given that p > 2.

I factor it out with differences of squares p^3 - 1 => (p - 1)(p^2 + p + 1)

And I don't really know what to do next. How do I involve p > 2 into the proof.

Isn't it utterly trivial that if `p>2` then `p-1` is neither 1 nor `p^3 - 1`? — John Coleman, Jan 06 '16 at 22:23
I'm voting to close this question as off-topic because it is not about programming. Would fit on [math.se]. — Reinstate Monica -- notmaynard, Jan 06 '16 at 22:32
You already showed that for most p it is composite by expressing it as the product of two terms. The only problem case as John Coleman points out is that if p = 2 then it degenerates to 1 times p^2 + p + 1 — HexedAgain, Jan 06 '16 at 22:33
okay please close this question sorry I didn't realize I was in coding — Michael Wu, Jan 06 '16 at 22:37

Maertin · Answer 1 · 2016-01-06T22:37:29.270

2

Its a composite number only if its a product of two numbers a and b both of which are greater than 1. If p = 2, then a would be 1.

If (p > 2) - then 
(p - 1) > 1 and 
(p^2 + p + 1) > 1.

edited Jan 06 '16 at 22:37

answered Jan 06 '16 at 22:26

Maertin

p has to be greater than 2 – Michael Wu Jan 06 '16 at 22:28
1

Yes, that's the point. p has to be greater than 2 for p-1 to be greater than 1. p^2 + p + 1 is always greater than 1 for p>2. – Maertin Jan 06 '16 at 22:31
If p > 2 - then (p - 1) > 1 and (p^2 + p + 1) > 1. – Maertin Jan 06 '16 at 22:35
okay I understand thank you very much – Michael Wu Jan 06 '16 at 22:36

score -1 · Answer 2 · answered Jan 06 '16 at 22:21

-1

Just use induction.

Base case p=3, p**3-1 = 8.

Inductive case: use your factorization.

answered Jan 06 '16 at 22:21

Darth Egregious

@iamnotmaynard it's close enough. You get the idea. – Darth Egregious Jan 07 '16 at 15:39

2 Answers2