Existence of solutions of quadratic equation in $\mathbb{Z}_p$ Let $p$ be a prime and $a,b,c$ integers such that $a$ and $b$ are not divisile by $p$. Prove that the equation $ax^2+by^2 \equiv c \pmod p$ has integer solutions.
Proof: 
Since $1^2\equiv (p-1)^2 \pmod p, \ 2^2\equiv (p-2)^2 \pmod p, \dots (\frac{p-1}{2})^2\equiv (\frac{p+1}{2})^2 \pmod p, $ $\Rightarrow$ we see that if $x\in \{0,1,2,\dots,p-2,p-1\}$ then $x^2 \pmod p$ takes $\frac{p-1}{2}+1$ values from different residue classes. Since $(a,p)=1$ then $ax^2 \pmod p$ takes also $\frac{p-1}{2}+1$ different values.
The same reasoning with $by^2 \pmod p$. Let's consider our modular equation in the following form: $ax^2\equiv c-by^2 \pmod p$.
Here we can apply pigeonhole principle. Let residue classes be the "boxes" and the number of boxes is equal to $p$. Since $ax^2 \pmod p$ takes $\frac{p+1}{2}$ different values and $c-by^2 \pmod p$ also takes $\frac{p+1}{2}$ different values. Since $\frac{p+1}{2}+\frac{p+1}{2}=p+1>p$ then some box contains one value from each. In other words, $ax_0^2\equiv j \pmod p, \ c-by_0^2\equiv j \pmod p$ for some $x_0, \ y_0,\  j$. 
Thus, $ax_0^2\equiv c-by_0^2 \pmod p$ $\Rightarrow$ $ax_0^2+by_0^2\equiv c \pmod p$.
Dear colleagues! What do you think about above reasoning? Is it correct?
It would be interesting to look at another solution with different approach.
 A: Here is "another solution with a different approach". I discard the trivial cases $p=2$ or $c\equiv 0$ mod $p$. Under the hypotheses, we can change notations and write the original congruence as an equation $x^2 - dy^2= e$ in $\mathbf F_p$, with $d, e \in \mathbf F_p^{*}$ . We consider 2 cases :  
1) If $d$ is not a square, the above equation is  equivalent to "$e$ is a norm from the quadratic extension $\mathbf F_p (\sqrt {d})$ ". But it is classically known that the norm map in a non trivial extension of finite fields  is surjective, so we are done.
2) If $d$ is a square, we can change notations again and consider the equation $x^2 -y^2=f$. But the quadratic form $X^2 -Y^2$ "represents $0$" in the sense that the equation  admits a non zero solution. It is then an easy exercise in linear algebra (see e.g. Borevitch-Chafarevitch's NT, Appendix, §1) to show that $X^2 -Y^2$ represents any element of the field, and we are done.
Geometrically, the above result means that a conic defined over a finite field always admits a rational point. The approach here is not quite elementary, but it allows to attack more general problems. Speaking of (non) elementary methods, the OP is a direct consequence of the Chevalley-Waring theorem which states that a homogeneous polynomial over a finite field $K$, of degree $d$, on $n$ variables, with $d>n$, always admits a non zero root. See e.g. P. Samuel's ANT, chap. 1, §7.A
