
Re: Is this well known?
Posted:
Jul 26, 2011 11:05 AM


quasi <quasi@null.set> wrote: >Danny73 <fasttrack2a@att.net> wrote: > >> 2^k +1, k = 16,+32,+32,+32,+32,+ >oo >> >> After each accumulated sum before each >> comma where k is an exponent of 2. >> >> Then (2^k +1) == 0 (mod (2^16 +1)) > > Yes, it's a triviality. > It follows from the fact that if n is odd, > > x^n + 1 is divisible by x + 1 > > by virtue of the polynomial identity (assuming n is odd) > > x^n + 1 = (x + 1) (x^(n1)  x^(n2) + ...  x + 1)
which follows by specializing f(x) = x^n, k = 1 in the
FACTOR THEOREM xk  f(x)f(k) for k in Z, f in Z[x]
Or, mod x+1: x = 1 => x^n = (1)^n = 1 by n odd.
Bill Dubuque

