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  "body": "\n## Statement\n\nLet $n$ be a positive integer.\nIf $2^n + 1$ is a [[prime number|0SN0MAEKYX]], then $n = 2^k$ for a natural number $k$.\nSo every prime of the form $2^n + 1$ with $n \\ge 1$ is a [[Fermat number|0R03B54V4Z]].\n\n## Remarks\n\nThe proof: if $n = k m$ with $m \u003e 1$ odd, then $x + 1$ divides $x^m + 1$ for $x = 2^k$. So $2^k + 1$ is a proper divisor of $2^n + 1$.\nFor example, $2^6 + 1 = 65 = 5 \\cdot 13$, and $5 = 2^2 + 1$.\nThe converse is false, because $F_5 = 2^{32} + 1$ is composite.\n\n## Explanations\n\n### L1 human\n\nAn odd factor of the exponent gives a factorization of $2^n + 1$.\nSo the only exponents that can give primes are the powers of $2$, which have no odd factor except $1$.\n",
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