# Thread: Elliptic Curve y^2 = x^3 +17; show N_p = p

1. Originally Posted by vernongetzler
Hat’s off. Well done, as we know that “hard work always pays off”, after a long struggle with sincere effort it’s done.
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vernon
Exercises

2. double posted by mistake...

3. Originally Posted by chiph588@
Let's assume $\displaystyle b_1,b_2\not|p$ and $\displaystyle g$ is a primitive root ($\displaystyle b_1\equiv0\mod{p}$ is trivial).
Why are you assuming that $\displaystyle b_1,b_2\not|p$? Do you actually mean $\displaystyle p\not|b_1,b_2$?

Then $\displaystyle b_1\equiv g^n,\;b_2\equiv g^m \mod{p}$, so $\displaystyle g^{3n}\equiv g^{3m}\mod{p}\implies 3n\equiv 3m\mod{\phi(p)}$.
Why $\displaystyle b_1\equiv g^n,\;b_2\equiv g^m \mod{p}$? What are n and m?

Since $\displaystyle (3,p-1)=1 \implies n\equiv m\mod{\phi(p)} \implies m=n+a(p-1)$

So $\displaystyle b_2\equiv g^m=g^{n+a(p-1)} \equiv 1\cdot g^n\equiv b_1\mod{p}$.
On the last line, I think you're using Fermat's little theorem, but how do you know that $\displaystyle p\not|g$?

Thanks for clarifying this!

4. This is my version of the definition of primitive root:
a is a primitive root mod m iff the order of a mod m is equal to phi(m) (where the order of a mod m is the smallest positive integer h such that a^h is congruent to 1 mod m.)

But from here I really don't see how we can get b1=g^n (mod p)...

5. Originally Posted by kingwinner
This is my version of the definition of primitive root:
a is a primitive root mod m iff the order of a mod m is equal to phi(m) (where the order of a mod m is the smallest positive integer h such that a^h is congruent to 1 mod m.)

But from here I really don't see how we can get b1=g^n (mod p)...
A primitive root will 'generate' every element in the set $\displaystyle \{1,2,\cdots,p-1\}$.

i.e. $\displaystyle \{g,g^2,\cdots, g^{p-1}\} = \{1,2,\cdots,p-1\}$

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