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The title must be confusing. Imagine we have this curve:

$y^2 = x^3 + 9x + 17$ over $\mathbb F_{23}$

And we know

[4]P = (19 , 20)

[8]P = (12 , 17)

If we only have the value of $[8]P$, Is it possible to calculate $2^{-1}X$ and $2^{-1}Y$ of $[8]P$ to get $[4]P$?

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  • 1
    $\begingroup$ Point halving: Point halving on elliptic curves of even order and article Point halving on elliptic curves of even order. I've corrected the notation, and even we say $x(P)$ for the x-coordinate of point $P$. This curve has an even order = 32, so it is applicable but not the way you look. Point doubling doesn't work in that way. $\endgroup$
    – kelalaka
    Dec 12, 2021 at 14:38
  • $\begingroup$ If you look at the addition formulas you will see that when $P_1 = P_2 = [2]P_1$ is not multiplication by 2. You can plug your numbers and do the arithmetic in the first link to find it without discrete log. $\endgroup$
    – kelalaka
    Dec 12, 2021 at 14:47
  • $\begingroup$ @kelalaka Thanks for your answer. Is point halving possible on elliptic curves of odd order? $\endgroup$
    – Lordi
    Dec 12, 2021 at 15:46
  • 2
    $\begingroup$ Be careful halving in even order may result in a double solution that prevents solving DLOG. In the odd case, let $n = 2k-1$ be the order then we can find the halve as; $[1/2]G = [k]G$ why? Because $[2k-1]G = \mathcal{O}$ then $[2k-1]G + G = G$ so $[k]G = [1/2]G$. This is a well-defined map for abelian groups of odd order. $\endgroup$
    – kelalaka
    Dec 12, 2021 at 16:17
  • $\begingroup$ Now, you can vote up and accept in Cryptography.SE. upvote if the answer is good, accept if the answer is satisfactory. $\endgroup$
    – kelalaka
    Dec 14, 2021 at 11:04

1 Answer 1

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Since 2 divides the group order (which is 32), there are two preimages. They can be found as roots of the multiplication-by-2 polynomial minus the target $x$ (which can be computed from division polynomials).

Example in Sage:

sage: E = EllipticCurve(GF(23), [9, 17])                                                                                                                                                                                                      
sage: E.multiplication_by_m(2)                                                                                                                                                                                                                
((x^4 + 5*x^2 + 2*x - 11)/(4*x^3 - 10*x - 1),
 (8*x^6*y - 8*x^4*y + 6*x^3*y + 3*x^2*y + 3*x*y + 6*y)/(-5*x^6 + 2*x^4 - 9*x^3 + 9*x^2 + 11*x + 4))

These are the two rational maps for computing $x$ and $y$ of the point $[2](x,y)$. We want $x$ to be equal 19, so:

sage: (E.multiplication_by_m(2)[0] - 19)
  .numerator()
  .univariate_polynomial()
  .roots(multiplicities=False)
[20, 10]

We can verify that $[2](20, *) = (19, *)$. Note that the sign of $y$ has to be chosen to match the output sign.

sage: P = E.lift_x(20)                                                                                                                                                                                                                        
sage: 2*P                                                                                                                                                                                                                                     
(19 : 3 : 1)
sage: 2*(-P)                                                                                                                                                                                                                                  
(19 : 20 : 1)

Can be repeated twice to get 4-roots, or use the multiplication-by-4 map directly (which is a bit less efficient).

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  • $\begingroup$ Is there a formal definition of multiplication_by_m $\endgroup$
    – kelalaka
    Dec 13, 2021 at 11:59
  • 1
    $\begingroup$ @kelalaka multiplication_by_m is a pair of functions $(f(x),y\cdot g(x))$, such that this pair equals to $[n]P$ when $P=(x,y)$. On the wikipedia page about division polynomials I linked, there are formulas for constructing $f(x)$ and $g(x)$, which are rational functions. $\endgroup$
    – Fractalice
    Dec 14, 2021 at 6:02
  • $\begingroup$ And. you may edit the question so that for future references it can be found easier. $\endgroup$
    – kelalaka
    Dec 14, 2021 at 11:03

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