Is it true that all numbers of the form $\sqrt[b_1]{a_1}+\sqrt[b_2]{a_2}+\cdots+\sqrt[b_k]{a_k}$ are not integers, where $a_i,b_i$ are positive integers such that none of the $k$ terms is an integer? I think this must be a theorem somewhere, probably a fundamental theorem in number theory, but not sure what it would be called.
3
-
1$\begingroup$ I'm pretty sure you want to require $a_i , b_i > 1$. $\endgroup$ – Eric Towers Nov 19 '15 at 20:29
-
3$\begingroup$ @EricTowers I required that none of the $k$ terms is an integer. $\endgroup$ – Alexi Nov 19 '15 at 20:32
2
$\begingroup$
$\endgroup$
Note that we may allow $a_i \in\Bbb{Q}_{>0} \setminus \{1\}$ by multiplying through by the least common denominator of the $a_i$.
Let $t_i = \sqrt[b_i]{a_i}$ and suppose $\sum t_i \in \Bbb{Q}$. Do you believe $\Bbb{Q}(t_1, \dots, t_k) = \Bbb{Q}(t_1 + \cdots + t_k)$? If so, $\Bbb{Q}(t_1, \dots, t_k) = \Bbb{Q}$, and every $t_i \in \Bbb{Q}$.
The fact I've used is 5.2(2) at Albu, T., Cogalois Theory: An Outline.
