Let $n \geq 1$ and consider the (Mersenne) number $M_n = 2^n-1$. Is it possible that $M_n = p^k$ for some prime $p$ and some (necessarily odd) $k > 1$? Thanks in advance.
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Why $k$ must be odd? – ajotatxe Jun 28 '14 at 20:40
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1Because for even $2^{2n}-1=(2^n-1)(2^n+1)$ and $2^n-1$ and $2^n+1$ are coprime. – agha Jun 28 '14 at 20:47
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1Yes but also because, if $k$ was even, then $p^k \equiv 1 \pmod{4}$ contrary to $2^n-1 \equiv 3 \pmod{4}$ for $n \geq 2$. – user152634 Jun 28 '14 at 20:50
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Similar to https://math.stackexchange.com/q/2352131/83175. – chux Jan 02 '21 at 22:50
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For an elementary solution, see this question. Some related questions around consecutive powers (provided with elementary solutions) are discussed in this conference paper. – Wembley Inter Feb 14 '25 at 13:12
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It is not possible. Catalan's conjecture (proved in 2002) states that the only solution of $x^a-y^b=1$ ($x$ and $y$ are positive integers and $a$ and $b$ are integers greater than $1$) is $3^2-2^3=1$.
If a Mersenne's number were the power of a prime, we'd have $$2^n-1=p^k$$ or $$2^n-p^k=1$$ contradicting Catalan's conjecture.
ajotatxe
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This is very nice: to the point it sounds a bit over the top :). Thanks a lot; I was not even aware of such a nice fact. – user152634 Jun 28 '14 at 20:59
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