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

Solution 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.

Solution 2:

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).

Solution 3:

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, $107(9):813–821$, $2000$.