Rigorously prove that $f(x) = x^k$ is the only rational function that satisfies a given equation.

Here is the problem.

A nonconstant rational function over the real numbers(a rational function is a function that can be expressed as $$\frac{p(x)}{q(x)}$$, with $$p(x), q(x)$$ as polynomial function) $$f(x)$$, is defined such that:

$$(f(x))^2 - a = f(x^2)$$ for all $$x$$ ( $$a$$ is a constant value)

Prove that $$f(x)$$ must be of the form $$x^k$$ for some constant $$k$$.

I have an idea of how to prove it, but I can't do so rigorously. I can prove that any polynomial with more than $$2$$ terms, or monomials, would not work. But I can't prove it won't work for any rational function.

• $x^k$ satisfies this equation only if $a=0$ Oct 5 '19 at 7:21
• Yes, so apparently there are no solutions when $a\ne0$—this is part of what needs to be proved. Oct 5 '19 at 7:28
• This is not true. For any positive integer $m$, $f ( x ) = x ^ m + x ^ { - m }$ gives a solution when $a = 2$. But these are the only additional ones. You can use the argument given in @GregMartin's answer to show that $f ( x )$ must be of the form $\frac { p ( x ) } { x ^ m }$ for some polynomial function $p$. After that, you can show that if $m = 0$ then $a = 0$ and $p ( x ) = x ^ n$ for some positive integer $n$, and if $m > 0$ then either $a = 0$ and $p ( x ) = 1$ or $a = 2$ and $p ( x ) = 1 + x ^ { 2 m }$. Aug 23 '21 at 8:48

If $$q(z)$$ is nonconstant, then $$q(z^2)$$ has at least one complex zero that $$q(z)$$ doesn't have. (Choose, for example, the zero $$z_1$$ of $$q(z^2)$$ with minimal nonzero argument.) This means that $$q(z)$$ must be constant, for otherwise the right-hand side is undefined at a point where the left-hand side is defined. This reduces to the case where $$f$$ is a polynomial, which you have (almost?) solved.
Edit: wnoise points out that this argument fails if $$q(z)$$ is a power of $$z$$. Perhaps the argument for polynomials extends to Laurent polynomials (where negative powers of $$z$$ are allowed)?
• @AaronyJamesys A "zero", or "root", of a function $g$ is a value $c$ for which $g(c)=0$.