Equation $ux^2+vy^2=w$ has at least one solution in $\Bbb{Z}/p\Bbb{Z}.$

Question

Let $p$ a prime number $>2.$

Using a morphism between $\Bbb{Z}/p\Bbb{Z} ^*\to\Bbb{Z}/p\Bbb{Z}^*$ and the fact that it's an abelian group of order $p-1$ and a field, we have $\frac{p+1}{2}$ squares in $\Bbb{Z}/p\Bbb{Z}.$

Now the next question it to prove that given $u\ne 0,v\ne 0,w\in\Bbb{Z}/p\Bbb{Z}$ the equation $$ux^2+vy^2=w$$ has at least one solution.

For this one, I don't have any idea, I don't really understand how to connect this to the first question. Any ideas ?

Answer 1

Let $S=\{ux^2 \mid x\in\mathbb{Z}/p\mathbb{Z}\}$ and $T=\{vy^2 \mid y\in\mathbb{Z}/p\mathbb{Z}\}$.

There are $|S| = \frac{p+1}{2}$ elements in the set $w-S = \{c \mid \exists s\in S : s+c=w\}$.

Since $|w-S| + |T| = p+1 > |\mathbb{Z}/p\mathbb{Z}|$, we have $(w-S)\cap T \neq \emptyset$.