Question

#5 cryptography

5) Find all solutions to x3 + x2 + 2:-3 in F5 6) Find a solution to the equation 13-5 in Z/30Z

Answer 1

Let F_5 = Z_5 = {0^bar , 1^bar , 2^bar , 3^bar , 4^bar}. X^3 + x^2 + 2^bar x = 3^bar x^3 + x^2 + 2^bar x - 3^bar = 0 Here, -3^bar = 2^bar (mod 5) x^3 + x^2 + 2^bar x + 2^bar = 0 x^2 (x + 1^bar) + 2^bar (x + 1^bar) = 0 (x^2 + 2^bar) (x + 1^bar) = 0 x^2 = -2^bar (or) x + 1^bar = 0 x^2 = 3^bar (or) x = -i [because -2^bar = 3^bar (mod 5)] = > x = 4^bar [because -1^bar = 4^bar (mod 5)] Now we have to find the elements in z_5 such that x^2 = 3^bar. \r\n 0^bar^2 = 0^bar , 1^bar^2 = 1^bar , 2^bar^2 = 4^bar , 3^bar^2 = 4^bar , 4^bar^2 = 1^bar In z_5 there is no element such that x^2 = 3^bar therefore x = 4^bar is the only solution in z_5 for the given equation. \r\n z/30 z