How can I show this?
For every prime number, $p$, there is at least one solution $(x, y)$ such that $x^2 + y^2 \equiv -1 \pmod p$.
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1Hint: for $p$ odd, there are $(p+1)/2$ squares. – Andrew Dudzik Dec 08 '15 at 18:18
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The same proof works for any $x^2+y^2\equiv k\pmod{p}$, $k\in\mathbb Z$. – user236182 Dec 08 '15 at 18:41
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In fact, it works for any $ax^2+by^2\equiv k\pmod{p}$, $a,b,k\in\mathbb Z, \gcd(ab,p)=1$. – user236182 Dec 08 '15 at 19:18
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The result is clear for $p=2$, so let $p$ be odd.
For $i=0$ to $\frac{p-1}{2}$, the numbers $i^2$ are distinct modulo $p$. Let $S$ be this set of numbers. Then $S$ has $\frac{p+1}{2}$ elements.
Similarly, let $T$ be the set of numbers of the shape $-j^2-1$, where $j$ ranges from $0$ to $\frac{p-1}{2}$. Then $T$ contains $\frac{p+1}{2}$ numbers that are distinct modulo $p$.
The sum of the cardinalities of $S$ and $T$ is $p+1$. Thus by the Pigeonhole Principle there exist $i$, $j$ such that $i^2\equiv -j^2-1\pmod{p}$. This completes the proof.
André Nicolas
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