Solving multivariable 4 grade equation

Question

Find the integer solutions of the following equation:

$$8x^4+y^4=6x^2y^2$$

I tried to reconfigure the equation using the notable identity $(a-b)^2=a^2+b^2-2ab$, but due to the numeric coeficients I was not able

Answer 1

You can factor

$8x^4-6x^2y^2+y^4$ as

$9x^4-6x^2y^2+y^4-x^4=(3x^2-y^2)^2-(x^2)^2=(4x^2-y^2)(2x^2-y^2)$

So the equation becomes

$(4x^2-y^2)(2x^2-y^2)=0$

At least one of these two factors has to be equal to 0

But $2x^2=y^2$ is impossible, because doubling a square changes the exponent of $2$ present in the prime factorization of the square from even to odd, which makes it a non-square integer. So:

$4x^2=y^2$

$y=\pm 2x$

The solutions are in the form $(k;\pm2k)$