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Nigel Smart's attack solves the discrete logarithm problem in linear time. It requires the curve, however, to be anomalous, i.e. to have a trace of Frobenius equal to one or, equivalently, to be of the same order as the underlying field, $\#E=p$.

Smart's paper is here: http://www.hpl.hp.com/techreports/97/HPL-97-128.html

I'm trying to understand why it doesn't work for other curves? I'm guessing that it has something to do with the properties of the p-adic logarithm but I can't quite put my finger on it.

Any hint or explanation will be appreciated.

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  • $\begingroup$ This was cross-posted on StackOverflow. Please don't do that. You should pick one site and stick with it. $\endgroup$ – mikeazo Nov 1 '17 at 0:45
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You can find an answer at section 5.4 of Washington's Elliptic Curves: Number Theory and Cryptography.

Following their notation (and skipping the technical details about reduction $\mod p$ and $E_r$ subgroups), let $N=\#E$ and assume $p\nmid N$. Let $\ell_1=\lambda_1(N\tilde P)$, $\ell_2=\lambda_1(N\tilde Q)$, so that $k\equiv \ell_2/\ell_1 \mod p$ (we want to prove this). Set also $\tilde K = k\tilde P-\tilde Q$. We have

$$k\ell_1 - \ell_2=\lambda_1(kN\tilde P-N\tilde Q)=\lambda_1(N\tilde K)\equiv N\lambda_1(\tilde K) \mod p.$$

In the last step you cannot continue, because the initial assumption makes $N$ invertible $\mod p$ and you can't say that this equals $0$, failing to prove that $k\equiv \ell_2/\ell_1 \mod p$. You can see that this proof does work if $p$ divides $N$.

To my understanding, this would mean that if $p$ divides the number of points, then you may also perform the attack, but I haven't read enough on it to confirm this.

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