Let's say I have a public generator $G$, an unknown, private $p$ and a public point $pG$ on an elliptic curve.

Given $pG$ it's easy to construct $-pG$ by just taking the negative. But can you construct $p^{-1}G$?


1 Answer 1


But can you construct $p^{-1}G$?

No, you can't if:

  • You're assuming that you can compute the inverse modulo an arbitrary base point (rather than a fixed one)

  • The CDH problem is hard.

That's because with an arbitrary 'inverse modulo' operation, you can compute Diffie Hellmans.

Here's one way, given $aG, bG$ (assuming point halving is easy):

  • If we compute the inverse of $G$ to the base point $aG$, we get the result $a^2G$

  • Similarly, we can compute $b^2G$ and $(a+b)^2G$

  • Compute $(a+b)^2G - a^2G - b^2G = 2abG$

  • Perform point halving on that, and that's $abG$, the CDH of $aG, bG$

Conversely, if the CDH problem is easy, then $p^{-1}G$ can be computed.

  • $\begingroup$ What about just over the base point G? Are there any conditions where it becomes possible to take the inverse point easily? $\endgroup$ Commented Aug 2, 2023 at 19:08
  • $\begingroup$ @mtheorylord: well, if CDH is easy, then you can. Otherwise, well, I don't see a reduction to a known hard problem (CDH, DDH), however the fact that, in an elliptic curve, there's nothing special about the generator point - that would tend to suggest that there isn't a way $\endgroup$
    – poncho
    Commented Aug 2, 2023 at 19:15
  • $\begingroup$ Could we find $p^2G$ given $pG$? Could we use an elliptic curve pairing $e(pG,pG)$ to get $p^2G$? $\endgroup$ Commented Aug 2, 2023 at 19:51
  • $\begingroup$ @mtheorylord: no, $e(pG, pG) = e(G,G)^{p^2}$. The pairing operations we use generate values, not in the elliptic curve group, but in a finite field, and those finite field values cannot be efficiently mapped back to elliptic curve points (at least, for any curve that anyone would consider secure) $\endgroup$
    – poncho
    Commented Aug 2, 2023 at 19:54
  • $\begingroup$ Ah, I see, thank you. $\endgroup$ Commented Aug 2, 2023 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.