Is this proof on absolute value ok?

Question

Prove that for any real numbers a and b, |a| = |b| if and only if $a^2=b^2$

let $a,b>0$

if |a| = |b|

This implies $a=b$

implies $a^2=b^2$

(i then do a similiar thing for $a,b<0$ and when $a<0 , b>0$)

let $a,b$ be real numbers

if $a^2=b^2$

implies $a^2-b^2=0$

implies $(a+b)(a-b)=0$

implies $a=-b or a=b$

So in both cases |a| = |b|

Answer 1

You could prove this a bit quicker by saying if $|a|=|b|$, then $a=\pm b$, so squaring both sides $a^2=b^2$, but what you have done works and is pretty rigorous