Answer to Question #337131 in Discrete Mathematics for bkay

Question

Show that if an integer n is not divisible by 3, then n2 – 1 must be a multiple of 3

Answer 1

If $n$ is not divisible by 3, then either $n=3k+1, k\in \Z$ or $n=3m+2,$ $m\in \Z.$

Then

$$n^2-1=(3k+1)^2-1=9k^2+6k$$

$$=3(3k^2+2k), k\in \Z$$

Let $p=3k^2+2k.$ We know that $p$ is an integer because $k$ is an integer. Therefore $n^2-1$ equals 3 times some integer. So $n^2-1$ is a multiple of 3.

Or

$$n^2-1=(3m+2)^2-1=9m^2+12m+3$$

$$=3(3m^2+4m+1), m\in \Z$$

Let $q=3m^2+4m+1.$ We know that $q$ is an integer because $m$ is an integer. Therefore $n^2-1$ equals 3 times some integer. So $n^2-1$ is a multiple of 3.

Therefore if an integer $n$ is not divisible by 3, then $n^2 – 1$ must be a multiple of 3.