Answer to Question #50077 in Math for Quamrul

Question

If P(x,n)= F(x,n)+F(x^2,n)+ F(x^3,n)+……+ F(x^n,n) where F(x, n)= x mod n, (i.e.
12 Mod 5 = 2 , means if we divide 12 by 5 we get a remainder of 2)
How many values of n are there so that, P(2014,n)=n ,where n<2012?

Answer 1

If P(2014,n)=n then F(x^a,n)=1 for all a from 1 to n.
If F(x^a,n)=1 for all a from 1 to n then F(x,n)=1.
So, F(2014,n)=1 and 2014 mod n = 1.
This means that (2014-1) is divisible by n, so, 2013 is divisible by n.
There are six numbers which are divisors of 2013 and less than 2012: 3,11, 33, 61, 183, 671 (and 1, but 2014 mod 1 = 0).
So, there are 6 values of n so that P(2014,n)=n,where n<2012.