If $a$ is an odd positive integer $>$ $2$ such that 3 $\not$ | $a$, then $12$ $|$ $(a^2 -$1 ))

Question

So I have the following proposal:

$a$ $\in$ $\mathbb{Z}_{+}$ : $a$ = $2k +1$ $>$ $2$ : 3 $\not$ | $a$ and $>2$ then $12$ $|$ $(a^2 -$1 ).

My initial idea was to show:

$a^2 -1\equiv 0 \mod 12$

Logically it follows:

$(2k+1)^2 -1\equiv 0 \mod 12$

$4k^2 + 4k\equiv 0 \mod 12$

However I do not know where to go from here. This proof also doesn't incorporate the condition 3 $\not |$ a . Any tips? Sorry for formatting, this is my first time using MathJax.

And also this and this. If a 50k+ user does not recognize a duplicate on the spot,... — Jyrki Lahtonen, May 25 '18 at 12:17

score 0 · Answer 1 · answered May 25 '18 at 12:14

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Note that $$a^2-1=(a+1)(a-1).$$ One of the factors is a multiple of $3$, both factors are even, in fact one of them must be a multiple of $4$. Hence $$ 24\mid a^2-1.$$

answered May 25 '18 at 12:14

Hagen von Eitzen

374,180

score 0 · Answer 2 · answered May 25 '18 at 12:17

"I do not know where to go from here. This proof also doesn't incorporate the condition 3 $\not |$ a" You're answering your own question there. $3\nmid a$ means exactly that $k$ is neither $0,3,6$ nor $9$ modulo $12$. Now you can just check the remaining cases.

It's a bit faster if you use the Chinese remainder theorem to split modulo 12 up into a simultaneous system using modulo 3 and modulo 4 separately, but it's not necessary.

If $a$ is an odd positive integer $>$ $2$ such that 3 $\not$ | $a$, then $12$ $|$ $(a^2 -$1 ))

2 Answers2