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.)
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.