Given two points $P$ and $Q = \sum_{i=1}^{n} x_i.P$ over $E_p(a, b)$ for $x_1,x_2,...,x_n \in \mathbb F_p$. The Elliptic Curve Factorization Problem (ECFP) is to find the points $x_1.P,x_2.P,...,x_n.P$

Is ECFP harder than ECDLP ? Can anyone provide me the proof ?


It's not harder.

  • Solve the DL problem to get $x$ in $Q=x.P$.
  • $Q = \sum_{i=1}^{n} x_i.P = (\sum_{i=1}^{n} x_i).P= x.P$

=> any set of $x_i$ that sums to $x$ (modulo the group order) is a solution to ECFP.

  • $\begingroup$ The question would be rather how it is used and whether you have the correct set of $x_i$ values. I see this similar to how Shamir's Secret Sharing is information-theoretically perfectly secure. $\endgroup$
    – Artjom B.
    May 4 '16 at 9:08
  • $\begingroup$ @CodesInChaos how will you find the correct set of $x_i$ values ? You will find $x$ by DL problem and then break $x$ into set of $x_i$. So don't you think it is harder than ECDLP ? $\endgroup$
    – Sandy
    May 4 '16 at 12:06
  • $\begingroup$ @Sandy What do you mean by "the correct set"? $\endgroup$ May 4 '16 at 12:08
  • $\begingroup$ @CodesInChaos suppose $x_1 = 2$ and $x_2 = 8$ then $x = 10$. So you have $P$ and $Q = 10.P$. By DL problem we can get back $x = 10$. But how to find $x_1$ and $x_2$ ? $\endgroup$
    – Sandy
    May 4 '16 at 12:12
  • 1
    $\begingroup$ @Sandy It's an Underdetermined system of linear equations. There are many equally valid valid solutions with nothing to distinguish one of them as the single correct one. But that doesn't make it a harder problem in the cryptographic sense. $\endgroup$ May 4 '16 at 12:16

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