Dear sir first of all very thansful for your nice cooperation. The above program give roungh result like
to be factored over the field with three elements.
The algorithm computes first
Since the derivative is non-zero we have w = f/c = x2 + 2 and we enter the while loop. After one loop we have y = x + 2, z = x + 1 and R = x + 1 with updates i = 2, w = x + 2 andc = x8 + x7 + x6 + x2+x+1. The second time through the loop gives y = x + 2, z = 1, R = x + 1, with updates i = 3, w = x + 2 and c = x7 + 2x6 + x + 2. The third time through the loop also does not change R. For the fourth time through the loop we get y = 1, z = x + 2, R = (x + 1)(x + 2)4, with updates i = 5, w = 1 and c = x6 + 1. Since w = 1, we exit the while loop. Since c ≠ 1, it must be a perfect cube. The cube root of c, obtained by replacing x3 by x is x2 + 1, and calling the square-free procedure recursively determines that it is square-free. Therefore, cubing it and combining it with the value of R to that point gives the square-free decomposition
- but the program given