Prove or disprove the following statement in modular arithmetic.
- If $a\equiv b \mod m$, then $ a^2\equiv b^2 \mod m$
- If $a\equiv b \mod m$, then $a^2\equiv b^2 \mod m^2$
- If $a^2\equiv b^2\mod m^2$, then $a\equiv b\mod m$
My proofs.
$ a\equiv b \mod m \implies (a-b) = mr, r\in\mathbb{Z}$ $ a^2-b^2 = (a+b)(a-b) = (a+b)mr = ms \text{ where } s = (a+b)\cdot r$ So the first statement is true
$ a\equiv b \mod m \implies (a-b) = mr, r\in\mathbb{Z}$ $a^2-b^2 = (a+b)(a-b) = (a+b)mr$ but $(a+b)\neq ms$ $\forall s\in\mathbb{Z}$ in general. So the second one is false.
$a^2-b^2 = m^2r, \exists r\in\mathbb{N}$ $a^2-b^2= (a+b)(a-b) $ Then I kind of got stuck here. I'm not sure how to continue it. Am I missing some properties I don't know? Or there is a algebra trick that could be applied here?