Prove that $ax^2 + by^2 \equiv c \pmod{p}$ has integer solutions 
Let $p$ be a prime number and $a, b, c$ integers such that $a$ and $b$ are not divisible by $p$. Prove that $ax^2 + by^2 \equiv c \pmod{p}$ has integer solutions

Well, this problem can be solved by Pigeonhole-principle.
Let $x, y = 0, 1, ..., p-1$. There are $2p$ such numbers. Any residue, except for $0$, can have at most two elements of the form $ax^2$ or of the form $c-by^2$. $(x_{1}-x_{2})(x_{1}+x_{2})=0$ because $ax_{1}^2 \equiv ax_{2}^2$ implies that $a_{1}^2 = a_{2}^2$, this can happen only if $x_{1} = \pm x_{2} $
We distinguish two cases. If $c-by_{0}^2 \equiv 0$ for some $y_{0}$, then $(0,y_{0})$ is a solution. Otherwise, the $2p-1$ numbers $ax^2, c-by^2, x, y=1, 2,..., p-1$ are distributed into $p-1$ 'holes, namely the residue classes $1, 2,..., p-1$. There of them must lie in the same residue class, so the solution is the pair $(x_{0},y_{0})$ such that $ax_{0}^2 \equiv c - by_{0}^2\pmod{p}$
Now, I'm looking for other solutions to prove it.
 A: First off, the answer is clear when $p=2$: $(0,0)$ is a solution when $c\equiv 0 \pmod  2$ and $(1,0)$ is a solution when $c\equiv 1   \pmod  2$.
Now assume $p\neq 2$. Consider the group morphism
$$f:\left(\frac{\mathbb{Z}}{p\mathbb{Z}}\right)^{\times}\rightarrow \left(\frac{\mathbb{Z}}{p\mathbb{Z}}\right)^{\times}: x\mapsto x^{2}.$$
Its kernel is $\{x\in \left(\frac{\mathbb{Z}}{p\mathbb{Z}}\right)^{\times}: x^{2}=1\}$, which
has order $2$ since $1\not\equiv -1 \   mod \ p$ (as $p\neq 2$).
It follows that the order of the image of $f$ is
$$\frac{\left|\left(\frac{\mathbb{Z}}{p\mathbb{Z}}\right)^{\times}\right|}{2}=\frac{p-1}{2}.$$
Along with $0$, we get that the number of squares modulo $p$ is $$\left|\left\{x^{2} : x\in\frac{\mathbb{Z}}{p\mathbb{Z}}\right\}\right|=\frac{p+1}{2}.$$
It follows that
$$\left|\left\{a^{-1}(c-by^{2})\pmod p: y\in\mathbb{Z}\right\}\right|=\frac{p+1}{2}.$$
As there are only $p-\frac{p+1}{2}=\frac{p-1}{2}$ not-squares modulo $p$, the set $\{a^{-1}(c-by^{2})\pmod p: y\in\mathbb{Z}\}$ must contain a square modulo $p$. That is, there exist $x,y\in\mathbb{Z}$ such that
$$x^{2}\equiv a^{-1}(c-by^{2})\pmod p,$$
or
$$ax^{2}+by^{2}\equiv c\pmod p.$$
