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.
 A: 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.
A: Another viewpoint is by using the composition of binary quadratic forms, to  reduce the case $ n $ odd to the case $ n = 1 $, in a `systematic' way.  We have the following identities writing a product
$$
(2a^2 + 3b^2)(2c^2 + 3d^2) = (3 b c + 2 a d)^2 + 6 (-a c + b d)^2
$$
and
$$ 
(2a^2+3b^2)(c^2 + 6y^2) = 2 (3 b c + a d)^2 + 3 (-2 a c + b d)^2 \, .
$$
So the $ n = 2m+1 $-fold product of
$$
  N = (2x_1^2 + 3y_1^2) \cdots (2x_n^2 + 3y_n^2)
$$
can be written as
$$
  N = 2X^2 + 3Y^2 \, 
$$
for some integer polynomials $ X = f(x_0,\ldots,x_n,y_0,\ldots,y_n) $ and $ Y = g(x_0,\ldots,x_n,y_0,\ldots,y_n) $.
So if this product is ever a positive square, we can obtain values $ X, Y $ to make $ 2X^2 + 3Y^2 $ a positive square.  Following other answers, we see this is impossible.
If $ 2X^2 + 3Y^2 = Z^2 $ is a positive square, we can assume $ \gcd(X,Y,Z) = 1 $ by dividing through if necessary.  Then reducing modulo 3 shows that $ -X^2 \equiv Z^2 \pmod{3} $.  If $ X \not\equiv 0 \pmod{3} $, we obtain $ -1 \equiv \Box \pmod{3} $.  Contradiction, as the squares modulo 3 are 0 and 1.  So $ 3 \mid X $ and $ 3 \mid Z $.  Hence $ 9 \mid 3Y^2 $, so $ 3 \mid Y $, contradicting $ \gcd(X,Y,Z) = 1 $.  So $ Z = 0 $ is the only possible solution, which immediately gives $ X = Y = 0 $ too.
A: 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.
A: This might not be the most efficient way to do this, but it should work.
First, show that $2x^2 + 3y^2$ must be of the form $2^{2j+1}(8k\pm 1)$ or $2^{2j}(8k\pm3)$ for some integers $j$ and $k$. This can be done by factoring out powers of 2 until one of $x$ or $y$ is odd, then reducing mod 8.
Now suppose that there are an odd number of such factors as in the problem.
If the number of factors of the form $2^{2j+1}(8k\pm 1)$ is odd, we are done, as the resulting product has 2 to an odd power in its prime factorization. Such a number cannot be a perfect square.
If the number of factors of the form $2^{2j}(8k\pm3)$ is odd, their product will be of the form $4^m(8k\pm3)$ for some integer $m$. This follows from $3^2 \equiv (-3)^2 \equiv 1 \mod 8$. The product of the factors of the form $2^{2j+1}(8k\pm 1)$ will take the form $4^m(8k\pm1)$. Thus, the total product will be of the form $4^m(8k\pm1)(8k\pm3)$. This can't be a perfect square, as $(8k\pm1)(8k\pm3) \not\equiv 1 \mod 8$.
A: I think a theoretical answer can be given to a generalization of your question, when replacing $(2,3)$ by any pair $(a,b)\in \mathbf N^2$ : can  the product $P_n =  \prod_j (a x_j^{2} + b y_j^{2}), 1\le j \le n$ odd, be a perfect square ? Since $P_n$ is homogeneous, we may as well consider solutions $(x_j , y_j) \in \mathbf  Q^2$ and use the fact that $a x_j^{2} + b y_j^{2}=a(x_j^{2} + \frac ba y_j^{2})= aN(x_j + \sqrt {\frac ba} y_j)$, where $N$ is the norm map of the imaginary quadratic field $\mathbf Q (\sqrt {-\frac ba})$. Since the norm is multiplicative, $P_n$ will be of the form $a^n (x^2 + \frac ba y^{2}) = a^{n-1}(a x^{2} + b y^{2})$. Since $n$ is odd, $P_n$ is a rational square iff $(a x^{2} + b y^{2})$ is, and we are brought back to the question whether the rational quadratic form $Q=z^2 - ax^2 - by^2$ represents $0$. The answer is given by the Hasse-Minkowski local-global principle (here, for a ternary form, it even goes back to Legendre): over $\mathbf R$, $Q$ represents $0$ because of its signature; over any  $p$-adic field $\mathbf Q_p, Q$ represents $0$ iff in $\mathbf Q_p^*/\mathbf Q_p^{*2}$, one has $\epsilon = (-1, -d)_p$, where $(.,.)_p$ is the Hilbert symbol, $d=ab$ is the discriminant, and $\epsilon$ is the product over $i<j$ of the $(a_i , b_j)_p$'s, with $a_i = -a, -b, 1$ (see e.g. Serre's "A course in arithmetic", chapter IV, thm. 6).
