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How can I determine if $x^2+y^2 = z$ has infinitely many rational solutions $(x,y)$ based on the value of $z\in \Bbb{Q}$.

I understand that $a^2+b^2=1$ has infinitely many rational solutions. As does $a^2+b^2=2$. However, $a^2+b^2=3$ has no rational solutions. When $z$ is an integer I know how to determine if it has rational solutions. However when $z$ is not an integer it is not clear when it has rational solutions.

For example, it is not immediately clear to me if the equation $$x^2+y^2 = \frac 43$$ has infinitely many rational solutions.

Ryan
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  • https://math.stackexchange.com/questions/1513733/solving-a-diophantine-equation-of-the-form-x2-ay2-byz-cz2-with-the-co/1514030#1514030 – individ May 23 '17 at 07:42
  • For any $z > 0$, If it has one solution, it has infinitely many because there are infinitely many $u,v \in \mathbb{Q}$ satisfy $u^2+v^2=1$ and hence $(xu-yv)^2 + (xv+yu)^2 = x^2+y^2 = z$. – achille hui May 23 '17 at 07:58
  • @achillehui: how do you get the last identity ? –  May 23 '17 at 08:08
  • If there is one solution we can write the formula. https://math.stackexchange.com/questions/1513733/solving-a-diophantine-equation-of-the-form-x2-ay2-byz-cz2-with-the-co/1514052#1514052 – individ May 23 '17 at 08:15
  • Or more common there. https://mathoverflow.net/questions/225781/fricke-klein-method-for-isotropic-ternary-quadratic-forms/225995#225995 – individ May 23 '17 at 08:17
  • @YvesDaoust $(x+iy)(u+iv) = (xu-yv)+(xv+yu)i$ – achille hui May 23 '17 at 08:22
  • @achillehui: where is $z$ gone, now ? –  May 23 '17 at 08:24
  • @achillehui: got it now, thanks. I was missing that is $x,y$ is the known solution. –  May 23 '17 at 08:35
  • @DanielW.Farlow I understand that $a^2+b^2=1$ has infinitely many rational solutions. As does $a^2+b^2=2$. However, $a^2+b^2=3$ has no rational solutions. When $z$ is an integer I know how to determine if it has rational solutions. However when $z$ is not an integer it is not clear when it has rational solutions. Thanks. – Ryan May 23 '17 at 13:53
  • @Ryan I wonder. I brought the formula and there is a solvability condition. What is not satisfied? – individ May 23 '17 at 14:27
  • @individ - Thank you for continuing to try to help me with this problem. Maybe I'm being slow, but I don't see how the solutions proposed on the other problem answer my question. You say "I brought the formula and there is a solvability condition." Can you please explain further? What formula? And what is the solvability condition? Perhaps if you can show me how you are able to determine if a circle of radius sqrt(4/3) has rational solutions this would help me. Thanks again. – Ryan May 23 '17 at 15:16

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