If a univariate polynomial is greater than another, is their difference a square?

Question

Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) \ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?

Answer 1

Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $x\in\mathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.

Answer 2

There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.

Answer 3

I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$

so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,

if that's what you mean..

Answer 4

Take your favorite positive polynomial $f$ which is not a square, $2f>f$ gives an immediate counterexample.