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Let $d$ and $n$ be square-free natural numbers. Is it true that $x^2+dy^2=n$ has a rational solution if and only if it has an integral solution? I know this is true for circles (i.e., when $d=1$) but I can't seem to be able to extend that proof to ellipses in general.

Can someone give me a proof (hopefully elementary), or a counterexample?

unity
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  • Do you want to show that if there is a rational solution then there is an integral solution? (The other direction is obvious) Or do you want to show that all rational solutions are integral? – user2566092 Jun 01 '15 at 21:51
  • The former. Clearly, if there is an integral solution, then there are rational solutions. If there is an integral solution, there is in fact a parametrization of all rational solutions, so there will be infinitely many points defined strictly over $\mathbb{Q}$, in addition to those defined strictly over $\mathbb{Z}$. – unity Jun 01 '15 at 21:56
  • http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 – individ Jun 02 '15 at 12:21

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$x^2+23y^2=41$ is easily checked to have no integral solutions, but $x=1/3$ and $y=4/3$ is a rational solution. This is related to factorization in the integers of $\mathbf{Q}(\sqrt{-23})$.

Jeroen
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  • Thanks! Could you elaborate what's the reason behind this? $\mathbb{Q}(\sqrt{-5})$ is also of class number $>1$, but every equation $x^2+5y^2=n$ that has a rational solution seems to also have an integral solution... – unity Jun 02 '15 at 00:46
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    It's a little subtle (which is why I didn't go into it in the post). First the 41 analysis. $-23$ is a square mod 41 so 41 splits in the integers $R$ of $\mathbf{Q}(\sqrt{-23})$ but it's readily checked that there are no elements of norm 41 in $R$ so there are no integral solutions. To check there are rational solutions one only needs to check locally at 2, 23, 41 and at the infinite place; there are clearly real solutions, we've just checked 41 splits in $R$ so there are local solutions at 41, and there are local solutions at 2 because 41 is 1 mod 8 so a square in the 2-adics. – Jeroen Jun 02 '15 at 18:41
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    (we don't need to check 23 because we know there will be an even number of places where local solvability fails). For 5 the story is a bit different. You're only interested in integral $n$ for the question to make sense. WLOG it's squarefree. If there's a rational solution then for local solvability at a prime $p\not=2,5$ dividing $n$ we need $-5$ to be a square mod $p$ so $p$ is in some set of congruence classes mod 20, and it looks like the same argument is going to work, but then a miracle occurs: local solvability at 2 forces another condition on $n$. Let's stick to the case $n>5$ prime... – Jeroen Jun 02 '15 at 18:57
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    --- then local solvability at 2 is another condition which coincidentally happens to coincide precisely with $n$ splitting completely in $\mathbf{Q}(\sqrt{-5},\sqrt{-1})$, and by a miracle this is the Hilbert class field of $\mathbf{Q}(\sqrt{-5})$. And primes that split completely in the HCF are principal! So we can find an integer solution. This is just some sort of low-dimensional coincidence because the Hilbert class field of $\mathbf{Q}(\sqrt{-5})$ is abelian over the rationals and won't happen in general. For general $n$ the parity condition coming from 2 must force things to work smlrly. – Jeroen Jun 02 '15 at 19:00
  • Thanks so much for taking the time to explain this. – unity Jun 02 '15 at 19:55