Quote:
Originally Posted by RomanM
Do You know that
^3-7))^2\equiv3^4\ mod\ x^3+1)
for all x in Z, abs(x)>=5??? |
If you want to check this as a polynomial congruence, you might first try finding
2*(x^2+1)^3 - 7 (mod x^3 + 1)
by polynomial division with quotient and remainder
2*(x^2+1)^3 - 7 = (x^3 + 1)*q(x) + r(x); q(x), r(x) polynomials
The remainder r(x) will be a polynomial of degree less than 3.
EDIT: There are even slicker and quicker ways, but I'm not telling.