<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] Case n=1
Let $z^2=2x^2+3y^2$
$z^2=2x^2 [3]$.
If x is not divisible by 3, then $z^2=2[3]$ which is impossible. Then $x$ is divisible by 3, and consequently $z$ is also a multiple of 3.
Let us look at the prime decomposition of $z$ such that $z^2=2x^2+3y^2$. 3 is a prime factor of $z$ and let us not $p$ its power: $z=3^p z'$
Let us note $x=3^a x'$ and $y=3^b y'$, then $$3^{2p} z'^2 = 2x'^2 3^{2a}+y'^2 3^{2b+1}$$ If we look at this equation modulo $3^{\min(2p, 2a, 2b+1)}$ we get a contradiction.
Indeed, if $2a=2p$ we obtain $z'^2 = 2x'^2+y'^2 3^{(2b+1)-2a}$ which gives $1=2[3]$. If $2a \neq 2p$, dividing by $3^{\min(2p, 2a, 2b+1)}$ leads to an equation $u=v+w$ with exactly 2 numbers among $u$, $v$ and $w$ which are divisible by 3, which is impossible.
This proves the case $n=1$.
General case
For the general case, let us note that the whole product is equal to 0 mod 3 iff one $x_i$ is a multiple of 3. ($2x^2+3y^2$ is 0 modulo 3 iif $2x^2=0[3]$). The product cannot be equal to 1 modulo 3 since we have an odd number of factors (and $2x^2+3y^2$ = 0 or 2). Then the same argument as before shows that $z$ is divisible by 3.
We have $z=3^p z'$ and $3^{2p}z'^2 = \Pi_{i=1}^n (2x_i^2+3y_i^2)$
The factors $2x_i^2+3y_i^2$ are of 2 different types:


*

*the ones with $x_i \neq 0 [3]$ and in that case $(2x_i^2+3y_i^2)= 2[3]$

*the ones with $x_i = 0 [3]$, then the power $p_i$ of 3 in the prime decomposition of $2x_i^2+3y_i^2$ is even or odd. We have two subcases:


*

*if  $p_i$ is odd then $(2x_i^2+3y_i^2) / 3^{p_i} = 1[3]$

*if  $p_i$ is even then $(2x_i^2+3y_i^2) / 3^{p_i} = 2[3]$



We must have an even number of odd $p_i$, because $2p=\sum_{i} p_i$. Then when we divide the equation by $3^{2p}=\Pi_i 3^{p_i}$ and we take it modulo 3 we have a product off an even number of factors that are 1 modulo 3  and an odd number of factors that are 2 modulo 3. Consequently the right hand side is 2 modulo 3, while the left hand side (a perfect square) is equal to 1 modulo 3. Hence we have a contradiction.
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