# Finding an infinite sequence of natural numbers for which any finite partial sum avoids being a perfect power.

$$\textbf{Question:}$$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.

• 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 – Chris Culter Jul 16 '20 at 5:12

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.