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$\textbf{Question:}$Is there an infinite set of positive integers such that no matter how we choose some elements of this set, their sum is not a perfect power?

$\textbf{My progress:}$I thought of solving an easier version,sum not a perfect $k-th$ power for fixed $k$.

In that case I came up with, $p^{d}$ where $d$ ranges over all positive integers not divisible by $k$ and $p$ some prime.But this idea didn't seem good enough to generalize that is I could do for any finite amount of numbers $k$ but not for all $k$ at the same time.

$\textbf{Edit:}$ I used the above idea.The construction is just: $p^dq^{d+1}$ for two different primes $p$ and $q$ and $d$ ranges over all positive integers.

Thanks to everyone who helped.

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    $\begingroup$ Hint 1: Fermat numbers F. Luca, Infinite sets of positive integers whose sums are free of powers, Rev. Colomb. Matem. 36 (2002), 67–70. revistas.unal.edu.co/index.php/recolma/article/view/33829 Hint 2: Perfect powers are pretty sparse. Dubickas, Artūras, and Šarka, Paulius. "Infinite sets of integers whose distinct elements do not sum to a power.." Journal of Integer Sequences [electronic only] 9.4 (2006) eudml.org/doc/53352 $\endgroup$ – Chris Culter Jul 16 '20 at 5:12
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For each positive integer $m$, let $a_m=2^{m}3^{m+1}$, and let $A=\{a_1,a_2,a_3,...\}$.

Claim:$\;$No nonempty finite subset of $A$ sums to a perfect power (with positive integer exponent greater than $1$) of a positive integer.

Proof:

Suppose instead that there exists a nonempty finite subset $B=\{b_1,...,b_k\}$ of $A$ such that $$b_1+\cdots+b_k=x^y$$ for some positive integers $x,y$, with $y > 1$.

Our goal is to derive a contradiction.

Without loss of generality, assume the sequence $b_1,...,b_k$ is increasing.

Let $n$ be such that $b_1=a_n=2^{n}3^{n+1}$.

Since $2{\,\mid\,}a_m$ for all $m$, it follows that $2{\,\mid\,}x$, and since $3{\,\mid\,}a_m$ for all $m$, it follows that $3{\,\mid\,}x$.

Let $r$ be such that $2^r{\,{\mid}{\mid}\,}x$, and let $s$ be such that $3^s{\,{\mid}{\mid}\,}x$.

By unique factorization, it follows that $2^{ry}{\,{\mid}{\mid}\,}x^y$ and $3^{sy}{\,{\mid}{\mid}\,}x^y$.

Since all terms of the sequence $b_1,...,b_k$ other than $b_1$ are divisible by $2^{n+1}3^{n+2}$, it follows that $2^n{\,{\mid}{\mid}\,}(b_1+\cdots +b_k)$ and $3^{n+1}{\,{\mid}{\mid}\,}(b_1+\cdots +b_k)$.

But then we must have $n=ry$ and $n+1=sy$,$\;$so $y$ is a common factor of $n$ and $n+1$, contradiction, since $y > 1$ and $\gcd(n,n+1)=1$.

This completes the proof.

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