Casus of x^3+1

[RomanM]

Do You know that

for all x in Z, abs(x)>=5???

[Dr Sardonicus]

Originally Posted by RomanM

Do You know that

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.

[Dobri]

Originally Posted by RomanM

Do You know that

for all x in Z, abs(x)>=5???

Yes, we know.