Thread: mod x^4+1
View Single Post
Old 2022-03-22, 19:43   #24
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

22·3·499 Posts
Default

I did come up with a "reduced" version of part of your condition Mod(Mod(1,n)+x,1+x^4)^n = Mod(1, n)*(1 - x) for n == 5 (mod 8)

First, when n == 5 (mod 8), lift(Mod(1 + x, 1 + x^4)^n) is a polynomial of degree less than 4, in which the coefficients of x^3 and x^2 are always equal, and the coefficients of x and 1 are always equal and opposite.

Let ak be the coefficient of x in lift(Mod(1 + x, x^2 - 2)^k). This is a "Fibonacci-like" sequence: a0 = 0, a1 = 1, and ak+2 = 2*ak+1 + ak.

Starting with a1 we have 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, ...

When n == 5 (mod 8) the coefficients of x^3 and x^2 in lift(Mod(1 + x, 1 + x^4)^n) are divisible by n when n divides a(n+1)/2.

Note that 5 divides a3 = 5 and 13 divides a7 = 169, but 21 does not divide a11 = 5741.

I have not checked for composite n == 5 (mod 8) which fulfill this condition. (As you have noted, to satisfy your condition, n also has to be a base-2 pseudoprime.)
Dr Sardonicus is offline   Reply With Quote