On 23 Jan 2012, David Wilson asked:

What n satisfy 2^n == 3n+2 (mod n^2 - 2n)

On 30 Jan 2012, Hans Havermann wrote:

At first glance, except for n=5, these appear to be the greater of twin primes (A006512). However, on closer inspection, some n (563, 645, 647, 1105, 1907, 2467, 2701, 2821, 4371, 4373, 4681, 6601, ...) are clearly not.

(Spoilers ahead) The clue is to observe that n^2 - 2n = n(n-2). First note that n is necessarily odd, which means that n and n-2 are relatively prime. By the Chinese Remainder Theorem, the equation is equivalent to the following system of equations: 2^n == 2 (mod n) 2^n == 8 (mod n-2) Since n-2 is odd, the second equation is equivalent to 2^{n-2} == 2 (mod n-2). Thus n is a solution if and only if each of n and n-2 is a prime or a Sarrus/Poulet pseudoprime. Jakob