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There exist two such question but I have noticed my question is fundamentally different as it asks for mapping between two different curves, rather two different prime field like this.

Given a finite field $\mathbb{F}_{103}$ we have two elliptic curves defined over it. The equations of the curves are $y^2 =x^3$ and $y^2 = x^3 + 5$. The genrators of cyclic groups are $G_1 = (2, 27)$ and $G_2 = (2, 42)$ respectively.

Given $3G_2 = (82 ,23)$ is there a way to find $3G_1$.

The answer is $3G_1 = (46 ,1)$. I am asking as sonetimes you do not know the multiplied scalar, in this case $3$, and in such case is there a way to find the corresponding point.

I think it would not be possible that such a method exist, because if it did, we would be able to map secure curves to arbitrary insecure curves an curve vased scheme would have broken.

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Little Sagemath

K =  GF(103)
E = EllipticCurve(K,[0,5])  ## Make 5 to 0 for the first curve
print(E)
print(E.order())
print(E.abelian_group())

  • first curve; ArithmeticError: $y^2 = x^3$ defines a singular curve.

  • Second curve ; Elliptic Curve defined by $y^2 = x^3 + 5$ over Finite Field of size 103. Additive abelian group isomorphic to Z/97

  • Third curve; Elliptic Curve defined by $y^2 = x^3 + 3 $ over Finite Field of size 103. Additive abelian group isomorphic to Z/62 + Z/2 embedded in Abelian group of points on Elliptic Curve defined by y^2 = x^3 + 3 over Finite Field of size 103

Therefore they cannot be isomorphic.

I think it would not be possible that such a method exist, because if it did, we would be able to map secure curves to arbitrary insecure curves an curve vased scheme would have broken.

Yes, otherwise you would break the secure curve.

Besides, isomorphism means they are the same, just the names are different. If you can break one then the other is already broken due to the isomorphism.

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  • $\begingroup$ What if first curve have been $y^2 = x^3 + 3$? Should I add this curve to question to know method? $\endgroup$
    – madhurkant
    Dec 24, 2023 at 15:46
  • $\begingroup$ updated, it has product of groups. $\endgroup$
    – kelalaka
    Dec 24, 2023 at 15:58

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