If $a ≡ b \pmod 7$, is $a^2 \equiv b^2 (\bmod 7)$? 
Let $a$ and $b$ be integers with $a ≡ b \pmod 7$. Is $a^2 ≡ b^2 \pmod 7$? Justify by giving a proof or a counterexample.

I actually have no clue how to even begin tackling this question. How would I go about it?
 A: Definition
$$\rm x\equiv y\pmod m \iff m|x-y.$$
HINT:
$$\rm a-b=7k \rightarrow (a-b)(a+b)=a^2-b^2=(7k)q.$$
For some $\rm k,q\in \Bbb Z$.
A: $$a\equiv b\pmod 7$$ means, by definition, that $a-b$ is a multiple of 7.
$$a^2\equiv b^2\pmod 7$$ means that $a^2-b^2$ is a multiple of 7.  
Is there some way you could show that if $a-b$ is a multiple of 7 then $a^2-b^2$ must also be a multiple of 7?
A: Actually, it's quite simple.
$$a^2\equiv b^2 \pmod7$$
$$a^2- b^2\equiv0 \pmod7$$
$$(a-b)(a+b)\equiv 0\pmod 7$$
Since $a-b\equiv0\pmod7$, the equivalence holds.
Or, basically, being $(a-b)$ a multiple of $7$, also $(a-b)(a+b)=a^2-b^2$ is a multiple of $7$.
Conclude by definition of congruence modulo $7$.
A: In general if $a = b \mod(n)$ then $a*c = b*c \mod(n)$ (but not necessarily vice versa).
Pf: $a = b \mod(n) \implies n|a - b \implies n|c(a - b)=ca - cb \implies ac = bc \mod (n)$.
Also $a = b \mod(n); b = c \mod(n) \implies a = c \mod(n)$.
Pf: $n|a - b; n|b - c \implies n|(a - b) + (b - c) = a - c.$
So  $a = b \mod(n) \implies ab = b^2 \mod(n)$.  $a = b \mod (n) \implies a^2 = ab \mod(n).$  So $a^2 = b^2 \mod(n)$.
