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Show that $$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$

Indeed,

First let's show

$7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $

we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid y^2$ then $ 7\mid x^2+y^2 $

  • Am i right and can we write $a\mid x \implies a\mid x^P ,\ \forall p\in \mathbb{N}^*$

Now let's show

$7\mid x^2+y^2 \Longrightarrow 7\mid x \text{ and } 7\mid y$

$7\mid x^2+y^2 \Longleftrightarrow x^2+y^2=0 \pmod 7 $

for

\begin{array}{|c|c|c|c|c|} \hline x& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline x^2& 0 & 1 & 4& 2 & 2 & 4 & 1 &\pmod 7\\ \hline y& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline y^2& 0 & 1 & 4& 2 & 2 & 4 & 1 & \pmod 7 \\ \hline \end{array}

which means we have one possibility that $x=y= 0 \pmod 7 $

  • Am I right and are there other ways?
Martin Sleziak
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Educ
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    This is exactly right, and it's probably the most direct method. To answer your one question, if $a|x$, then $x = ay$ for some $y \in \mathbb{Z}$. Then $x^n = a^ny^n = a(a^{n-1}y^n)$ where $a^{n-1}y^n \in \mathbb{Z}$. We conclude $a|x \Longrightarrow a|x^n$. Another way to see this is to simply work in the field $\mathbb{Z}_7$. If $x \equiv 0 \pmod{7}$, then $x^n \equiv 0^n \equiv 0 \pmod{7}$. – Kaj Hansen Jun 21 '14 at 09:35
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    I see an approach to the second implication involving the fact that every divisor of $x^2+y^2$ is a sum of two squares if $x$ and $y$ are relatively prime. Presumably it is considered too high-powered, but on the other hand it makes the generalization to primes other than $7$ obvious. – Karl Kroningfeld Jun 21 '14 at 10:56
  • See also: How to show that $7\mid a^2+b^2$ implies $7\mid a$ and $7\mid b$? – Martin Sleziak Jul 26 '17 at 07:07

3 Answers3

7

Let $x,y \in \mathbb{F}_p$ be with $x^2+y^2=0$. If $x=0$, then $y=0$. Now assume $x \neq 0$. Let $z:=y/x$, then $z^2=-1$. If $p=2$, this means $z = 1$. If $p > 2$, this means that $z$ has order $4$ in $\mathbb{F}_p^*$, which happens iff $4|p-1$ i.e. $p \equiv 1 \bmod 4$. Hence, for every odd prime $p$ with $p \not\equiv 1 \bmod 4$ the quadratic form $x^2+y^2=0$ has only the trivial solution.

(While your method for $p=7$ is fine, try it with $p=67$!)

Martin Brandenburg
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Your example is just a special case of this theorem,

Theorem

If $p\equiv 3\pmod 4$ Then whenever $p\mid x^2+y^2$, we have $p\mid x$ and $p\mid y$.

Proof

Assume that $p\mid x^2+y^2$ with $p\not\mid x,y$ then $$x^2+y^2\equiv 0\pmod p$$ $$x^2\equiv -y^2\pmod p$$ Which means $$x^{2\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}y^{2\frac{p-1}{2}} \pmod p$$ $$x^{p-1}\equiv - y^{\frac{p-1}{2}}\pmod p$$ That’s because $(p-1)/2$ is odd, but let’s see what we’ve got, using Fermat’s little theorem $$1\equiv -1\pmod p$$ Hence $p=2$ but that’s clearly a contradiction. $\quad \quad \quad \quad \quad \quad\square$

PNT
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$►$ If $x=7x_1$ and $y=7y_1$ then $x^2+y^2=7(x_1^2+y_1^2$).

$►$ If $x^2+y^2\equiv0\pmod7\iff x^2\equiv -y^2\pmod7$ then because of $\mathbb F_7^2=\{1,4,2,0\}$ and $(-1)\mathbb F_7^2=\{6,3,5,0\}$ the only possibility for $x^2\equiv -y^2\pmod7$ is that both $x^2$ and $y^2$ are equal to $0$ modulo $7$ so $x$ and $y$ are equal to $0$ modulo $7$ (since $7$ is prime). Then $x\equiv y\equiv0\pmod7$.

Piquito
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