2 Feb
2012
2 Feb
'12
10:26 a.m.
What n satisfy 2^n == 3n+2 (mod n^2 - 2n)
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.
I hope someone will submit this to the OEIS! Neil