Post
Topic
Board Development & Technical Discussion
Re: x^3+7=0 ?
by
Yves Cuicui
on 23/02/2014, 22:07:16 UTC
This just to finalize this topic.

Because P=9xu+7, if a cubic root exists it can be computed by r1=a^((P+2)/9).
The other two solutions are:
r2=0x3fffffffffffffffffffffffffffffffffffffffffffffffffffffffbfffff0c . r1
r3=0x1c71c71c71c71c71c71c71c71c71c71c71c71c71c71c71c71c71c71c555554e9 . r1


Then it is easy to see that -7 has no cubic root because ((-7)^((P+2)/9))^3 <> -7

Then there is no points with y=0

my 2 cents
Thanks to you all