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I've had lots of practice adding points for my crypto class. However I've run into a situation where I need to subtract two points for decryption:

Pm + kPb - nb(kG) = Pm

Where Pm is the plaintext, Pb is participant b's public key, nb is participant b's private key, G is a base point in the elliptic group Ep(a,b), and k is a random positive integer chosen by participant a.

I am able to compute nb(kG) and kPb, but I'm unsure how to subtract the two. How is this done?

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Adding the inverse of nbkG, i.e., -nbkG, to the point (Pm+kbP)? –  DrLecter Oct 27 '13 at 0:25
    
How do I calculate an inverse point? –  ConditionRacer Oct 27 '13 at 1:02
    
You may use this as a strating point for arithmetics on elliptic curves. –  DrLecter Oct 28 '13 at 8:43
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1 Answer

The inverse of a point $P = (x_P,y_P)$ is its reflexion across the $x$-axis : $P' = (x_P,-y_P)$.

If you want to compute $Q-P$, just replace $y_P$ by $-y_P$ in the usual formula for point addition.

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Thanks! Should -yp be taken mod P? –  ConditionRacer Oct 29 '13 at 15:20
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