<s>[INST] 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 $x,y$
 [/INST] 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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