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Prove the equation $$\left(2x^2+1\right)\left(2y^2+1\right)=4z^2+1$$ has no solution in the positive integers

My work:

1) I have the usually problem

$$\left(nx^2+1\right)\left(my^2+1\right)=(m+n)z^2+1$$

in the positive integers. Initially I use case $\gcd(m,n)=1$

2) Let $m=n=2$. It is this case. I need to prove that $(x^2-y^2)^2+(x^2+y^2) $ is not perfect square for any $x,y$

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    $\begingroup$ Seems to be hard. I tried infinite descent, but I failed. Also considering the prime factors of $2x^2+1$,$2y^2+1$ and $4z^2+1$ led to nowhere. I am curious whether someone can solve this ... $\endgroup$ – Peter Jul 10 '18 at 17:28
  • $\begingroup$ If it's any help, it's easy to prove the impossibility of the special case $x = y$. $\endgroup$ – Connor Harris Jul 10 '18 at 17:42
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    $\begingroup$ Could you clarify the relevance, in (2), of the expression $(x^2 - y^2)^2 + (x^2 + y^2)$, which equals $x^4 - 2x^2y^2 + y^4 + x^2 + y^2$? With its 4th powers, it seems unrelated to your initial problem statement. $\endgroup$ – Adam Bailey Jul 14 '18 at 12:30
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    $\begingroup$ In the case of $(2)$ (equation in the title) it can be shown that $x,y,z$ must be divisible by $3$, hence there is no solution in positive integers when $$36m^2n^2+2m^2+2n^2$$ cannot be a positive perfect square. But this apparently does not help much. $\endgroup$ – Peter Jul 15 '18 at 13:42
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    $\begingroup$ Found similar question: Math Overflow Link $\endgroup$ – cvogt8 Jul 26 '18 at 7:59
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As it says in the comments this is answered on Math Overflow. The only solution is indeed the trivial one: $(0,0,0)$. This can be found in Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations.

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    $\begingroup$ My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote? $\endgroup$ – Mason Dec 13 '18 at 16:39
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    $\begingroup$ You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 4 at 16:07
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The question is concerned only with $x,y,z\ge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $\mod x$.

$1\cdot (2y^2+1)\equiv 4z^2+1\mod x$.

Let $z\equiv a\mod x$. Then $2y^2+1=4(nx+a)^2+1$.

Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2\Rightarrow 2=\frac{y^2}{(nx+a)^2}\ $ which would make $\sqrt{2}$ rational. Hence there are no positive integers that satisfy the equation.

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    $\begingroup$ There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$ $\endgroup$ – jjagmath Dec 12 '18 at 18:45
  • $\begingroup$ I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $\mod x \dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this. $\endgroup$ – Mason Dec 17 '18 at 20:42

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