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Based on Differentia-addition on P I can understand (Xp,Zp) which is the base point, (Xq,Zq) which comes from Doubling, but I don't know what is the equation used to get P-Q to get X-,Z-.

So for example if I have P1(8,3,1) so it means (X⊖,Z⊖) = (8,1) and the value of (Xq,Zq) is from doubling? what about (Xp,Zp)?

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    $\begingroup$ Does the text I added to the question address your concern? Or, are you looking to use differential addition in a context other than the Montgomery Ladder? $\endgroup$
    – poncho
    Commented Mar 2, 2023 at 18:43
  • $\begingroup$ Actually I am looking for deferential Addition in Montgomery curve hyperelliptic.org/EFD/g1p/auto-montgom-xz.html $\endgroup$ Commented Mar 3, 2023 at 14:32

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I don't know what is the equation used to get P-Q to get X-,Z-.

Actually, if you're using the Montgomery Ladder algorithm, it's the base point we're multiplying.

At each step of the iteration, we have the points $P, P+G, G$ (where $P = zG$, where $z$ is the part of the multiplier we've already entered), and depending on whether the next bit in the multiplier is a 1, we want (if it is a 0) $2P, 2P+G, G$ or (if it is a 1) $2P+G, 2P+2G, G$.

So, the first step is to take $(X_P, Z_P) = P+G$, and $(X_Q, Z_Q) = P$, and $(X_\ominus, Z_\ominus ) = G$; it is easy to see that the precondition $(X_P, Z_P) - (X_Q, Z_Q) = (X_\ominus, Z_\ominus )$ holds; the addition algorithm then gives us $(X_\oplus, Z_\oplus ) = (X_P, Z_P) + (X_Q, Z_Q) = 2P+G$. And, all we need to do is compute either $2P$ or $2P+2G$ - either case is just a doubling of a value we already have.

And, to start the process, we start with $P=0$ (hence our triple is $(0, G, G)$, which fulfills the precondition, and then we can start by shifting the multiplier bits in, in msb to lsb order...

Hence, in this application, $X_\ominus, Z_\ominus )$ is always $G$ (the base point we are multiplying by).

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  • $\begingroup$ Actually I am using Montgomery curve differential addition and want to know what is the equation used to get the value of (X⊖,Z⊖) ? $\endgroup$ Commented Mar 2, 2023 at 4:45
  • $\begingroup$ @CiscoSaeed: so, you have $(X_\oplus, Z_\oplus)$, and want to know how to get $(X_\ominus, Z_\ominus)$? $\endgroup$
    – poncho
    Commented Mar 2, 2023 at 13:25
  • $\begingroup$ the issue is how to get (X⊖,Z⊖) to use it in equation without calculation (X⊕,Z⊕) $\endgroup$ Commented Mar 2, 2023 at 13:54
  • $\begingroup$ @CiscoSaeed: If you're just given $(X_P, Z_P), (X_Q, Z_Q)$, you can't. The representation of the points $P, Q$ are missing the $y$ coordinates, that is, the sign information, and so there's nothing distinguishing $P-Q$ from $P+Q$, that is $(X_\ominus, Z_\ominus)$ from $(X_\oplus, Z_\oplus)$. While you could devise a procedure to compute both, there would be no way to know which is which. $\endgroup$
    – poncho
    Commented Mar 2, 2023 at 14:05
  • $\begingroup$ But i don't understand as in paper to use it incase to reduce multiplication so if i used addition so what is the difference? And what is the addition equation? $\endgroup$ Commented Mar 2, 2023 at 14:55

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