Express $x^4 + y^4 + x^2 + y^2$ as sum of squares of three polynomials in $x,y$ I don't know any identity that'd help me simplify it. I know of Brahmagupta's identity and tried using but no good. Any hints? 
Edit: So far I've tried various things, 
$(x^2+1/2)^2 + (y^2 +1/2)^2 -1/2$
This doesn't get me anywhere.
I tried solving in x as a quadratic so as to get an insight or something but that didn't help too.
I even tried writing general polynomials of degree too and then getting the solution by comparing coefficients, but that wasn't elegant and was too computational.
 A: let real constant
$$  t = \sqrt{\sqrt 8 - 2} $$
so that
$$  \frac{t^4}{4} + t^2 = 1. $$
Then
$$ \left(-x^2 + \frac{t^2}{2} y^2 \right)^2 +  \left( t y^2 + x \right)^2 +  \left( txy - y \right)^2 = x^4 + y^4 + x^2 + y^2 $$
ADDED: apparently this was asked and answered six years ago, and I commented  there. 
A: As an application of Legendre's three squares theorem, this problem has no solution in $\mathbb Z[x,y]$.
In fact, suppose $$f(x,y)=x^4+y^4+x^2+y^2=(h_1(x,y))^2+(h_2(x,y))^2+(h_3(x,y))^2$$ so one has $$f(1,3)=92=(h_1(1,3))^2+(h_2(1,3))^2+(h_3(1,3))^2$$
But $$92=4(2\cdot8+7)$$
This is impossible; for the above theorem, $92$ can not be representable as a sum of three squares (that one or two of the  $h_i(1,3)$ be zero  is easily discarded).
A: More of a comment: Hilbert proved that this was possible in 1888 (a positive definite quartic in two variables can always be written as a sum of squares of three polynomials).  For instance, see the discussion of the proof in [1].  But it is a highly nontrivial result, and the proof is not especially constructive.
[1] W. Rudin. Sums of squares of polynomials. Amer. Math. Monthly, $107(9):813–821$, $2000$.
A: Just an observation, one considers a writing
$$f(x,y)= (x,y,x^2, xy, y^2) M (x,y,x^2, xy, y^2)^t$$ where $M$ is a symmetric $5\times 5$ matrix. Assume $M$ is positive semidefinite. Diagonalizing $M$ will produce a writing of $f$ as a sum of squares. To have three squares, we need  that the rank of $M$ is at most $3$. So now the problem is to find a positive semidefinite matrix $M$ of rank at most $3$ and satisfying the equality. 
