# Polynomial satisfying a functional equation [closed]

I am currently stuck with the following question:

Let $$q$$ be a polynomial of degree $$n+1$$ with distinct positive zeros $$x_0, ... , x_n$$. Find a polynomial $$p \in P_n$$ that satisfies the functional equation $$q(x)p(x) + q(-x)p(-x) = 1$$ for all $$x$$ in $$\Re$$. Is such a polynomial unique?

Would appreciate your kind explanations! Thank you!

• You can solve the problem by looking at the smallest (or largest coefficient first) and then working your way up (or down). For example, by setting $x=0$ you get $p(0) = 1/2q(0)$. So you can try to work the coefficients of $p$ one by one. For an abstract argument, consider $a_i$ to be the coefficients of $p$ (you have $n+1$ of them). Then $F(x) = q(x)p(x)$ has degree $2n+1$. When looking at $F(x) + F(-x)$ all monomials of odd degree cancel. There remains $n+1$ even degree monomials. So $n+1$ linear constraints and $n+1$ variables. There is a solution (use distinct roots for uniqueness). – ARG May 9 at 10:55

Nice exercise, though I doubt that it can be considered as a research question. Put $$y=x^2$$, and write $$q(x)= a(y)+xb(y)$$. I claim that the polynomials $$a$$ and $$b$$ are coprime: if a non-constant polynomial $$f$$ divides $$a$$ and $$b$$, then $$f(x^2)$$ divides $$q(x)$$, but that implies that some of the roots $$x_i$$ of $$q$$ are negative. Note also that since the roots are positive $$y$$ does not divide $$a$$. Therefore there exist polynomials $$c,d$$ such that $$a(y)c(y)+yb(y)d(y)=\dfrac{1}{2}$$. Put $$p(x)=c(x^2)+xd(x^2)$$; then $$q(x)p(x)=\dfrac{1}{2}+ x g(x^2)$$, with $$g= ad+bc$$, so $$q$$ answers the question.
As usual you can replace $$c(y),d(y)$$ by $$c(y)-yb(y)e(y), d(y)+ a(y)e(y)$$ for some polynomial $$e$$, so the solution is far from unique.
• I don't think so. For one thing, there are many solutions for $p$, with different zeroes. – abx May 7 at 15:20