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Find an infinite set of positive integers $S$ such that them sum of the elements of any finite subset of $S$ is not a perfect square.

I've had one person tell me that $\{3\cdot 4^k : k\in \mathbb{N}\}$ works, but I don't see why. What are some other sets that work? Why does this one work?

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    $\begingroup$ $\{2^{2k+1} : k\in \mathbb{N}\}$ would be another simple example $\endgroup$ – Sil Mar 31 at 16:04
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Let $a^2=\sum_{i=1}^n3*4^{k_i}$ with $k_i<k_{i+1}$, then dividing by $4^{k_1}$ we get ( setting $\frac{a}{2^{k_1}}=u$ ): $u^2=\sum_{i=1}^n3*4^{k_i-k_1}=3+\sum_{i=2}^n3*4^{k_i-k_1}$, and thus $u^2=3\mod{}4$ which is a contradiction since $u^2=0$ or $1\mod{}4$.

In general you can set $A=\{ab^i,i\in\mathbb{Z}\}$ where $a$ is not a quadratic residue $\mod{}b$ or equivalently $a\ne{}x^2\mod{}b$ for every $x\in{}\mathbb{Z/}b\mathbb{Z}$.

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