$Q$. (a) Let $E$ be the elliptic curve given by $y^2 = x^3+2015^2+2015^3$. Prove that $E(mathbb Q)_{text{tors}}$ is the trivial group. (b) Now let $E$ be the elliptic curve $y^2 = x^3+x+2$. Determine $E(mathbb Q)_{text{tors}}$.

What’s a good method for part (a)? In the attempt below, at some point I’ve to show that $E(mathbb F_7)_{text{tors}} cong mathbb Z/ 11 mathbb Z$. During this do I’ve to check that every point I find is torsion? Or is there some fact that allows to know the points on my list are indeed torsion without any checking? In addition I can’t finish part (b) after showing $E(mathbb F_5) cong mathbb Z/ 4 mathbb Z cong E(mathbb F_3).$

(a) The discriminat $Delta = -5^6 cdot 13^6 cdot 31^7$. This means $E$ has good reduction for all primes $P ne 2,5,13,31.$ In particular, for ${E}(mathbb F_3):$ $y^2= x^3+x+2$ we have $E(mathbb F_3)_{text{tors}} cong mathbb Z/ 4 mathbb Z, $ as the torsion points in this case are $4(1,1) = 4(1, 2) = 2(2,0) = mathcal{O}$.

Since the next good prime is $7$ we can reduce ${E}(mathbb F_7):$ $y^2= x^3+x+6$, which has points:

$left{(1 , 1), (1 , 6), (2 , 3), (2 , 4), (3 , 1), (3 , 6), (4 , 2 ), (4 , 5), (6 , 2), (6 , 5)right} $

Then I find the order of every point on this list and conclude that:

and $left{(1 , 1), (1 , 6), (2 , 3), (2 , 4), (3 , 1), (3 , 6), (4 , 2 ), (4 , 5), (6 , 2), (6 , 5)right} cup left{mathcal Oright} cong mathbb Z/ 11 mathbb Z.$

Therefore $E(mathbb F_7)_{text{tors}} cong mathbb Z/ 11 mathbb Z. $ Then the order of $E(mathbb Q)_{text{tors}}$ divides $gcd(4,11) = 1$ therefore $E(mathbb Q)_{text{tors}}$ is the trivial group.

(b) If ${E}/ mathbb Q:$ $y^2= x^3+x+2$ then $E(mathbb F_3) cong mathbb Z/ 4 mathbb Z$ as above, but also similarly $E(mathbb F_5) cong mathbb Z/ 4 mathbb Z$. I can’t finish this but I’m guessing it somehow follows it from part (a). If so, how?