Prove if $2\mid(x^2-1) $, then $4\mid(x^2-1)$

Question

I have no idea where to start. Any hint(s) or suggestions? Prove if $2\mid(x^2-1) $, then $4\mid(x^2-1)$

Answer 1

Suppose $2\mid (x^2-1)$. Then $2\mid (x-1)(x+1)$.

By Euclid's lemma, either $2\mid (x+1)$ or $2\mid (x-1)$.

But $x+1$ is even if and only if $x-1$ is even. Thus, both must be even.

The product of two even numbers must be divisible by four, so $$4\mid (x^2-1)$$

Answer 2

One way is that the assumption implies that $x$ must be odd, so $x=2n+1$ and thus $$x^2=4n^2+4n+1$$

Answer 3

$x-1$ and $x+1$ are separated by $2$ so they are either both odd or both even. The first case means that the product will not be divisible by two, so the second case must hold true.

Hence $2|(x-1)(x+1) \implies 2|(x+1)$ and $2|(x-1)$ so $4|(x-1)(x+1)$.

Answer 4

$x^2\equiv1\pmod2\iff x\equiv1\pmod2,x=2a+1$ where $a$ is any integer

$$\implies x^2=(2a+1)^2=8\frac{a(a+1)}2+1\equiv1\pmod8$$