Besicovitch proved in 1940 in the paper 'On the linear independence of fractional powers of integers' https://doi.org/10.1112/jlms/s1-15.1.3 that if $\alpha$ is a non-integer rational number then $S_\alpha=\{n^\alpha:n \text{ is a squarefree positive integer}\}$ is linearly independent over the rational.
It is an open problem whether $S_\alpha$ is linearly independent over the rational or not when $\alpha=\sqrt{2}$. I was wondering if we can at least say that there are 'very few' relations within the terms of $S_{\sqrt{2}}$ in the following sense:
There exists a constant $c>\frac{3}{4}$ such that for every (large) positive integers $R$ and $K$, we have the following:
$\text{card} \Big\{\big\{\displaystyle{\sum_{n=1}^{K} a_n s_n}\big\}: 0\leq a_n \leq R \text{ are integers} \Big\} \geq c (R+1)^K$, where $\{x\}$ denotes the fractional part of $x$, and $(s_n)$ is the sequence obtained by arranging the terms of $S_{\sqrt{2}}$ in an increasing order.
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1$\begingroup$ The claim that you attribute to Besicovitch is false. For example, if $\alpha=1/2$, then $1^\alpha$ and $4^\alpha$ are linearly dependent over $\mathbb{Q}$. $\endgroup$– GH from MOCommented Oct 3 at 21:38
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1$\begingroup$ The linked paper discusses linear independence of $a^{r_1},\ldots,a^{r_s}$ for fractional $r_i$. $\endgroup$– Andrei SmolenskyCommented Oct 3 at 21:42
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$\begingroup$ Thanks @GHfromMO for the comment. I updated the statement of Besicovitch's result and the questions as well. $\endgroup$– Sovanlal MondalCommented Oct 3 at 21:57
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2$\begingroup$ The linear independence is definitely not true for all irrational $\alpha$. For instance there is one irrational solution to $2^x+3^x=6^x$. $\endgroup$– WojowuCommented Oct 3 at 22:01
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3$\begingroup$ You are right. In general, the linear independence is not be true for all $\alpha$. However, it is known to be true for all but countably many $\alpha$. However, What I am interested in is that for a fixed $\alpha$, can we find a S of integers which has positive density and $\{s^\alpha: s\in S\}$ is linearly independent over $\mathbb{Q}$. I guess it is worth clarifying in the question. @Wojowu $\endgroup$– Sovanlal MondalCommented Oct 3 at 22:54
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