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There are many questions on this site asking about integer solutions to the generalized Pell equation $x^2 - dy^2 = n$ for $d$ and $n$ integers and $d$ squarefree. What is known about the existence of solutions when we allow rationals? Do we only have rational solutions when we have integer solutions or is the situation more complex?

This question is relevant but only concerns the usual Pell equation $x^2 - dy^2 = 1$.

  • The answer by Phillip Gibbs to the linked question does not depend on the right side being $1$. It may not be as specific as you are looking for, it is a procedure. – Ross Millikan Oct 11 '21 at 02:11
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    It becomes integer $u^2 - d v^2 = n w^2 ; , ; ;$ where $\gcd(u,v,w) = 1,$ with $x= u/w$ and $y = v/w.$ Conditions for $u^2 - d v^2 = n w^2 $ are the subject of Legendre's https://math.stackexchange.com/questions/987948/clarification-of-legendres-theorem-re-ax2by2-cz2 ALSO the complete collection of solution triples is determined by triples of binary quadratic forms. This is page 47, Theorem 4, in Mordell's Diophantine Equations. Meanwhile, I like the treatment of Legendre in Cassels, Rational Quadratic Forms. I see, Mordell writes Theorem 3 on page 46 about Legendre ternary – Will Jagy Oct 11 '21 at 02:15
  • The solution by Phillip Gibbs starts with the trivial solution (1,0) then generates more, but we don't have a trivial solution here @RossMillikan –  Oct 11 '21 at 02:17
  • @WillJagy thanks for the link/references. Is it known when (if at all) rational solutions imply the existence of integer solutions? I guess that was the gist of my question, but you've given me plenty of reading material to start on. –  Oct 11 '21 at 02:43
  • It is easy enough to decide whether a number $n$ is integrally represented by at least one binary quadratic form of discriminant $4d.$ However, as soon as there is more than one form per genus...famous example, we can express $p = x^2 + 27 y^2$ when $p \equiv 1 \pmod 3$ with the extra condition that $2$ be a cube $\pmod p$ See Hudson and Williams for similar examples, and the book Primes of the Form $x^2 + n y^2$ by Cox; HW: http://zakuski.utsa.edu/~jagy/Hudson_Williams_1991.pdf – Will Jagy Oct 11 '21 at 02:56
  • @WillJagy (I thought by HW you meant you were assigning me homework!) Thanks again for the references, I have read parts of Cox but not Hudson and Williams and the linked example is interesting. However I think you misunderstand my question. I understand we can skirt the 'rational solution' in practice and just check if there is an integer solution. But if I tell you that I have rational solutions to $p = x^2 + ny^2$ for some $p, n$ can we say anything about the existence of integer solutions? My guess is no. –  Oct 11 '21 at 03:12
  • Rational $x^2 + n y^2 = p $ says, with some restrictions on the common denominator of $x,y,$ that the prime is represented integrally by some $ax^2 + bxy + c y^2,$ where $b^2 - 4ac = -4n.$ Earlier example: every prime $p \equiv 1 \pmod 3$ is rationally represented by $x^2 + 27 y^2.$ However, they are all integrally represented by either $x^2 + 27 y^2$ or $4x^2 + 2 xy + 7 y^2,$ but not by both. Recommend you run a small program, find the primes up to 1000 that are integrally represented by $x^2 + 27 y^2,$ similar list for $4 x^2 + 2xy + 7 y^2.$ No overlap! – Will Jagy Oct 11 '21 at 15:55

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