Score:2

If $e(aP, bP) = e(P, P)^{ab}$ then how can we solve $e(P^a, P^b)$?

va flag

I'm a bit confused regarding the bilinear pairing operation. Let's say I have a Public key of a receiver $P_r = P^x$ and I want to create a symmetric key using KEM with a pairing operation. If I chose $R = rP$ and compute $V = (Pr, P)^r $which results into $V = (P^x, P)^r = (P^x, rP) = (P^x, R)$. Here, I am confused about how can I solve $(P^x)$?

So, basically, my question is If $e(aP, bP) = e(P, P)^{ab}$ (an additive elliptic group to a multiplicative elliptic group) then how can we solve $e(P^a, P^b)$? Moreover, $e(P^a, P^b)$ will be an additive group or multiplicative group?

cn flag
This reads like a confusion about additive vs multiplicative notation.
va flag
Thank you for the reply. Could you please elaborate a bit more? I believe I need to read more about it but I'm not really sure what concept I'm missing.
kelalaka avatar
in flag
@Maeher half question that was not defined the groups. the map is from additive to multiplicative groups.
kelalaka avatar
in flag
Could you [edit] your question so that it proivedes coxtext like the bilinear pairing go from which group to which group?
va flag
I have edited it. Thanks!
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