Find all integers n such that 2n - 1 is divisible by n Solution Clearly we must have n 0. In fact, n = 0 and n = 1 are solutions. Now consider n > 1, where we will argue reductio ad absurdum. Suppose n divides 2n - 1. Since n > 1, we can consider p, the smallest prime factor of n. Note that 2n - 1 is odd, so n is odd, and p is odd. Hence p > 2. Since p is the smallest prime factor of n, the greatest common divisor of n and p - 1 is 1. Hence, by a standard corollary to Euclid\'s algorithm, there exist integers x and y such that nx + (p - 1)y = 1. We may therefore write: 2 = 21 = 2nx + (p-1)y = 2nx × 2(p-1)y = (2n)x × (2p- 1)y. Now consider this equation, modulo p. We have 2n 1 (mod 2n - 1), and so 2n 1 (mod n), and 2n 1 (mod p). Also, by Fermat\'s Little Theorem, since the greatest common divisor of 2 and p is 1, we have 2p-1 1 (mod p). Putting these two results together, we get 2 1x × 1y 1 (mod p); a contradiction. (Note that this is true even though one of x and y must be negative.) Hence our original supposition, that there is a p which is the smallest prime factor of n, which divides 2n - 1, is false. Therefore the only integers n such that 2n - 1 is divisible by n, are 0 and 1..

1. Find all integers n such that 2n - 1 is divisible by n
Solution
Clearly we must have n 0. In fact, n = 0 and n = 1 are solutions. Now consider n >
1, where we will argue reductio ad absurdum. Suppose n divides 2n - 1. Since n > 1, we can
consider p, the smallest prime factor of n. Note that 2n - 1 is odd, so n is odd, and p is odd.
Hence p > 2. Since p is the smallest prime factor of n, the greatest common divisor of n and p - 1
is 1. Hence, by a standard corollary to Euclid's algorithm, there exist integers x and y such that
nx + (p - 1)y = 1. We may therefore write: 2 = 21 = 2nx + (p-1)y = 2nx × 2(p-1)y = (2n)x × (2p-
1)y. Now consider this equation, modulo p. We have 2n 1 (mod 2n - 1), and so 2n 1 (mod n), and
2n 1 (mod p). Also, by Fermat's Little Theorem, since the greatest common divisor of 2 and p is
1, we have 2p-1 1 (mod p). Putting these two results together, we get 2 1x × 1y 1 (mod p); a
contradiction. (Note that this is true even though one of x and y must be negative.) Hence our
original supposition, that there is a p which is the smallest prime factor of n, which divides 2n -
1, is false. Therefore the only integers n such that 2n - 1 is divisible by n, are 0 and 1.