Prove that $a^{b} \equiv 3 \,( \text{mod}\, 4)$ implies $a,b$ odd. So far I've shown that 
$a^{b} \equiv 3 \,( \text{mod} \, 4) \implies a^{b} \,\text{odd}\implies a \, \text{odd}$. I also know that since $a^{b} \equiv 3 \, (\text{mod} \, 4)$, there exists prime $p \mid a^{b}$ such that $p \equiv 3 \, (\text{mod} \, 4)$.
Any help would be appreciated, thanks!

As you've shown, $a$ must be odd as any even value can't be congruent to $3$ modulo $4$. Thus, either $a \equiv 1 \pmod 4$, which doesn't work as $a^b \equiv 1 \pmod 4$ for any $b$, or $a \equiv 3 \pmod 4$. For the second case, note $a^2 \equiv 3^2 \equiv 1 \pmod 4$. Thus, for any even $b = 2c$ for some integer $c$, you have $a^b \equiv (3^2)^c \equiv 1^c \equiv 1 \pmod 4$.
This means $b$ must be odd. Note this gives that $b = 2d + 1$ for some $d$, so $a^b \equiv (3^2)^d(3) \equiv 3 \pmod 4$.
