Sum of Two Squares of a Quartic

Question

I am trying to write a quartic in sum of two squares, can anyone help me with the following polynomial: $$6y_0^{4}+6y_0^{2}y_1^{2}+y_1^4+4y_0y_1^{2}y_2+4y_0^{2}y_2^{2}+ 6y_1^{2}y_2^2+6y_2^{4},$$ one easily can be find: $$y_1^4+4y_0y_1^{2}y_2+4y_0^{2}y_2^{2} =(y_1^2 +2y_0y)^2,$$ I am stuck by completing the other one, any help and guidance are highly appreciated.

Answer 1

I don't think this is possible, here's why

First, the remaining parts all have coefficients of 6, which makes me think there's either no way to do this or that your first choice of the sum of two squares isn't what they were looking for.

This could mean it's still possible to find 2 other squares, but I claim that isn't the case as well.

We have a $y_0^4$ term, which in the absence of division (which would create other problems) can only be formed by the sum of two squares if each square can be written as

$$(a_1y_0^2+...)^2+(a_2y_0^2+...)$$

Where $a_1$ and $a_2$ are integers. However, this would produce

$(a_1^2+a_2^2)*y_0^4$ as the only $y_0^4$ term. So $a_1^2+a_2^2=6$. However, there is no integer values of $a_1, a_2$ where this is true, so I think it's impossible