Consider the elliptic curve $y^2 = x^3 – k^2x + k^3$ where $k$ is a non-zero integer.
I've come across a question that asks to show that if $(x,y)$ is a rational torsion point on the elliptic curve, then:
$$\vert{y}\vert \le 5\vert{k}\vert^3$$ and $$\vert{x}\vert \le 3\vert{k}\vert^2$$
Now by the Nagell-Lutz Theorem, $x$ and $y$ must be integers, and if $y$ is non-zero, then $y^2$ divides the discriminant of the curve, which I believe is $-4k^6 + 27k^6 = 23k^6$.
I therefore believe this should mean $y^2$ divides $k^6$ and so $\vert{y}\vert \le \vert{k}\vert^3$.
I'm therefore wondering why a factor of 5 has been included in the result of the question – have I missed something?
Further I'm not sure how to find the bound on $x$. I'm assuming I need to use the bound on $y$ but this seems difficult.
Any help much appreciated.
Best Answer
For $y$, $y^2|-23k^6$, then $|y|^2\leq |23k^6|\leq|25k^6|$, so $|y|\leq 5|k^3|$.
For $x$, we know $|x^3 - k^2x + k^3|\leq 25|k|^6$, thus $|x^3 - k^2x| \leq|k^3|+25|k|^6$, which is also $|x||x ^2- k^2|\leq|k^3|+25|k|^6$.
If $|x|>3k^2$, then $|x||x ^2- k^2|>3k^2(9k^4-k^2)$, so $3k^2(9k^4-k^2)<|k^3|+25|k|^6$, we now get $2k^6-3k^4<|k^3|$.
Now there are two possibilities.
If $k>0$, then $2k^3<3k+1$, so $k=1$. Now $y^2=x^3-x+1$ where $\Delta=-23$ and $y=1$ or $-1$, hence $x^3-x+1=1$, then we know $x=0,1,-1$ which doesn't satisfy $|x|>3|k|^2$.
If $k<0$, then $2k^3>3k+1$, this is always impossible too since $k$ is an integer.