Primes of form Power of Two plus One/Proof 1
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Theorem
Let $n \in \N$ be a natural number.
Let $2^n + 1$ be prime.
Then $n = 2^k$ for some natural number $k$.
Proof
Suppose $n$ has an odd divisor apart from $1$.
Then $n$ can be expressed as $n = \paren {2 r + 1} s$.
So:
\(\ds 2^n + 1\) | \(=\) | \(\ds 2^{\paren {2 r + 1} s} + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^s}^{\paren {2 r + 1} } + 1^{\paren {2 r + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^s + 1} \paren {2^{2 r s} - 2^{\paren {2 r - 1} s} + 2^{\paren {2 r - 2} s} - \cdots - 2^s + 1}\) | Sum of Odd Positive Powers‎ |
and so $2^n + 1$ is not prime.
Hence $2^n + 1$ can be prime only if $n$ has only even divisors.
That is, if $n = 2^k$ for some natural number $k$.
$\blacksquare$