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I have to prove that if $p$ is a prime number of the form $p = 4n - 1$, $n\in N$ and $x^2+y^2\equiv 0\pmod{p}$, then $x\equiv 0\pmod{p}$ and $y\equiv 0\pmod{p}$.

I have gone about this as follows and I hope you will correct me if I made any errors:

First of $x^2 + y^2$ is divisible by p so we can write it as being equal to some multiple $k$ of $p$ $+$ a remainder of $0$.

Then let's assume the opposite that neither $x$ or $y$ are congruent to $p$ ergo they aren't divisible by $p$ so we can represent them as being equal to some multiples $l$ and $m$ of $p$ (respectively) and with remainders (nonzero of course) $a$ and $b$ (respectively).

So we get:

$$ x^2 + y^2 = kp + 0$$ $$ x = lp + a $$ $$ y = mp + b$$

If we take the second and third expression and square them we get: $$ x^2 = l^2p^2 + 2lpa + a^2 $$ $$ y^2 = m^2p^2 + 2mpb + b^2$$

Then we add them up:

$$ x^2 + y^2 = l^2p^2 + 2lpa + a^2 + m^2p^2 + 2mpb + b^2$$

Factorize:

$$ x^2 + y^2 = \left(l^2p + 2la + m^2p + 2mb\right)p + a^2 + b^2$$ We see that $k = l^2p + 2la + m^2p + 2mb$, which gives us:

$$ x^2 + y^2 = kp + a^2 + b^2$$

At the beginning it is stated that the remainder $a$ is equal to zero since $x^2 + y^2$ is divisible by $p$ so get to the conclusion that the sum of two nonzero, positive numbers, more precisely the sum of two squares $a^2 + b^2$ has to be $0$. This is a contradiction.

The contradiction was brought on by the assumption that $x$ and $y$ aren't divisible by $p$ therefore we conclude that $x$ and $y$ must be divisible by $p$.

Martin Sleziak
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n4869
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  • Actually this is a direct proof, not a proof by contradiction – which is preferable. – Bernard Sep 18 '15 at 00:37
  • I'm asking because this was a question on the last year's Elementary Number Theory exam at my Uni, I was primarily wondering if my arguments are correct and if I would get full points on this question ? – n4869 Sep 18 '15 at 00:41
  • what is the Legendre symbol $(-1|p)?$ – Will Jagy Sep 18 '15 at 00:44
  • 1 if $p$ is a prime number of the form: $p = 4k +1$ and -1 if $p$ is a prime number of the form $p = 4k+3$. Get it now... Thanks. Could've saved me lots of time if thought of it... – n4869 Sep 18 '15 at 00:46
  • At first sight it is correct, albeit a bit long. As for the points problem unfortunately I have no competence to tell you about it. – Bernard Sep 18 '15 at 00:52
  • You did not use the fact that $p = 4n - 1$, so the proof must be flawed. – corindo Sep 18 '15 at 01:27
  • Could you then point me to the flaw ? I know I didn't use it, I also found that strange, so I posted here. – n4869 Sep 18 '15 at 01:35
  • You want to prove that $a = b = 0$, but you cannot assume that at first. To use contradiction you have to suppose that $a$ and $b$ can be nonzero and lead to a contradiction. – corindo Sep 18 '15 at 01:38
  • You assume that $k = l^2p + 2la + m^2p + 2mb$, which is not rooted. – corindo Sep 18 '15 at 01:40
  • You say " At the beginning it is stated that the remainder $a$ is equal to zero..." which is false. You had $x=l p+a$ with non-zero $a$, "of course".And if you do not use the fact that $p+1$ is divisible by $4$ you cannot do it . – DanielWainfleet Sep 18 '15 at 01:40

1 Answers1

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I.e. you want to prove:

$p\equiv 3\pmod{4},\ p\mid x^2+y^2\implies p\mid x,y$.

If $p\mid x$, then $p\mid y$, and vice versa.

So for contradiction assume $\gcd(p,x)=\gcd(p,y)=1$. See Modular Inverse.

$x^2\equiv -y^2\pmod{p}\iff (xy^{-1})^2\equiv -1\pmod{p}$

This can give a contradiction in several ways:

$1)\ $ Just remember Quadratic Reciprocity.

$2)\ $ Raise both sides by $\frac{p-1}{2}$ (which is odd):

$\implies \left(xy^{-1}\right)^{p-1}\equiv (-1)^{\frac{p-1}{2}}\equiv -1\pmod{p}$,

which contradicts Fermat's Little Theorem.

$3)\ $ Square both sides: $\left(xy^{-1}\right)^4\equiv 1\pmod{p}$.

Then $\text{ord}_p\left(xy^{-1}\right)\mid 4$. But it cannot be $1$ or $2$, because $\left(xy^{-1}\right)^2\equiv -1\pmod{p}$, therefore $\text{ord}_p\left(xy^{-1}\right)=4$, so $4\mid p-1$ by Fermat's Little Theorem (see below).

Theorem: If $\text{ord}_m(a)=d$ and $a^k\equiv 1\pmod{m}$, then $d\mid k$.

Proof: For contradiction, assume $k=dl+r$ for some $l,r\in\Bbb Z^+,\, 0<r<d$.

$a^{k}\equiv \left(a^d\right)^la^r\equiv 1^la^r\equiv a^r\equiv 1\pmod{m}$, contradiction.

user236182
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