quasi <firstname.lastname@example.org> wrote: >Danny73 <email@example.com> 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^(n-1) - x^(n-2) + ... - x + 1)
which follows by specializing f(x) = x^n, k = -1 in the
FACTOR THEOREM x-k | 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.