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 2021-12-15, 18:47 #1 RomanM   Jun 2021 3·17 Posts Casus of x^3+1 Do You know that $(2*(x^2+1)^3-7))^2\equiv3^4\ mod\ x^3+1$ for all x in Z, abs(x)>=5???
2021-12-17, 17:01   #2
Dr Sardonicus

Feb 2017
Nowhere

2·17·173 Posts

Quote:
 Originally Posted by RomanM Do You know that $(2*(x^2+1)^3-7))^2\equiv3^4\ mod\ x^3+1$ 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.

Last fiddled with by Dr Sardonicus on 2021-12-17 at 18:57 Reason: As indicated

2021-12-17, 19:35   #3
Dobri

"Καλός"
May 2018

22×5×17 Posts

Quote:
 Originally Posted by RomanM Do You know that $(2*(x^2+1)^3-7))^2\equiv3^4\ mod\ x^3+1$ for all x in Z, abs(x)>=5???
Yes, we know.

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