Can we solve the equation $nP = Q$, where $P$, $Q$ is a rational poins on elliptic curve ($P$ is unknown), and $n$ is integer?
2026-04-03 22:55:40.1775256940
Dividing elliptic curve point on integer
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1
If $E$ is an elliptic curve the set $E(\Bbb C)$ of its complex points can be identified via Weierstrass theory with the quotient $\Bbb C/\Lambda$ where $\Lambda$ is some lattice, i.e. $\Lambda=\Bbb Zz_1\oplus\Bbb Zz_1$ for some complex numbers $z_1$, $z_2$ which are linearly independent over $\Bbb R$.
Thus, if $Q\in E$ is a point the equation $nP=Q$ has always $n^2$ solutions over the complex numbers, which under the above mentioned identifications can be written as $$ P=\frac1nQ+\frac{n^{-1}\Lambda}\Lambda. $$ If $Q$ is a rational point (meaning that it has rational coordinates in some Weierstrass model of $E$) you cannot expect that any of the points $P$ is rational too, though. In general, the coordinates of $P$ will be in some number field of degree $n$ over $\Bbb Q$.
This is also clear from the Mordell-Weil theorem. It says that the group $E(\Bbb Q)$ of rational points of $E$ is finitely generated, i.e. abstractly isomorphic to a group of the form $T\oplus\Bbb Z^r$ where $T$ is a finite abelian group. Now, no finitely generated abelian group is divisible. Concretely, if $Q$ happens to be, for instance, one of the generators of the free part, no point $P\in E(\Bbb Q)$ has the property that $nP=Q$ for any $n>1$.