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.

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.
