<s>[INST] Prove that the product is never a perfect square 
Prove that for nonnegative integers $x_1,\ldots,x_{2011}$ and $y_1,\ldots,y_{2011}$ the product $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_{2011}^2+3y_{2011}^2)$$ is never a positive perfect square.

I thought about generalizing this question to any odd subscript $n$ instead of $2011$. Thus, $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_n^2+3y_n^2)$$ is never a perfect square. For $n = 1$ we have $2x^2+3y^2 = z^2$ and I want to show the only solution is $x = y = z = 0$. If $x$ is even, then $y$ must be even by taking modulo $4$. If $x$ is odd, then $y$ must be odd. I didn't see how to continue from here.
 [/INST] Modulo $3$, the equation for $n=1$ reduces to $2x^2=z^2 \pmod3$.
If $x$ were not divisible by $3$ then we would find $2=u^2 \pmod3$ for some $u$. But $2$ is not a quadratic residue $\pmod 3$ so this is impossible.
But if $x$ were divisible by $3$ this would make $z$ divisible by $3$, and thus $z^2$ divisible by $3^2 = 9$. This would in turn force $y$ to be divisible by $3$ otherwise $x^2+3y^2$ would be divisible by $3$ but not $9$.
Dividing such a solution by $9$ would give another solution $(x/3,y/3,z/3)$ with $z$ divisible by a smaller power of $3$, and eventually we reach the same case as above (infinite descent), or conclude that $x,y,z$ are infinitely divisible by $3$ and are thus equal to $0$.
So the only solution to $2x^2 + 3y^2 = z^2$ over the integers is $(x, y, z) = (0, 0, 0)$.

For the case of arbitrary odd $n$, we proceed by induction on $n$.
Now suppose we had a solution to the equation $(2x_1^2 + 3y_1^2)...(2x_{n}^2 + 3y_{n}^2)(3k + 1) = z^2$ for $n$ odd. Note that we have loosened the equation by a factor of $3k + 1$ for any $k \in \mathbb{Z}$. Also note that the base case $n = 1$ for this equation is proven exactly the same way as above because the factor $3k + 1$ does not affect the residues modulo $3$. So proceed with assuming we proved that there are no nonzero solutions for $n-2$.
As long as we have some $i$ such that both $x_i$ and $y_i$ are divisible by $3$, replace $(x_i, y_i, z)$ with $(x_i/3, y_i/3, z/3)$ to obtain another valid equation (dividing by $9$ on both sides). If the process never terminates this means $z = 0$ and thus one of the $(x_i, y_i)$ must be $(0, 0)$ as well.
Otherwise we have reached a point where every factor is divisible at most by $3$ but not by $9$ (this would require both $x_i$ and $y_i$ divisible by $3$).
We consider two cases for the number $N$ of $i$ for which $2x_i^2 + 3y_i^2$ is divisible by $3$.
1. Case $N = 0$
Suppose we have a solution to $(2x_1^2 + 3y_1^2)...(2x_{n}^2 + 3y_{n}^2)(3k + 1) = z^2$ with none of the terms divisible by $3$. Then the equation reduces to $-1 = (-1)^n = 2^n = (zx_1^{-1}...x_n^{-1})^2 \pmod 3$ which is impossible because $-1$ is not a residue $\pmod 3$.
2. Case $N \geq 1$
We must then have $x_i$ divisible by $3$ for some $i$. Because of the process above, this means $y_i$ is not divisible by $3$. Since $z^2$ is divisible by an even power of $3$, there must be another factor $2x_j^2 + 3y_j^2$ divisible by $3$ (exactly once). Now divide the equation by $9$ on both sides to replace $z$ by $z/3$ and $(2x_i^2 + 3y_i^2)(2x_j^2 + 3y_j^2)$ by $(6(x_i/3)^2 + y_i^2)(6(x_j/3)^2 + y_j^2)$. Both of these factors are of the form $3k + 1$ because $y_i^2, y_j^2 = 1 \pmod 3$.
We have therefore obtained an equation of the form $(2x_1^2 + 3y_1^2)...(2x_{n-2}^2 + 3y_{n-2}^2)(3k + 1) = z^2$. Such an equation has been assumed to have no nonzero solutions, and so we are done by induction.
</s>