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  "body": "\n## Proof\n\nWe prove the contrapositive: if $n$ is not a power of $2$, then $2^n + 1$ is not prime.\n\n1. Let $n$ be a positive integer. By the [[fundamental theorem of arithmetic|084HH1NDCB]], $n = 2^k m$ with $k \\ge 0$ and $m$ a positive odd integer.\n2. Assume $n$ is not a power of $2$. Then $m \\ne 1$, so $m \\ge 3$.\n3. Put $x = 2^{2^k}$, so that $2^n + 1 = x^m + 1$ and $x \\ge 2$.\n4. For odd $m$, $(x + 1)(x^{m-1} - x^{m-2} + \\dots - x + 1) = x^m + 1$. Indeed, the product is $\\sum_{i=0}^{m-1} (-1)^{m-1-i} (x^{i+1} + x^i)$, which telescopes to $x^m + (-1)^{m-1} = x^m + 1$.\n5. So $x + 1$ [[divides|096PW7Q014]] $2^n + 1$.\n6. We have $x + 1 \\ge 3 \u003e 1$. Also $x + 1 \u003c x^m + 1$, because $x \\ge 2$ and $m \\ge 3$ give $x^m \u003e x$.\n7. So $2^n + 1$ has a divisor strictly between $1$ and itself, and it is not a [[prime number|0SN0MAEKYX]].\n8. By contraposition, if $2^n + 1$ is prime, then $n = 2^k$ for a natural number $k$.\n9. Then $2^n + 1 = 2^{2^k} + 1$, which is the [[Fermat number|0R03B54V4Z]] $F_k$. So every prime of the form $2^n + 1$ with $n \\ge 1$ is a Fermat number.\n\n## Idea\n\nAn odd factor $m \u003e 1$ of the exponent lets us factor $y^m + 1$ by $y + 1$, because $-1$ is a root of $y^m + 1$ when $m$ is odd. This gives a proper divisor. So only exponents with no odd factor, the powers of $2$, can give primes.\n",
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