I will post later .

Standard result. note the reduction to the case m=n and then the result follows from a problem in PFTB.

A solution straight from the straight from the book. Let $\alpha^{\frac 1n}+\beta^{\frac 1m}=k$ where $k$ is rational Look at the polynomials $$p(x)=x^n-\alpha$$, $$q(x)=(k-x)^m-\beta$$. These are polynomials over rationals and have $(x-\alpha^{\frac{1}{n}})$ as a common root, which is easily seen to be of single multiplicity. Also if $\gamma \neq \alpha$ is another common root then $\gamma=\omega_n \alpha^{\frac{1}{n}}$ where $\omega$ is a nth root of unity. But then $$\omega_m\beta^{\frac{1}{m}}=(k-\omega_n \alpha^{\frac{1}{n}})$$and by comparing the real parts, we get that both $\omega_m,\omega_n$ have real part equal to $1$, a contradiction. Hence $(x-\alpha^{\frac{1}{n}})$ is only common root , and hence the gcd of $p(x),q(x)$. Since these polynomials are rational, their gcd is also rational (consider euclidian algorithm to get the gcd). Hence $\alpha^{\frac{1}{n}}$ is rational and we are done. Thank you SFTB for saving my fate in the TST

This was actually probably a trivial problem Click to reveal hidden text

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Can I ask you what book is PFTB? Im seeing it a lot and I dont know it

godfjock wrote: Can I ask you what book is PFTB? Im seeing it a lot and I dont know it I too was faced with the same question and found out that it is : Problems from the book

kapilpavase wrote: Also if $\gamma \neq \alpha$ is another common root then $\gamma=\omega_n \alpha^{\frac{1}{n}}$ where $\omega$ is a nth root of unity. Why?

Because equation $$x^n=\alpha\wedge x\in\mathbb C$$is equivalent to $$x\in\left\lbrace \sqrt[n]{\alpha}\cdot\left(\cos\frac{2k\pi}{n}+\sin\frac{2k\pi}{n}\right):\ k\in\lbrace 1,2,...,n\rbrace\right\rbrace$$