$\{ax^2 + by^2 | x,y \in \mathbb{Z}_p\} = \mathbb{Z}$

Question

I want to show that for an odd prime $p$ and integers $m,n, \gcd(mn,p) =1$ that: $$\{mx^2 + ny^2 | x,y \in \mathbb{Z}_p\} = \mathbb{Z}_p$$. I'm thinking about using the fact that there are $(p-1)/2$ quadratic residues and nonresidues in the multiplicative group mod $p$, but I'm not sure how to formalize or proceed.

Answer 1

You presumably want $\cdots=\Bbb Z_p$ where the $\Bbb Z_p$ denotes the integers modulo $p$.

This is a textbook proof. Let $a\in \Bbb Z_p$. One needs to solve $$mx^2=a-ny^2$$ in $\Bbb Z_p$. Now $mx^2$ takes $\frac12(p+1)$ values as $x$ varies, and so does $a-ny^2$ as $y$ varies...