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7

These are "red flags". No one knows of a specific exploit, only some possible reasons to be concerned that one might exist. Since no one knows of a specific attack, we can't possibly know how much speedup such a hypothetical attack might allow. Basically, you're asking for speculation where there is not enough information to allow meaningful speculation, ...


3

In your particular case the order of the point divides $p-1$, this means that the embedding degree of your curve is 1. You should be able to apply the MOV attack to transfer your instance of ECDLP into an instance of DLP over $\mathbb{F}_{p}^*$. This would allow you to use the Index Calculus to solve your problem. As the Index Calculus is subexponential, ...


3

The conversion formula from twisted Edwards to Montgomery form is: $$x_{mont} = \frac{X_{mont}}{Z_{mont}}= \frac{1+y_{ed}}{1-y_{ed}} = \frac{1+\frac{Y_{ed}}{Z_{ed}}}{1-\frac{Y_{ed}}{Z_{ed}}} = \frac{Z_{ed}+Y_{ed}}{Z_{ed}-Y_{ed}}$$ If you want the affine $x_{mont}$, you need to compute the inversion. But if you just need $X_{mont}$ and $Z_{mont}$ you can ...


1

I don't know your exact scenario. However you have two options to encrypt data using elliptic curve cryptography (ECC). I'd recommend going with the first option I present. Use elliptic curve integrated encryption scheme (ECIES). ECIES basically performs ElGamal-like encryption on a key. The key is generated at random and encrypted like in ElGamal (replace ...


1

From what you say, I assume that you are talking about the Crypto 3 challenge from HackingWeek. As Ruggero explained, the curve is vulnerable to both the MOV attack and the older FR attack that works similarily, using Weil or Tate pairings (respectivly). A simple sage code for the FR-attack would be: q = 134747661567386867366256408824228742802669457 Zq = ...



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