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-   -   Casus of x^3+1 (https://www.mersenneforum.org/showthread.php?t=27409)

RomanM 2021-12-15 18:47

Casus of x^3+1
 
Do You know that
[TEX](2*(x^2+1)^3-7))^2\equiv3^4\ mod\ x^3+1[/TEX]
for all x in Z, abs(x)>=5???

Dr Sardonicus 2021-12-17 17:01

[QUOTE=RomanM;595306]Do You know that
[TEX](2*(x^2+1)^3-7))^2\equiv3^4\ mod\ x^3+1[/TEX]
for all x in Z, abs(x)>=5???[/QUOTE]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.

[b]EDIT:[/b] There are even slicker and quicker ways, but I'm not telling.

Dobri 2021-12-17 19:35

[QUOTE=RomanM;595306]Do You know that
[TEX](2*(x^2+1)^3-7))^2\equiv3^4\ mod\ x^3+1[/TEX]
for all x in Z, abs(x)>=5???[/QUOTE]
Yes, we know. :smile:


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