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  "body": "\n## Statement\n\nLet $n$ be a natural number.\nIf $2^n - 1$ is a [[prime number|0SN0MAEKYX]], then $n$ is a prime number.\n\n## Remarks\n\nThe proof: if $n = ab$ with $1 \u003c a \u003c n$, then $2^a - 1$ divides $2^n - 1$, because $x - 1$ divides $x^b - 1$ for $x = 2^a$. And $1 \u003c 2^a - 1 \u003c 2^n - 1$.\nFor example, $2^6 - 1 = 63$ is divisible by $2^2 - 1 = 3$ and by $2^3 - 1 = 7$.\nThe converse is false: $11$ is prime, but $2^{11} - 1 = 23 \\cdot 89$.\n\n## Explanations\n\n### L1 human\n\nTo look for primes of the form $2^n - 1$, you only need to try prime exponents $n$.\nA factorization of $n$ always gives a factorization of $2^n - 1$.\n",
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