Rational solutions for $x^2+y^2=3$ Are there any rational numbers $x$ and $y$ such that $x^2+y^2=3$. I think there are no rational solutions, but I haven't been able to prove it.
 A: If you had solutions over the rationals, clearing denominators would give you a solution to $x^2+y^2 = 3z^2$ over the integers with $gcd(x,y,z) =1$. Now looking at the quadratic residues modulo 3, the only way $3|(x^2+y^2)$ is if $3|x$ and $3|y$. But then $9| (x^2+y^2)$, so $3|z$ and $gcd(x,y,z) \ne1$.
A: This is going to sound a little peculiar: the quadratic form $x^2 + y^2$ is one of only a handful of binary forms with this strong property: for a given integer $n,$ there is a solution to $x^2 + y^2 = n$ in the rationals $\mathbb Q$ if and only if there is a solution in the integers $\mathbb Z.$
This can be proved using Nate's observation and a good deal of care. It is generally referred to as the Davenport-Cassels theorem, but actually goes back to a French person named Aubry in 1912. The argument, for the sum of three squares rather than two, is in Serre's little book.
For an introduction in one place, see the pdf
[15] Euclidean quadratic forms and ADC forms I. (pdf), Acta Arithmetica 154 (2012), 137-159.
at http://alpha.math.uga.edu/~pete/papers.html
EEDDIITT: I should have said that there are just a handful of binary forms for which the Aubry argument works. Evidently there might be as many as 400 or so forms that have the rational-if-and-only-if-integral property, collected by Voight, but for the majority of these more difficult proofs are required.
Well, why not: the entire collection of positive forms for which Aubry works, up to the maximum possible (ten variables) was enumerated by Gabriele Nebe using Lie algebra notation, see pdf
Even lattices with covering radius strictly smaller than sqrt{2}.
Beiträge zur Algebra und Geometrie, Vol. 44, No. 1, 2003, 229-234
Note that I forgot the case R=A2A1A1 in Theorem 7. This root system yields two further lattices, one with covering radius =sqrt{2} and one with c.r. strictly smaller than sqrt{2}
at http://www.math.rwth-aachen.de/~Gabriele.Nebe/pl.html
A: For variety, you're asking if there is an element $z$ in the number field $K = \mathbb{Q}(i)$ such that $N(z) = 3$.
However, $3$ is inert in $K$; the only prime ideal in $\mathcal{O}_K$ lying over $3$ has degree $2$. Therefore, the factorization of the norm of any element of $K$ must have $3$ raised to an even power.
And, thus, no element of $K$ can have norm $3$.
A: As shown in this answer, $n$ is the sum of the squares of two integers if and only if each prime factor of $n$ which is $3\pmod4$ appears with even exponent.
Suppose that we find
$$
\left(\frac pr\right)^2+\left(\frac qr\right)^2=3\tag{1}
$$
then
$$
p^2+q^2=3r^2\tag{2}
$$
and the number of factors of $3$ in $3r^2$ is odd. Thus, $(1)$ cannot be true.
