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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.

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    $\begingroup$ $x^k$ satisfies this equation only if $a=0$ $\endgroup$
    – Martund
    Oct 5 '19 at 7:21
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    $\begingroup$ Yes, so apparently there are no solutions when $a\ne0$—this is part of what needs to be proved. $\endgroup$ Oct 5 '19 at 7:28
  • $\begingroup$ 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 } $. $\endgroup$ Aug 23 '21 at 8:48
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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)?

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  • $\begingroup$ Can you explain what you mean by complex zero $\endgroup$ Oct 5 '19 at 7:41
  • $\begingroup$ @AaronyJamesys A "zero", or "root", of a function $g$ is a value $c$ for which $g(c)=0$. $\endgroup$
    – J.G.
    Oct 5 '19 at 7:49
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    $\begingroup$ > If q(z) is nonconstant, then q(z^2) has at least one complex zero that q(z) doesn't If q(z) = z, then it has a zero only at z = 0. This remains true for q(z^2). $\endgroup$
    – wnoise
    Oct 5 '19 at 8:37

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