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I seem to recall that a general homogenous real polynomial $P$ of even degree in $n$ variables, $n\geq 3,$ cannot always be expressed as $P(x_1,\dotsc,x_n)=\sum_j a_j Q_j^2(x_1,\dotsc,x_n)$ where $a_j \in \mathbb{R},$ and the $Q_j^2$ are homogenous of the same degree as $P.$ (Please, correct me if I am wrong).

Now, what if we know that $P$ is symmetric? Is it still true that a general polynomial cannot be expressed as above? And if it can, what if we require that the $Q_j$ themselves are symmetric?

Question: If $P$ is symmetric and homogenous of even degree, can it be expressed as a sum/difference of squares of homogenous and symmetric polynomials?

I know that if $P$ is symmetric, and $P(x_1+t,x_2+t,\dotsc,x_n+t)=P(x_1,\dotsc,x_n),$ then there is always such a representation as the one above, but can one lose the translation-invariant condition and still have sum-and-difference of squares representation?

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Every element of a ring in which $2$ is a unit is the difference of two squares: $P=(P+1/4)^2-(P-1/4)^2$. – Emil Jeřábek Feb 26 '13 at 12:39
Ah, let me rephrase that question a bit, I realized that I have some conditions on Q – Per Alexandersson Feb 26 '13 at 12:56
@Per: you mean the degree of Q is 1/2 that of P...So you are looking at a higher degree version of Gauss's decomposition of quadratic forms? – Abdelmalek Abdesselam Feb 26 '13 at 18:24
@Abdelmalek Abdesselam: Yes, I believe so. (Thank you very much for the input on one of my previous questions by the way.) – Per Alexandersson Feb 26 '13 at 20:09
At the risk of stating the even more obvious, the degree of $Q^{2}$ is twice the degree of $Q$... – George Melvin Feb 26 '13 at 22:26

My college provided me with a simple example: $$x^2 + xy + y^2$$ cannot be expressed as a sum/difference of symmetric, homogenous polynomials of degree 1, for obvious reasons.

(Each homogenous, symmetric polynomial of degree 1 in two variables are of the form a(x+y). Thus, all possible $Q_j$ are of the form $a'(x+y)^2$ which, of course, cannot give the polynomial above.)

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