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Let $p$ be an odd prime.

Show that the equation $x^2 + y^2 = p$ has a rational solution (i.e., with $x,y \in \mathbb{Q}$) if and only if it has an integer solution (with $x,y \in \mathbb{Z}$), i.e., if and only if $p \equiv 1 \pmod 4$.

Not sure how to begin. Is the converse not obvious, since an integer is a rational number?

Tony
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  • Use modular arithmatic. $x^2\equiv 0,1\pmod 4$, $y^2\equiv 0,1\pmod 4$, so $p$ can never be $3\pmod 4$. –  Mar 05 '16 at 17:35
  • For a more general question, see http://math.stackexchange.com/questions/1445058/which-rationals-can-be-written-as-the-sum-of-two-rational-squares. – lhf Mar 05 '16 at 17:37
  • Converse is obvious. You don't have to show that. – fleablood Mar 05 '16 at 17:37
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    A little googling and I see this is true for any integer that is a sum of two rational squares. It doesn't need to be an odd prime. – fleablood Mar 05 '16 at 17:42

2 Answers2

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If the equation $x^2 + y^2 = p$ has a rational solution, then $x^2 + y^2 = pz^2$ has an integer solution, and so does $x^2 \equiv -1 \bmod p$. This implies that $x^4 \equiv 1 \bmod p$ has a solution and so there is an element of order $4$ in $U(p)$. By Lagrange's theorem, $4$ divides $p-1$. As it is well known, this implies that $x^2 + y^2 = p$ has an integer solution.

lhf
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  • Let $x=\frac{a}{b}$ and $y=\frac{c}{d}$ then $x^2+y^2=\frac{a^2}{b^2}+\frac{c^2}{d^2} = \frac{a^2d^2+b^2c^2}{b^2d^2}$ How did you get $pz^2$? – Tony Mar 05 '16 at 17:48
  • @Tony, it is obvious. $\left(\frac xz,\frac yz\right)$ is an solution to that equation then. –  Mar 05 '16 at 18:07
  • How can you assume the denominator is the same? – Tony Mar 05 '16 at 18:11
  • And how did you get $x^2 \equiv -1 \pmod p$? – Tony Mar 05 '16 at 18:19
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Perhaps the following can clear up some doubts in lhf's answer:

Suppose there are $\;a,b,c,d\in\Bbb Z\;$ such that with reduced fractions we have

$$\frac{a^2}{b^2}+\frac{c^2}{d^2}=p\iff (ad)^2+(bc)^2=p(bd)^2$$

Let $\;q\;$ be a prime dividing $\;b\;$ , then from $\;b^2\left(pd^2-c^2\right)=(ad)^2\;$ we get that $\; q\mid(ad)^2\implies q\mid ad\;$ . Since $\;\frac ab\;$ is a reduced fraction we get that $\;q\mid d\;$, and by symmetry of the argument we get that any prime dividing $\;d\;$ also divides $\;b\;$ and from here that $\;b=\pm d\;$ , and if we decide that $\;a,b,c,d\in\Bbb N\;$ then in fact $\;b=d\;$, so we have the equality

$$a^2+c^2=pb^2\implies \left(\frac ac\right)^2=-1+\left(\frac bc\right)^2p$$

so putting $\;x:=\frac ac\;$ , the above means $\;x^2=-1\pmod p\;$ , and this means the cyclic group $\;\Bbb F_p^*\;$ has an element of order four , so

$$4\mid|\Bbb F_p^*|=p-1\iff p=1\pmod4$$.

DonAntonio
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