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Problem 640

AIME late

Number theory Difficulty 5.1 Prove it

21. Prove that there are infinitely many composite numbers of the form $2^{n}-1$ , where $n$ is an odd natural number.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

21. Let $n=3(2 k+1)$ , where $k \in N$ . Then $2^{3(2 k+1)}-1=$ $=\left(2^{2 k+1}\right)^{3}-1=\left(2^{2 k+1}-1\right)\left(2^{4 k+2}+2^{2 k+1}+1\right)$ . Thus, for any natural $k$ we get a composite number.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.