# Exponentiating a 'polynomial' with non-negative real powers to produce a polynomial with integer powers

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.