Best proof method with modulo If $a$ is an integer, then $a ≡ 3 \pmod 7$ if and only if $(a^2 + 5a) ≡ 3 \pmod 7$. How could I prove this, or what type of proof method should I use to prove this. I know I can't use direct and I can't use contradiction. 
 A: $\Rightarrow)$ If $a\equiv 3 \pmod 7$ then in the expression $a^2+5a(\text{mod}7)$ you can simply replace $a$ with $3$.

 Indeed, $3^2+5\cdot 3\equiv 9+15\equiv 24\equiv 3\pmod 7$

$\Leftarrow)$ If no convenient choices seem available, you can always prove by brute force.  Here we might try approaching via contrapositive.
The contrapositive implication is "If $a\not\equiv 3\pmod 7$ then $a^2+5a\not\equiv 3\pmod 7$"

 There are six cases to check, and all relatively quick.  E.g. for $a\equiv 0\pmod7$ you have $a^2+5a\equiv 0^2+5\cdot 0\equiv 0\not\equiv 3\pmod7$

A: The number $7$ is so small that "try everything" is probably the best strategy. 
For this kind of problem (and larger moduli) I like completing the square. Note that $5\equiv -2\pmod{7}$. So
$$a^2+5a\equiv a^2-2a\equiv (a-1)^2-1\pmod{7}.$$
It follows that $a^2+5a\equiv 3\pmod{7}$ if and only if $(a-1)^2\equiv 4\pmod{7}$. 
But this congruence holds if and only if $a-1\equiv \pm 2\pmod{7}$. That gives us the solutions $a\equiv 3\pmod{7}$ and $a\equiv -1\pmod{7}$.
A similar strategy is useful for the congruence $a^2+5a\equiv c\pmod{p}$, where $p$ is an odd prime.
