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  "body": "\n## Proof\n\nWe prove the contrapositive: if $n$ is not prime, then $2^n - 1$ is not prime.\n\n1. Let $n$ be a natural number that is not a [[prime number|0SN0MAEKYX]]. Then $n = 0$, $n = 1$, or $n$ is composite.\n2. If $n = 0$, then $2^n - 1 = 0$. If $n = 1$, then $2^n - 1 = 1$. Neither $0$ nor $1$ is prime.\n3. Let $n$ be composite. Then $n = ab$ with integers $a$ and $b$ such that $1 \u003c a \u003c n$ and $1 \u003c b \u003c n$.\n4. For every number $x$ and every integer $b \\ge 1$, $(x - 1)(1 + x + x^2 + \\dots + x^{b-1}) = x^b - 1$. Indeed, the product telescopes: $\\sum_{i=0}^{b-1} (x^{i+1} - x^i) = x^b - 1$.\n5. Put $x = 2^a$. Then $2^n - 1 = x^b - 1 = (2^a - 1) \\cdot s$, where $s = 1 + 2^a + \\dots + 2^{a(b-1)}$ is a positive integer.\n6. So $2^a - 1$ [[divides|096PW7Q014]] $2^n - 1$.\n7. Since $1 \u003c a \u003c n$, we have $1 \u003c 2^2 - 1 \\le 2^a - 1 \u003c 2^n - 1$.\n8. So $2^n - 1$ has a divisor strictly between $1$ and itself, and it is not prime.\n9. By contraposition, if $2^n - 1$ is prime, then $n$ is prime.\n\n## Idea\n\nIf $n = ab$, then $2^n - 1 = (2^a)^b - 1$, and $y - 1$ always divides $y^b - 1$. This gives a proper factor $2^a - 1$ of $2^n - 1$. So a composite exponent never gives a prime.\n",
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        "locator": "Theorems about Mersenne numbers",
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