Would you like to prove this fact about polynomials?

Question

Let P be a polynomial with real coefficients such that P(x) >= 0 for every real x. Show that there are polynomials R and S such that P(x) = R(x)^2 + S(x)^2 for every complex x.

Answer 1

you can write
P=d product((x-a_i)^k_i)) poduct((x-b_j)^2 + c_j^2)
Then I give you 3 hints:
1) d must be positive
2) all k_i are even numbers
3) use the formula (a^2+b^2)(c^2+d^2) =
= (ac-bd)^2 + (ad+bc)^2

Answer 2

When complex numbers are equal it means that their real parts are equal and their imaginary parts are equal. Hmm, p>=0 for every real x means that all real roots must be repeated roots because that's the only way that p can equal 0 and not be negative in a small neighborhood around it. Repeated real roots maybe can be factored. Or maybe the p has complex roots!