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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 $p^{-1}G$?

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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.

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  • $\begingroup$ 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

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