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Prove that if $x^2+y^2 \equiv 0\pmod{p}$ where $p$ is a prime and $x,y$ are not both divisible by $p$, then $p \equiv 1 \pmod{4}$.

I tried using that $x^2 \equiv -y^2 \pmod{p}$ and conjectured that $-1$ must a quadratic residue modulo $p$, but I am not sure how that would help.

Puzzled417
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4 Answers4

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Since $p \not \mid y$, there exists integer $y'$ such that $yy' \equiv 1 \pmod p$. Then multiply both side of the congruence with $y'$ to get that:

$$(xy')^2 \equiv -(yy')^2 \equiv - 1 \pmod p$$

So hence $-1$ is a quadratic residue modulo $p$. To prove that $p \equiv 1 \pmod 4$ from this note that:

$$1 \equiv x^{p-1} \equiv (x^2)^{\frac{(p-1)}{2}} \equiv (-1)^{\frac{(p-1)}{2}} \pmod p$$

This implies that $\frac{(p-1)}{2}$ is even and hence $p \equiv 1 \pmod 4$

Stefan4024
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    I also like this! +1 – Andres Mejia Jun 29 '16 at 15:15
  • I am confused how you got $x^2 \equiv -1 \pmod{p}$. If you mean for $x$ to be $xy'$ you would have to show that $xy'$ is coprime with $p$. – Puzzled417 Jun 29 '16 at 15:24
  • @Puzzled417 If you're reffering to the second equation $x$ is some number such that $x^2 \equiv -1 \pmod p$, and it's existence was proved from the first part. Anyway it's trivial that $(xy, p') = 1$. As $p \not \mid x, y'$ from the condition and the fact that $y'$ has a modular inverse – Stefan4024 Jun 29 '16 at 15:50
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No. The way the question is worded, $p=2$ also works.

Anyway, $y \neq 0 \pmod p,$ this means $y$ has a multiplicative inverse $\pmod p,$ for no better reason than $\gcd(y,p)=1$ and we have integers with $ys+pt=1.$

$$x^2 + y^2 \equiv 0 \pmod p,$$

$$x^2 \equiv -y^2 \equiv 0 \pmod p,$$

$$ \frac{x^2}{y^2} \equiv -1 \pmod p,$$

$$ \left( \frac{x}{y} \right)^2 \equiv -1 \pmod p.$$

Will Jagy
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by definition, there exists some $k$ so that $pk=x^2+y^2$. wlog, assume that $k$ is prime (otherwise look at its factorization etc.)

If $p$ does not divide $x$ or $y$, then $p^2$ does not divide $x^2+y^2$. But then, use this theorem (the top answer).

Since $p^1$ is an odd power, we deduce that $p \equiv 1 \mod 4$

Community
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Andres Mejia
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Not an Elementary Proof

If you know anything about finite fields, then you can show that the group of units of a finite field $\mathbb{K}$ is cyclic (otherwise, you can show that there exists a positive integer $m<|\mathbb{K}|$ such that $x^m=x$ for all $x\in\mathbb{K}$, meaning that $|K|\leq m$, a contradiction). If $C_n$ is the cyclic group of order $n\in\mathbb{N}$ and $d$ is a natural number such that $d\mid n$, then there exists an element $x\in C_n$ of order $d$.

Combining these two observations, you get that the group of units of the field $\mathbb{F}_p$ of $p$ elements is the cyclic group $C_{p-1}$. Hence, if a positive integer $d$ satisfies $d\mid p-1$, then there exists $x\in\mathbb{F}_p$ such that $x^d\equiv 1\pmod{p}$ but $x^j\not\equiv 1\pmod{p}$ for all positive integer $j<d$. In particular, if $p\equiv 1\pmod{4}$, then $d=4$ leads to $x^4\equiv 1\pmod{p}$ but $x^2\not\equiv1\pmod{p}$, for some $x\in\mathbb{F}_p$. Then, [...]. I leave the rest to you.

Batominovski
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