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I was reading some question on this site and stream of thought led me to the creation of another question that could be trivial for someone but I am unable even to start solving it. I wanna share this question with you in hope that someone will be able to answer it.

Let $S_2$ be the set of square roots of all positive integers that are not of the form $a^2$ , $S_3$ the set of cube roots of all positive integers that are not of the form $a^3$, $...$ , $S_n$ the set of $n$-th roots of all positive integers that are not of the form $a^n$...

Now let us define set $S$ as $S=\bigcup_{n=2}^\infty S_n$.

The question is:

Does there exist a natural number $k$ and rational numbers $r_1,r_2,...,r_k$ (all different from zero) such that for some $k$ different elements of the set $S$, denote them as $s_1,s_2,...,s_k$, which are not in the same set of roots, for instance if $s_i$ is in the $S_4$ then $s_j$ is in the $S=(\bigcup_{n=2}^\infty S_n) \setminus S_4$, we have $\sum_{i=1}^{k}r_is_i=0$?

(All roots in this question are unique real roots.)

A. P.
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  • You are asking for something much stronger than a linear dependence relation (which certainly exists) if by "some $k$ elements" of the set $S$ you require the first $k$ elements. However the weaker version is almost trivial, since $2\sqrt{3} - \sqrt{12} = 0$. – hardmath Jun 19 '15 at 22:40
  • @hardmath There is additional assumption now, that those roots are in different constituent sets of the set $S$, that is more interesting version, to avoid trivialities as in your example. – A. P. Jun 19 '15 at 22:53
  • So now all $k$ roots must be of a different exponent. – A. P. Jun 19 '15 at 22:54
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    That refinement does not really avoid the near-trivialities, since $\sqrt{3} = \sqrt[4]{9}$. – hardmath Jun 19 '15 at 22:58
  • @hardmath Those are not different elements of the set $S$, they are in different constituent sets but they do not satisfy requirement that they are different elements of $S$ so that is not valid counterexample. – A. P. Jun 19 '15 at 23:03
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    What I'm saying is that $2\sqrt[4]{9} - \sqrt{12} = 0$ is still a dependence relation, conforming to your new requirement that the roots belong to different "index" constituent sets, $S_4$ and $S_2$ in this case. – hardmath Jun 19 '15 at 23:09
  • @hardmath I would like that you post equation $2\sqrt[4]{9} - \sqrt{12} = 0$ as an answer, I will accept it, if answers are allowed to be so short. – A. P. Jun 19 '15 at 23:11
  • Not equation, that is identity. – A. P. Jun 19 '15 at 23:13
  • How about modifying the question so the answer is affirmative and non-trivial. – orangeskid Jun 19 '15 at 23:18
  • @orangeskid I was writing a comment to hardmath of the type "may I edit the question so many times until you no more have counterexamples" but did not add it, I think that I can do it with just one more edit, haha! If he wants I will do it. – A. P. Jun 19 '15 at 23:20
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    @Ante P.: Yeah, you can do this: fix an order $N$ for the roots. Now consider distinct natural numbers that are ($N$-th power )-free. Their $N$-th roots will be linearly independent over $\mathbb{Q}$. True, and takes some care to prove. It's a good start for a more general result. – orangeskid Jun 19 '15 at 23:26
  • @orangeskid Interesting, I actually had in mind a version that has roots of every order but is also (at least it looks to me) non-trivial. – A. P. Jun 19 '15 at 23:31
  • @Ante P. : I would prove first for $N=2$, square roots. Even this one is interesting already. In the end, this statement with general $N$ might probably imply any valid one for different orders, since you can bring them to a common order. – orangeskid Jun 19 '15 at 23:34
  • @orangeskid Look – A. P. Jun 19 '15 at 23:52

1 Answers1

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We can devise examples, for arbitrarily large $k$, where the roots $s_i$ are drawn from different constituent classes, so that $\sum_{i=1}^k r_i s_i = 0$ with all coefficients $r_i$ nonzero integers.

A simple example with just two terms is $2\sqrt[4]{9} - \sqrt{12} = 0$.

hardmath
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