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Let $E_1$ and $E_2$ be two elliptic curves defined over $F_{p1}$ and $F_{p2}$ respectively and with same number of points $\#E_1(F_{p1}) = \#E_2(F_{p2})$. Since number of points of two curves are same, then there should exist one to one map between the curve points, can we find such map between points of curves $E_1$ and $E_2$ which can translate points from one curve to another?

an example of such curves:

$E_1$: $y^2=x^3+a_1x+b_1$ mod $p_1$

$E_2$: $y^2=x^3+a_2x+b_2$ mod $p_2$

$a_1 = 37290831958$

$b_1 = 275568094926$

$p_1 = 316038962341$

$a_2 = 246174478060$

$b_2 = 172100671463$

$p_2 = 316040145563$

$\#E_1(F_{p1})=\#E_2(F_{p2})=316039975837$

Edit:

For simplicity, let's assume number of points on curves is a prime n. Such that $|P| = |Q| = n $ $\forall P(x_1,y_1)\in E_1$ and $\forall Q(x_2,y_2) \in E_2$

There exists a proof that two cyclic groups of same order are isomorphic: https://proofwiki.org/wiki/Cyclic_Groups_of_Same_Order_are_Isomorphic

AlphaCentauri
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  • The theorem you are referring to requires the elliptic curves to be defined over the same field. Indeed, if $p_1$ and $p_2$ are different the idea of a morphism doesn't even make sense. – Mummy the turkey Oct 15 '23 at 23:24
  • @warrenmoore it's a remarkable fact about elliptic curves that a pair of elliptic curves $E_1/\mathbb{F}_q$ and $E_2/\mathbb{F}_q$ admit a morphism (of group varieties over $\mathbb{F}_q$) if and only if their point sets have the same order. The map need not even be bijective on points! – Mummy the turkey Oct 15 '23 at 23:28
  • I think our math is not advanced enough to answer this question. – AlphaCentauri Oct 16 '23 at 01:14
  • @WarrenMoore Two cyclic groups of same order are isomorphic here's the proof:

    https://proofwiki.org/wiki/Cyclic_Groups_of_Same_Order_are_Isomorphic

     – AlphaCentauri Oct 16 '23 at 03:30
  • Maybe I should've made my point more clear, sorry. The question mentioned finding a "one-to-one" mapping for arbitrary $p_1$ and $p_2$, which is not always possible. Whilst I understand there is only one cyclic group of a given order, elliptic curves over finite fields don't only have to be cyclic groups - they can also be the product of two cyclic groups.

    For your given example $316039975837$ is prime so yes the groups are isomorphic, but they need not be in general.

     – Warren Moore Oct 16 '23 at 08:32

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