TITLE: Torsion group of the following elliptic curve
QUESTION [6 upvotes]: Let $p_1=2, p_2 = 3,\ldots,$ be the prime numbers, and define $n_i = \prod_{j=1}^i p_j$. Moreover, let $E_i $ be the elliptic curve defined by $y^2 = x^3 + n_i$. 
Can one compute the torsion group $E_i(\mathbb Q)_{tors}$ in terms of $i$? Or at least its cardinal?
I expect it to be trivial. I checked this for all $i\leq 1000$.
I was thinking about considering the reduction of $E_i$ modulo $p_{i+1}$ and showing that $E_i(\mathbb F_{p_{i+1}}) = \{ 0\}$. That would imply the torsion group is trivial.

REPLY [5 votes]: Suppose that $(x,y)$ is a torsion point on the curve given by $Y^2=X^3+n$, where $n\ne\pm1$ is a squarefree integer. We can't have $y=0$. So assume $y\ne0$. Then, by Nagell-Lutz, $y^2$ divides $-27n^2$, so $y$ divides $9n$. If $p>3$ is a divisor of $y$, then (note that $n$ is square free) $y^2=x^3+n$ implies that $p$ divides $x^3$, but $p^2$ does not, which is absurd. So $y$ is, up to a sign, a power of $3$, which yields the same contradiction unless $y=\pm1$.
So suppose that $y=1$ (thanks to Jeremy Rouse for pointing out that this case was missing). Then $2(x,y)=(x',y')$ with $y'=(-27n^2 + 18n + 1)/8\ne0$. Again by Nagell-Lutz, $y'$ divides $9n$, which fails for size reasons except for a few small values of $n$ which can be checked by hand.