💡 Fermat Prime Condition
If $2^n + 1$ is prime, then $n$ must be a power of $2$.
Proof: Suppose $n$ has an odd divisor $k > 1$, so $n = km$ for some $m$.
Using the factorization for odd $k$: $x^k + 1 = (x+1)(x^{k-1} - x^{k-2} + \\cdots + 1)$:
$2^n + 1 = 2^{km} + 1 = (2^m)^k + 1 = (2^m + 1)(\\cdots)$.
Since $2^m + 1 > 1$, this shows $2^n + 1$ is composite.
Therefore $n$ can h...
From: numbers-geometry
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Interactive courses covering the mathematics that powers modern technology, from foundational algebra to the cryptography securing the internet.