Even perfect numbers have the form $(2^n - 1)2^{n-1}$ where $2^n - 1$ is a prime number. This restricts the possible values of $n$ to $34, 35, 36, 37$. Since $n$ must be prime, only $37$ has to be checked. But how did Fermat prove that $2^{37} - 1$ is not prime?
How did Fermat handle Frenicle's challenge to find a perfect number between $10^{20}$ and $10^{22}$?
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2$\begingroup$ Your description is for even perfect numbers. I added that. $\endgroup$– Gerald EdgarCommented Jul 26, 2023 at 16:55
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$\begingroup$ It's divisible by $223$ $\endgroup$– J. W. TannerCommented Jul 26, 2023 at 18:31
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Probably (I cannot easily verify...) Fermat knew a basic modular arithmetic device, that for prime $p$ dividing $b^n-1$, either $p|b^d-1$ for a proper divisor of $n$, or $p=1\mod n$. (This is related to Fermat's little theorem.) Thus, for prime exponent $37$, we have $p=1\mod 37$. This cuts down the search space by a factor $18$ or so, ...
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$\begingroup$ Then he was lucky it wasn't a Carmichael number. $\endgroup$– SpencerCommented Aug 1, 2023 at 14:14