melese said:
(HUN,1979) Prove the following statement: If a polynomial $f(x)$ with real coefficients takes only nonnegative values, then there exists a positive integer $n$ and polynomials $g_1(x),g_2(x),...,g_n(x)$ such that $f(x)=g_1(x)^2+g_2(x)^2+\cdots+g_n(x)^2$.
A related question of my own, but I don't have/know the answer:
If $\deg({f})=d$, then what is the smallest possible value of $n$.
For example: I know that it's $2$, when $d=2$ and $1$ when $d=0$.
መለሰ
I thought of this in terms of linear algebra and vector spaces. Consider the set
\begin{matrix}
g_0(x)=\sqrt{c_0} & g_0^2(x)=c_0 \\
g_1(x)=(1+c_1x) & g_1^2(x)=1+2c_1x+c_1^2x^2\\
g_2(x)=\sqrt{c_2}x & g_2(x)=c_1x^2\\
g_3(x)=(x+\sqrt{c_3}x^2) & g_3(x)=x^2+c_3x^3+c_3^2x^4\\
g_4(x)=\sqrt{c_4}x^2) & g_4(x)=c_4x^4 \\
g_5(x)=(x^2+\sqrt{c_5}x^3 & g_5(x)=x^4+c_5x^5+c_5^2x^6 \\
\vdots & \vdots \\
\end{matrix}
This gives a matrix that looks like this
\begin{bmatrix}
c_0 & 0 & 0 & 0 & 0 & \dots \\
1 & 2c_1 & c_1^2 &0 & 0 & \dots \\
0 & 0 & c_2 & 0 & 0 & \dots \\
0 & 0& 1 & 2c_3 & c_3^2 & \dots \\
\vdots &\vdots &\vdots &\vdots &\vdots & \ddots \\
\end{bmatrix}
As long as the number of rows $$ k$$ is odd the set will be linearly independent and thus span polynomial space of degree $$k-1$$
Just to be clear now we can pick [math]c_0,c_1,c_2,...[/math] so that we can get any polynomail
For example if $$f(x)=3x^2-8x+10$$, then
We solve
\begin{matrix}
c_0+1=10 \\
2c_1=-8 \\
c_1^2+c_2=3
\end{matrix}
This gives
$$c_0=9, \quad c_1=-4, \quad c_2=-13$$
and $$9g_0^2-4g_1^2-13g_2=9+1+2(-4)x+(-4)^2x-13x^2=3x^2-8x+10$$
I think this will help settle your other question as well.