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Old 2021-12-15, 18:47   #1
RomanM
 
Jun 2021

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Default 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???
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Old 2021-12-17, 17:01   #2
Dr Sardonicus
 
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Quote:
Originally Posted by RomanM View Post
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
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Old 2021-12-17, 19:35   #3
Dobri
 
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Quote:
Originally Posted by RomanM View Post
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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