# Is this set of numbers independent/Sidon?

Given a subset $$E\subset \mathbb N\setminus \{0\}$$ of integers, we say that $$E$$ is independent if for any choice of $$k\in \mathbb N$$, $$\epsilon_i\in\{-1,0,1\}$$ for $$i=1,\dots,k$$, if $$\Sigma_{i=1}^k\epsilon_in_i=0$$ for $$n_i\in E$$, then $$\epsilon_i=0$$ for any $$i$$.

My Question is the following. Is the set $$\{5^n3^m\}_{n,m\in \mathbb N}$$ a finite union of independent sets? (one can take any two primes $$p,q\neq 2$$.

Moreover, is this set a Sidon set? (Sidon sets are the following - the definition from: Rudin, Walter, Trigonometric series with gaps, J. Math. Mech. 9, 203-227 (1960); DOI: 10.1512/iumj.1960.9.59013, jstor. ZBL0091.05802, MR116177.)

(b) A set $$E$$ is said to be a Sidon set if there is a finite constant $$B$$ such that $$\sum_{-\infty}^\infty |\hat{f}(n)| \le B \|f\|_\infty$$ for every $$E$$-polynomial $$f$$.

where $$\hat{f}(n)$$ is the $$n$$'th Fourier coefficient of $$f$$.

Thanks.

• I had just worked out that $\{2^n3^m\}$ is not independent (since $3+6+9 = 18$) when you edited the question. May 30, 2021 at 17:01
• Also $1+3+5-3^2=0$. May 30, 2021 at 17:02
• Thanks. I am modifying the question a bit more now, to so that it will ask whether this set is Sidon. May 30, 2021 at 17:06

It was proved by Pisier (Arithmetic characterizations of Sidon sets) that a set $$\Lambda \subset \mathbf{Z}$$ is Sidon if and only if there is a constant $$c$$ such that any finite subset $$A \subset \Lambda$$ contains an independent set $$B$$ of cardinality $$\ge c|A|$$. Together with Fedor Petrov's answer, this implies that, in a Sidon set, the number of elements in $$[0,x]$$ is $$O(\log x)$$. In particular, $$\{3^n 5^m\}_{n,m}$$ is not Sidon.

No, it is not.

By pigeonhole principle, an independent set $$E$$ has $$O(\log x)$$ elements not exceeding $$x$$: otherwise two subsets have equal sums (since there are $$2^{|E\cap [1,x]|}$$ subsets, and only at most $$x^2$$ possible sums), and the difference of these two sums gives a vanishing linear combination of elements of $$E$$ with coefficients $$0,1,-1$$.

Thus the same holds for a finite union of independent sets. But the set $$\{5^n3^m\}$$ has at least $${\rm const}\cdot (\log x)^2$$ elements not exceeding $$x$$ (choose arbitrary $$n,m$$ between 1 and $$\log x/10000$$).

• Thanks. Does it mean also that the set is not Sidon? by what I understand, Sidon sets are finite unions of something called "$k$ independent sets", where one only requires that there are no trivial sums of $k$ elements (I believe this is a result by Bourgain). In this case the number of subsets of size $k$ is considerably smaller than 2^{\abs{E\cap[1,x]}}, so the set may still be Sidon May 31, 2021 at 13:31

The set has a Sidon-like property. There is a constant $$s,$$ such that any difference $$d$$ appears at most $$s$$ times, i.e. the number of solutions of $$x+y=d$$ is at most $$s,$$ where $$x,y$$ are from your set. It would be nice to find a simple proof of this.

Also, the number of solutions of $$\Sigma_{i=1}^k\epsilon_in_i=0$$ in coprime $$n_i$$ elements is bounded by a constant $$s_k$$ depending on $$k$$ only. You can find references in the classical paper by Erdős, P.; Stewart, C. L.; Tijdeman, R.: Some Diophantine equations with many solutions. Compositio Math. 66 (1988), no. 1, 37-56. https://users.renyi.hu/~p_erdos/1988-31.pdf and in the more recent paper of Kálmán Győry https://doi.org/10.48550/arXiv.1901.11289