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Oct 5 at 9:19 vote accept Duality
Jul 11, 2023 at 2:20 vote accept Duality
Oct 5 at 9:19
Jul 3, 2023 at 15:47 comment added R. van Dobben de Bruyn The point is that I choose $x$ first and then choose $y$ and $D$ simultaneously. This is implicitly also what's going on in the other answer. But the other answer doesn't explain why the different values of $(x,y)$ cannot accidentally all give the same value of $D$.
Jul 3, 2023 at 12:49 comment added R. van Dobben de Bruyn Also recall that the equation $DY^2 = X^3+17X$ is equivalent to $y^2 = x^3+17D^2x$.
Jul 3, 2023 at 12:44 comment added R. van Dobben de Bruyn Any positive rational number $z$ (in this case $z = X^3+17X$) can be uniquely written as $z = DY^2$ where $D \in \mathbf Z$ is squarefree and $Y \in \mathbf Q$. Indeed, $D$ has to be the product of all primes $p$ with $v_p(z) \equiv 1 \pmod 2$, and then $z/D$ is a square in $\mathbf Q$. (If we did not care that $D$ is squarefree and integral, we would just take $Y=1$.) I'm just saying that every class in $\mathbf Q^\times/\mathbf Q^{\times 2}$ has a unique squarefree integer representative.
Jul 3, 2023 at 10:15 comment added Duality How you take $Y$ as rational number ? For $X=m/4n$, I couldn't find the reason why $(X,Y)$ is a rational point of $E$.
Jul 2, 2023 at 21:43 history edited R. van Dobben de Bruyn CC BY-SA 4.0
Small clarification.
Jul 2, 2023 at 21:36 history answered R. van Dobben de Bruyn CC BY-SA 4.0