Sum of two irrational radicals is irrational?

Solution 1:

Let $a'$, $b'$, $m'$ and $n'$ be positive integers such that $\sqrt[m']{a'}$ and $\sqrt[n']{b'}$ are irrational. Let $a$, $b$, $m$ and $n$ be the minimal positive integers such that $\sqrt[m]{a}=\sqrt[m']{a'}$ and $\sqrt[n]{b}=\sqrt[n']{b'}$, so that their minimal polynomials are $f_a=X^m-a$ and $f_b=X^n-b$, respectively. Note that $m,n>1$ as $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational.

Suppose $\sqrt[m]{a}+\sqrt[n]{b}=q\in\Bbb{Q}$. Then $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are roots of $f_b(q-X)$ and $f_a(q-X)$, respectively, which shows that $f_a$ divides $f_b(q-X)$ and $f_b$ divides $f_a(q-X)$, respectively. In particular we see that $m\leq n$ and $n\leq m$, so $n=m$, and hence $f_a=cf_b(q-X)$ for some nonzero constant $c\in\Bbb{Q}$. Then $$X^m-a=f_a=cf_b(q-X)=c(q-X)^n-cb=c(q-X)^m-cb,$$ which immediately shows that $q=0$ because $m,n>1$. It follows that $c=\pm1$ and $a=cb$. Because $a$ and $b$ is positive it follows that $c=1$ and so $$\sqrt[m]{a}+\sqrt[n]{b}=2\sqrt[m]{a}=2\sqrt[n]{b},$$ which is irrational because $\sqrt[n]{b}$ is, by assumption.