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

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 $x,y$

• 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 ... – Peter Jul 10 '18 at 17:28
• If it's any help, it's easy to prove the impossibility of the special case $x = y$. – Connor Harris Jul 10 '18 at 17:42
• 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. – Adam Bailey Jul 14 '18 at 12:30
• 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. – Peter Jul 15 '18 at 13:42
• Found similar question: Math Overflow Link – cvogt8 Jul 26 '18 at 7:59

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.

• 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? – Mason Dec 13 '18 at 16:39
• You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him. – GNUSupporter 8964民主女神 地下教會 Feb 4 at 16:07

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.

• There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$ – jjagmath Dec 12 '18 at 18:45
• 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. – Mason Dec 17 '18 at 20:42