20211215, 18:47  #1 
Jun 2021
2^{2}·11 Posts 
Casus of x^3+1
Do You know that
for all x in Z, abs(x)>=5??? 
20211217, 17:01  #2 
Feb 2017
Nowhere
12326_{8} Posts 
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. Last fiddled with by Dr Sardonicus on 20211217 at 18:57 Reason: As indicated 
20211217, 19:35  #3 
"刀比日"
May 2018
247_{10} Posts 
