There are nonnegative polynomials that are not sums of squares. For example Motzkin gave the example $x^4y^2+x^2y^4+z^6-3x^2y^2z^2$ in 1967.


Is there a real polynomial $f\in{\mathbb{R}}[x\_1,\dotsc,x\_n]$ in several indeterminates that is not a sum of squares but $f^N$ is a sum of squares for some odd integer $N>0$?


This question is interesting in the following sense. The notion of writing nonnegative polynomials $f$ as a sum of squares is to give an algebraic proof of the inequality $f\ge 0$. As per Motzkin's example, we know that this is not always possible. One way to resolve this is to follow Artin and use denominators. Another way (which I learnt from D'Angelo) is to show that $f^{N}$ is a sum of squares for some odd $N$.


This question is me wondering whether such a technique of consider the radical of sum of squares is vacuous.



When is a power of a nonnegative polynomial a sum of squares?
Motzkin's original proof shows that $x^4y^2 + x^2y^4 + z^6 - a x^2y^2z^2$ is psd and
not sos for any $a$ in the interval $(0,3]$. If you take $a = .02$ say, it is reasonably
simple, though messy, to show that $(x^4y^2 + x^2y^4 + z^6 - .02x^2y^2z^2)^3$ is a 
sum of squares; in fact, it's a sum of binomial squares $(x^b y^c z^d - x^e y^f z^g)^2$, 
where $b+c+d=e+f+g=9$. The idea is to look at any monomial with a negative coefficient
and make it into the middle term of this square, in a way that the other two terms
are still in the Newton polytope. For example, one term in the given cube is $-.06x^10y^6z^2$, which is "handled" by $.03(x^6y^3 - x^4y^3z^2)^2$. It's sort of messy to work out, but I've convinced myself (at least) that it's true. 


