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Math 2.0

Number theory Proposition

If 2^n + 1 is prime then n is a power of 2

Builds on Fermat number Prime number

Statement

Let $n$ be a positive integer. If $2^n + 1$ is a prime number, then $n = 2^k$ for a natural number $k$ . So every prime of the form $2^n + 1$ with $n \ge 1$ is a Fermat number.

Remarks

The proof: if $n = k m$ with $m > 1$ odd, then $x + 1$ divides $x^m + 1$ for $x = 2^k$ . So $2^k + 1$ is a proper divisor of $2^n + 1$ . For example, $2^6 + 1 = 65 = 5 \cdot 13$ , and $5 = 2^2 + 1$ . The converse is false, because $F_5 = 2^{32} + 1$ is composite.

Reading path

To understand this page from the ground up, read these 4 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.

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Divisibility definition
Fermat number definition
Prime number definition
If 2^n + 1 is prime then n is a power of 2 statement

Further reading

Web https://en.wikipedia.org/wiki/Fermat_number, lead section

About this page

Tag	`0VN4H8M12X`
URI	`m2://0VN4H8M12X`
Kind	statement, proposition
Writing	W0 stub
MSC	11A41
Status	This copy of the site has no build data. Run `m2 status 0VN4H8M12X`.
Uses	Fermat number, Prime number

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