Prove equations in modular arithmetic 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?

HINT $\: $ for $\rm (3),\ \ m^2\ |\ a^2 - b^2\ \Rightarrow\ m\ |\ a-b\ $ fails if $\rm\: m > 1 = a - b\:.\:$ Then $\rm\:a^2-b^2 = 2\:b+1\:$ so any odd number with a square factor $\rm\:m^2 \ne 1\:$ yields a counterexample.
