The following is something one of my teachers and I discussed but did not make any progress on, along with my own generalizations.

Let $p(x)=x^\alpha$ with $\alpha \in \mathbb{R}^+$. If we let $q(x) = (p(x))^\frac{a}{\alpha}$ with $a \in \mathbb{N}$, we have a polynomial of degree $a$.

Now let $p(x)=x^{\alpha_1} + x^{\alpha_0}$, $\alpha_1 > \alpha_0 > 0$. Does there exist a function $f(a,b,c)$ such that $q(x)=(p(x))^{f(a,\alpha_0,\alpha_1)}$ is a polynomial of degree $a$ with only integer powers? I have been able to solve cases for rational exponents which leads me to believe that such a function exists for rational $\alpha_0,\alpha_1$ (and possibly, that it depends only on $a$ and $\alpha_1$) but have been unable to find real power $\beta > 0$ such that $q(x) = (p(x))^{\beta}$ is a polynomial for non-trivial irrational $\alpha_1$ or $\alpha_0$, leading me to believe no such function exists when either power is irrational.

My question: Does there exist a function $f:\mathbb{R}^3 \to \mathbb{R}$ such that $(x^{\alpha_1} + x^{\alpha_0})^{f(a,\alpha_0,\alpha_1)}$ is a polynomial of degree $a$? How might I go about finding it, or disproving its existence? If no function exists for all powers, what are the (preferably non-trivial, because I can think of many) conditions the powers need to satisfy for one to exist? Further, might a similar function exist for $p(x) = x^{\alpha_k} + x^{\alpha_{k-1}} \dots x^{\alpha_{0}}$, with similar restrictions on $\alpha_k, \alpha_{k-1} \dots \alpha_0$?

Edit 1: quick afterthought, can this be proven via induction, where one inducts on the number of terms? I don't think so, but never a bad idea to try.

Edit 2: Perhaps one can forget about the function and simply seek to prove: given $p(x)=x^{\alpha_{k}} + x^{\alpha_{k-1}} \dots + x^{\alpha_{0}}$ with $\alpha_{k} > \alpha_{k-1} \dots > \alpha_{0} > 0$, for any $a \in \mathbb{N^+}$ there exists a $\beta \in \mathbb{R}$ ($\mathbb{C}$?) such that $(p(x))^\beta = q(x)$ where $q(x)$ is a polynomial of degree $a$. One can possibly induct over $k$ assuming that for all such $p(x)$ with more than $0$ and no more than $k$ terms that our proposition holds. The case for the monomial is trivial, and some clever algebra might help to prove the case for the 'polynomial' with $k+1$ terms using our assumption.


No, there is no such function for irrational $\alpha$ that will work for non-monomial $p$. One way to prove it is to find the Taylor series. The outline of this method is:

Consider $q(x) = (x^a + x^b)^c$, where $a$ and the ratio $b/a$ are irrational (pardon my simplification of the notation to avoid having to do a lot of greek letters and subscripts). Since we want $q$ to be well-defined where we are taking the Taylor series, we'll do it at $x = 1$.

  • $q(1) = 2$
  • $q'(x) = c(x^a + x^b)^{c-1}(ax^{a-1} + bx^{b-1})$, so $q'(1) = c2^{c-1}(a+b)$
  • $q''(x) = c(c-1)(x^a + x^b)^{c-2}(ax^{a-1} + bx^{b-1}) + c(x^a + x^b)^{c-1}(a(a-1)x^{a-2} + b(b-1)x^{b-2}$, so $q''(1) = c(c-1)2^{c-2}(a+b) + c2^{c-1}(a(a-1) + b(b-1))$
  • ...

This should be enough to see what happens: If $a$ is not an integer, then the progression $a(a-1)...$ will never be $0$, and while it takes some number theory to fully demonstrate it, if $a$ is irrational, the remaining terms will never cancel it out.

The Taylor series of a polynomial is the polynomial itself, which terminates (and thus is guaranteed to converge to the polynomial). Since $q$ has a non-terminating Taylor series, it cannot be a polynomial.

Of course, where the going got tough here, I just waved my hands. But this is the idea.


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