$a^2 \equiv b^2$ mod 4 equivalence classes.

Question

so we have the relation $a^2 \equiv b^2$ mod 4. And to find equivalence classes we say b or a = 0 so $a^2=4k$ so $a=+-2\sqrt{k} $ so all even numbers. But when we get to a=1 then $a^2=4k+1$ after plugging stuff in i see it is all the odd numbers, but how to prove it? Thanks in advance. Keep in mind i do not want to check if it is true for odd/even i want to prove it form zero.

Answer 1

whatever the parity of $a$ and $b$ we have $a-b=(a+b)-2b$ this shows that $$ 2|(a+b) \Leftarrow\Rightarrow 2|(a-b) \tag{1} $$ keeping this in mind throughout, we notice that if $$ a^2 \equiv b^2 \mod 4 $$ then $$ 4|(a^2-b^2)=(a+b)(a-b) $$ from which $$ 2|(a-b) $$ or $$ a \equiv b \mod 2 $$

given (1), these steps are reversible