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Next: About this document ... Up: Lecture 30: Using Elliptic Previous: Pollard's Method in Action!

Motivation for the Elliptic Curve Method

Fix an integer $ B$. If $ N=pq$ with $ p$ and $ q$ prime and neither $ p-1$ nor $ q-1$ a $ B$-power-smooth number, then the Pollard $ (p-1)$-method is extremely unlikely to work. For example, let $ B=20$ and suppose that $ N=59\cdot 101 = 5959$. Note that neither  $ 59-1=2\cdot29$ nor $ 107-1=2\cdot 53$ is $ B$-power-smooth. With $ m=\lcm(1,2,3,\ldots,20)=232792560$, we have

$\displaystyle 2^m - 1 \equiv 5944\pmod{N},$

and $ \gcd(2^m-1,N)=1$, so we get nothing.

As remarked above, the problem is that $ p-1$ is not $ 20$-power-smooth for either $ p=59$ or $ p=101$. However, notice that $ p-2=3\cdot 19$ is $ 20$-power-smooth! If we could somehow replace the group $ (\mathbb{Z}/p\mathbb{Z})^*$, which has order $ p-1$, by a group of order $ p-2$, and compute $ a^m$ for an element of this new group, then we might easily split $ N$. Roughly speaking, this is what Lenstra's elliptic curve factorization method does; it replaces $ (\mathbb{Z}/p\mathbb{Z})^*$ by an elliptic curve $ E$ over $ \mathbb{Z}/p\mathbb{Z}$. The order of the group $ E(\mathbb{Z}/p\mathbb{Z})$ is $ p+1\pm s$ for some nonnegative integer $ s<2\sqrt{p}$ (any $ s$ can occur). For example, if $ E$ is the elliptic curve

$\displaystyle y^2 = x^3 + x + 54
$

over $ \mathbb{Z}/59\mathbb{Z}$ then $ E(\mathbb{Z}/59\mathbb{Z})$ is cyclic of order $ 57$. The set of numbers $ 59+1\pm s$ for $ s\leq 15$ contain numbers with very small power-smoothness.

I won't describe the elliptic curve factorization method until the next lecture. The basic idea is as follows. Suppose that we wish to factor $ N$. Choose an integer $ B$. Choose a random point $ P$ and a random elliptic curve $ y^2=x^3+ax+b$ ``over $ \mathbb{Z}/N\mathbb{Z}$'' that goes through $ P$. Let $ m=\lcm(1,2,\ldots,B)$. Try to compute $ mP$ working modulo $ N$ and using the group law formulas. If at some point it is necessary to divide modulo $ N$, but division is not possible, we (usually) find a nontrivial factor of $ N$. Something going wrong and not being able to divide is analogous to $ a^m$ being congruent to $ 1$ modulo $ p$.

More details next time!


next up previous
Next: About this document ... Up: Lecture 30: Using Elliptic Previous: Pollard's Method in Action!
William A Stein 2001-11-27