Prove that if $x$ is odd then $x^2 -1$ is divisible by $8$.

Solution 1:

An easier way to approach this would be to observe: If x is odd, either x-1 or x+1 is divisible by 4. Then their product is divisible by 8.

Solution 2:

Are you familiar with the mod operator?

We want to show

$x^2 - 1 \equiv 0\mod8$

$x^2 \equiv 1\mod8$ for $x = 2k+1$ for some $k$

$(2k+1)^2 \equiv 1 \mod 8$

$4k^2 + 4k + 1 \equiv 1 \mod 8$

$4k^2 + 4k \equiv 0 \mod 8$

$4 (k^2 + 4k) \equiv 0 \mod 8$

This is what we want to show.

If $k$ is odd then $k^2$ is odd. And $k^2 + k$ is even. If $k$ is even then $k^2 + k$ is even

Hence $k^2 + k$ is always of the form $2a$, And hence

$4 (k^2 + 4k) \equiv 4(2a) \equiv 8a \equiv 0 \mod 8$