<s>[INST] Proving an homogenous sextic product never a square Can anyone prove that $$2(a^2+b^2)(a^4+a^2b^2+b^4) \ne \square$$ for $a,b$ positive integers?
 [/INST] I am a little confused. First, assume that $a, b$ are relatively prime (dividing through by the gcd divides the lhs by a square). Second, note that $a^2+b^2$ is relatively prime to $2(a^4 + b^2a^2+b^4)$ (since the expression within the last set of parentheses is $(a^2+b^2)^2 - a^2 b^2.$) So, if the thing has any hope of being a square, $a, b$ are two sides of a pythagorean triangle, and $2(a^4 + b^2a^2+b^4)$ is a square. Since, $a, b$ are two sides of a pythagorean triangle, one is even, the other is odd, so the expression within the last paren is odd, but twice an odd number is not a square.
UPDATE in the case, $a, b$ are both odd, $2(a^2 + b^2)$ is divisible by $4,$ but the expression in the last paren is $3$ mod $4,$ so again, not a square.
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