This is a $y^2=x^3+7$ elliptic curve points - $Q,G_1,G_2,G_3. k_1,k_2,k$? - secret exponents:

  • $k_1*G_1( x_1,y_1) = Q(X,Y)$

  • $k_2*G_2( x_2,y_2) = Q(X,Y)$

  • $k*G_3( x_3,y_3) = Q(X,Y)$

How to find a $k$?" ??

$Q,G_1,G_2,G_3. k_1,k_2$ - are known

$k$? - need to find.

  • 3
    $\begingroup$ If no relations known about $k_1,k_2,k$ or $G_1, G_2, G_3$, then $[k]G_3=Q$ is independent and so it is generally hard to recover $k$. Unless the curve is weak, but this depends on the underlying field. $\endgroup$ – Fractalice Feb 10 at 8:48
  • 1
    $\begingroup$ This question needs additional information on how these equations have arrived. This not an ordinary DLog. $\endgroup$ – kelalaka Feb 10 at 9:23
  • $\begingroup$ Base points G1,G2 and secret exponents k1,k2 are bruteforced, G - information from property of a curve secp256k1. $\endgroup$ – Donald Feb 10 at 9:25

This is known: $$k_1 \cdot G_1 = k_2 \cdot G_2 = k \cdot G_3 = Q$$

Additionally, let $G$ be a base point, such that we can have:

$$G_1 = k \cdot k_2 \cdot G$$ $$G_2 = k_1 \cdot k \cdot G$$ $$G_3 = k_1 \cdot k_2 \cdot G$$

This reduces to a case of tri-party Diffie-Hellman. So without further information, it's unsolvable.

  • $\begingroup$ What further information need for this ? Tnx for your comment @DannyNiu $\endgroup$ – Donald Feb 10 at 9:35
  • $\begingroup$ I'm not sure what further information is needed, but if it's a question from a text book, there may be some context you missed to put here, or if there is no further context, then it's an error in the book. $\endgroup$ – DannyNiu Feb 10 at 9:37
  • $\begingroup$ how to find a k from k_1*G_1=k_2*G_2=k*G_3=Q ??? $\endgroup$ – Donald Feb 10 at 9:46
  • $\begingroup$ @Donald as I said in the answer, it reduces to a case of 3-party DH. You can't solve it without side information. $\endgroup$ – DannyNiu Feb 10 at 10:34
  • $\begingroup$ DannyNiu how I can add to this your answer crypto.stackexchange.com/a/88141/81643 G = G_o*k_3 ? Because I don't know exact k for G, but I can get k_o for get G from (k_o * G_o=G) and this <>Q !!! ??? $\endgroup$ – Donald Feb 10 at 19:58

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