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About one hour ago I asked a question which at first sight looked non-trivial to me but it is really trivial. Shame on me, whether I want it or not.

Now I have, solely for fun, another question which is slightly modified and maybe little less trivial.

Let $S_2$ be the set of square roots of all prime numbers, $S_3$ the set of cube roots of all prime numbers, ... , $S_n$ the set of $n$-th roots of all prime numbers, ...

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

The question is:

Does there exist natural number $k$ and rational numbers $r_1,r_2,...,r_k$ (all different from zero) such that there are k numbers from the set $S$, let us denote them as $s_1,s_2,...,s_k$ such that we have $\sum_{i=1}^{k}r_is_i=0$

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

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A. P.
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    See http://math.stackexchange.com/a/30707/30402 – Erick Wong Jun 20 '15 at 01:21
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    @ErickWong Thank you. It looks that Mordell proved a more general result in this paper. So it would be nice that you post this paper in an answer to the question so that we have more answers on this site that are accepted. I think that it is not fair that I post the answer because you pointed me in the right direction. – A. P. Jun 20 '15 at 11:33
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    @AnteP.: It would be very fair to post an Answer to your own Question. Credit Erick's comment for pointing you in the right direction, but you've given the problem and Mordell's paper a lot of thought, and I'm sure you can do a good job of explaining for the benefit of future Readers! – hardmath Jun 22 '15 at 15:33

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Oh well, I will answer the question just because it seems to me that it is better to have as many as possible questions on the site that are answered. I would like to thank Erick Wong who pointed me in the right direction with his comment.

In this paper Louis Joel Mordell proves much more general result in a theorem which he states in this way:

A polynomial $P(x_1,x_2,...,x_s)$ with coefficients in $K$ and of degrees in $x_1,x_2,...,x_s$ less than $n_1,n_2,...,n_s$, respectively, can vanish only if all its coefficients vanish provided that the algebraic number field $K$ is such that there exists no relation of the form $x_1^{m_1}x_2^{m_2}\cdot \cdot \cdot x_s^{m_s}=a$, where $a$ is a number in $K$, unless $m_1 \equiv 0 \pmod {n_1}, m_2 \equiv 0 \pmod {n_2},..., m_s \equiv 0 \pmod {n_s}$

A. P.
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This is not really an answer, only a hint, but may be too long to fit as a comment.

Assume such a relation exists. Let us assume $m$th roots are the highest appearing in this relation (i.e. no $n$th roots with $n>m$ appear in this relation).

So this means we have primes numbers $p_1,p_2,\ldots, p_r$ and integers $a_1,a_2,\ldots,a_r$ such that the algebraic number $\alpha$ defined by $$\alpha=a_1\sqrt[m]{p_1}+a_2\sqrt[m]{p_2}+\cdots+a_r\sqrt[m]{p_r}$$ belongs to the field generated by lower radicals.

My guess is $\alpha$ is of degree $m^r$. And this relation will lead to a conclusion that $m^r$ divides a number factorizable as a product of numbers less than $m$. I am stuck there.

P Vanchinathan
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  • Thank you for your effort. It looks that here we have an answer to the question that is more general than mine. – A. P. Jun 20 '15 at 11:36