Suppose $p$ a prime and $a$ and $b$ are not divisible by $p$. I’d like to show that $a^2 \equiv kb^2 \pmod p$ implies that there exists a solution to $x^2 \equiv k \pmod p$ but I don’t know what modular arithmetic properties to use. I tried writing the following:
$$a^2=pq_1 + r$$ for $0\leq r < p$ and $$kb^2 = pq_2 + r$$ imply that $$a^2-kb^2 = p(q_1-q_2)$$
But I’m not sure where to go from here.
Assuming $b$ is invertible mod $p$ (i.e. $\gcd(b,p) = 1$), you can rewrite the first equivalence as $a^2 (b^{-1})^2 \equiv k \bmod p$ and then $x = a b^{-1}$ is a solution to your second equivalence. As already mentioned in a comment, if $b$ is not invertible mod $p$ (i.e. if $p$ divides $b$), then what you are trying to prove is not always true.