Solving x^2^k+1+x+a=0 in F_2^n with (n,k)=1

03/18/2019
by   Kwang Ho Kim, et al.
0

Let N_a be the number of solutions to the equation x^2^k+1+x+a=0 in n where (k,n)=1. In 2004, by Bluher BLUHER2004 it was known that possible values of N_a are only 0, 1 and 3. In 2008, Helleseth and Kholosha HELLESETH2008 have got criteria for N_a=1 and an explicit expression of the unique solution when (k,n)=1. In 2014, Bracken, Tan and Tan BRACKEN2014 presented a criterion for N_a=0 when n is even and (k,n)=1. This paper completely solves this equation x^2^k+1+x+a=0 with only condition (n,k)=1. We explicitly calculate all possible zeros in n of P_a(x). New criterion for which a, N_a is equal to 0, 1 or 3 is a by-product of our result.

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