Score:0

Multiplicative inversion of a generated point?

br flag

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 $p^{-1}G$?

Score:3
my flag

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.

mtheorylord avatar
br flag
What about just over the base point G? Are there any conditions where it becomes possible to take the inverse point easily?
poncho avatar
my flag
@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
mtheorylord avatar
br flag
Could we find $p^2G$ given $pG$? Could we use an elliptic curve pairing $e(pG,pG)$ to get $p^2G$?
poncho avatar
my flag
@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)
mtheorylord avatar
br flag
Ah, I see, thank you.
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