Prove that for any prime $p$ there exist natural numbers $a,b$ for which $ p$ divides $a^2+b^2+1$

Question

Prove that for each prime $p$ there exist natural numbers $a,b$ for which $p$ divides $a^2+b^2+1$

Answer 1

Suppose $p>2$, put $A=\{0,1,\ldots,\frac{p-1}{2}\}$ and consider the functions $$\begin{align*}f&:A\longrightarrow \mathbb{Z}_p & & g:A\longrightarrow \mathbb{Z}_p\\ &\ \ \ \ a\longmapsto a^2 \mod p & & \ \ \ \ \ \ b\longmapsto -b^2-1 \mod p\end{align*}$$

You can easily prove that both functions $f,g$ are injective. Therefore, $|f(A)|=|g(A)|=|A|=\frac{p+1}{2}$, so $f(A),g(A)$ cannot be disjoint because $f(A)\cup g(A)\subseteq \mathbb{Z}_p$ which has $p$ elements. Thus, there are $a,b\in A$ such that $a^2\equiv -b^2-1 \mod p$, that is, $a^2+b^2+1\equiv 0 \mod p$ and we conclude that $p$ divides $a^2+b^2+1$