Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Claim: $$12\mid(p^{2}-1) \space \forall\text{ primes }p>3$$

Attempt at proof:

$$p>3\space\Rightarrow\space p\text{ is odd} $$ $$p^2-1=(p-1)(p+1)\space\Rightarrow2^2\mid(p^{2}-1)$$

How do I go on to show that $3\mid(p^2-1)$ which would complete the proof?

share|cite|improve this question
As 3 doesn't divide $p$, it has to divide either $(p-1)$ or $(p+1)$ – Stefan Nov 6 '12 at 16:30
+1 for showing your work and showing effort! – amWhy Nov 6 '12 at 17:10

We know, $3\mid p(p-1)(p+1)\implies 3\mid (p-1)(p+1)$ as $p>3$

As $p$ is odd$=2a+1$(say), $p^2-1=(2a+1)^2=8\frac{a(a+1)}2+1-1\implies 8\mid (p^2-1)$

So, $lcm(3,8)\mid (p^2-1)\implies 24\mid (p^2-1)$

We know, any prime$>3,$ can be written as $6r\pm1$ where $r$ is a positive integer.

So, $p^2-1=(6r\pm 1)^2-1=36r^2\pm 12r=24r^2+24\frac{r(r\pm1)}2\implies 24\mid (p^2-1)$

Observe that, this will be true for any number of the form $6r\pm1$, not necessarily prime.

share|cite|improve this answer
the second part is very helpful, thanks! – Dexter Nov 6 '12 at 16:45
@Dexter, my pleasure. – lab bhattacharjee Nov 6 '12 at 17:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.