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if 2^n+1 is prime then prove that n isa power of 2

7 years ago

If$n$is a positive integer but not a power of 2, then$n = rs$where$1 \le r < n$,$1 < s \le n$and$s$is odd.

By the preceding lemma, for positive integer$m$,

$(a-b) \mid (a^m-b^m)$

where$\mid$means "evenly divides". Substituting$a = 2^r$,$b = -1$, and$m = s$and using that$s$is odd,

$(2^r+1) \mid (2^{rs}+1),$

and thus

$(2^r+1) \mid (2^n+1).$

Because$1 < 2^r+1 < 2^n+1$, it follows that$2^n+1$is not prime. Therefore, by contraption$n$must be a power of 2.