Express $x^4 + y^4 + x^2 + y^2$ as sum of squares of three polynomials in $x,y$

Question

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.

Answer 1

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.

Answer 2

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, $107(9):813–821$, $2000$.

Answer 3

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).

Answer 4

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.