My friend and I were discussing this, and while it looks obviously true, weren't very successful in getting anywhere with it. I'm not sure whether I lack the tools to solve this or whether I am missing something obvious. Any insight would be much appreciated.
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3Try proving the contrapositive. – Thomas Davis May 31 '18 at 23:56
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2This was a Putnam 1971 problem. Here is a solution: https://mks.mff.cuni.cz/kalva/putnam/psoln/psol716.html (Posting as a comment as the dupe answer mentions putnam 1971) – Aryabhata Jun 01 '18 at 02:43
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Partial solution: if $s$ is not an integer, then it has to be transcendental.
It cannot be a rational number. Otherwise say $s={p\over q}$ with $q>1,\gcd(p,q)=1$, and $N=2^s\in\Bbb N$ $$q\ |\ v_2(N^q)=v_2(2^{p})=p$$ which contradicts $q>1,\gcd(p,q)=1$. Here $v_2$ stands for the $2$-adic order function.
Now since $s$ is not a rational, it follows from Gelfond-Schneider theorem that it cannot be an algebraic number either. Otherwise $2^s$ would be transcendental.
One can observe that so far only the fact that $2^s\in\Bbb N$ has been used. This condition alone is not sufficient to rule out transcendental numbers as was pointed out by Grant B., e.g. $2^{\log_2(3)}=3$.
Arnaud Mortier
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1$2^s\in\mathbb{N}$ is not sufficient, as $s=\log_2(3)$ is a counterexample. However, $2^s\in\mathbb{N}$ and $3^s\in\mathbb{N}$ together should be sufficient. – Grant B. Jun 01 '18 at 01:26
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