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. 



Torsion group of the following elliptic curve
Exercise 10.19 of Chapter X of Silverman's Arithmetic of elliptic curves gives a classification of the torsion subgroup of $E : y^{2} = x^{3} + D$ if $D$ is a sixth-power free integer. The torsion subgroup is trivial unless $D$ is a square, a cube, or $-432$.


One way to prove this is to show that if $p \equiv 2 \pmod{3}$ is prime and $p \nmid 6D$, then $\# E(\mathbb{F}\_{p}) = p+1$. The idea then is to use Dirichlet's theorem on primes in arithmetic progressions. (A similar argument for the curves $y^{2} = x^{3} - n^{2} x$ is given in Koblitz's "Introduction to Elliptic Curves and Modular Forms" Proposition I.9.17.)


