If $a\equiv b \text{ mod}(c^2) \Longrightarrow a \equiv b \text{ mod}(c)$ proof

Question

It works for the examples I tried but I'm not sure how to prove. I tried assuming that $a-b\equiv k \text{ mod}(c)$, where $k \neq 0$ but am not sure how to then show that $a-b$ can't be congruent to $0$ $\text{ mod}(p^2)$. Or maybe the contrapositive isn't the direction to go in. I was thinking of using the divisibility property of modular arithmetic but that would just get me to $\frac{a}{c}\equiv \frac{b}{c} \text{ mod}(c)$.

Thanks

Answer 1

Don't overthink it, work straight from the definition: $a \equiv b \pmod{c^2} \iff c^2 \mid a - b$. But $c \mid c^2$ so $c \mid a - b$, and thus $a \equiv b \pmod{c}$.

Answer 2

$$a\equiv b \text{ mod}(c^2) \Longrightarrow c^2|(b-a)\implies c|(b-a) \implies a \equiv b \text{ mod}(c)$$

Answer 3

$c|c^2$ and more genereally $a \equiv b \mod mn \implies a\equiv b \mod m$.

This really should be obvious as $a\equiv b \mod mn \implies a = b + kmn$ for some integer $k$ so $a = b+m(kn)$ so $a \equiv b \mod m$.

Or in other words $a \equiv b \mod c \implies a = b + kc^2$ for some $c$ so $a = b + (kc)c$ so $a \equiv b \mod c$.