prove that $x^2+7y^2=4p$ has an integer solution if and only if $u^2+7v^2=p$ has an integer solution

Question

I need to prove this for $p$ prime. Clearly $x^2+7y^2=4p$ implies $x$ and $y$ are odd, but I'm not sure where to go from there. I'm far too tired to think I've been working for 16 hours straight (not just on this) please help

Answer 1

If you're too tired to think clearly, that's a sign you should sleep and come back to it later.

Clearly $x^2+7y^2=4p$ implies $x$ and $y$ are odd...

Well, no, it doesn't.

Hint: Show that if $x$ and $y$ are both odd, $8$ divides $x^2+7y^2$.

Also, the statement should specify $p$ is an odd prime; $p=2$ is a counterexample otherwise.