Let $P(x,y) = x^2y^4 + x^4y^2 + 1 - 3x^2y^2$

a.) Prove that $P(x,y)$ is non-negative

b.) Prove that $P(x,y)$ cannot be expressed as a sum of squares of polynomials.

## Wednesday, September 23, 2009

### Polynomial not sum of squares

Labels:
Algebra,
AM-GM,
definite positive,
polynomial,
Solved,
sum of squares

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a) By AM-GM inequality, $x^2y^4+x^4y^2+1\ge3x^2y^2$. So $x^2y^4+x^4y^2+1-3x^2y^2\ge0$.

ReplyDeleteb) Assume $P(x,y)=\sum (P_i(x,y))^2$. Each of $P_i(x,y)$ has degree at most 3, otherwise the degree of $P(x,y)$ would be greater than 6. So $P_i(x,y)$ is of the form $a_i+b_ix+c_iy+d_ix^2+e_iy^2+f_ixy+g_ix^2y+h_ixy^2+i_ix^3+j_iy^3$. But $x^3,y^3,x^2,y^2,x,y$ cannot appear, otherwise $x^6,y^6,x^4,y^4,x^2,y^2$ would appear in $P(x,y)$. Therefore $P_i(x,y)=a_i+f_ixy+g_ix^2y+h_ixy^2$. This implies $\sum f_i^2=-3$, a contradiction.

Johan Gunardi's solution is same as mine, so I will consider this problem solved.

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