What's the torsion subgroup of $y^2 = x^3+2015^2x+2015^3$? 
$Q$. (a) Let $E$ be the elliptic curve given by $y^2 = x^3+2015^2x+2015^3$. 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 that I find is torsion? Or is there some fact that allows me to know the points on my list are indeed torsion without any checking? In addition I can't finish part (b) after showing $E(\mathbb F_5)  \cong \mathbb Z/ 4 \mathbb Z \cong E(\mathbb F_3).$
[EDIT: I've realised that part (b) follows from Nagell-Lutz -- in fact $E(\mathbb Q)_{\text{tors}} \cong  \mathbb Z/ 4.$ However, I'm wondering if it can be done with reduction? Does the fact that $E(\mathbb F_5)  \cong \mathbb Z/ 4 \mathbb Z \cong E(\mathbb F_3)$ imply that $E(\mathbb Q)_{\text{tors}} \cong  \mathbb Z/ 4?$]
(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 $\left\{(1 , 1), (1 , 6), (2 , 3), (2 , 4), (3 , 1), (3 , 6), (4 , 2 ), (4 , 5), (6 , 2), (6 , 5)\right\} \cup \left\{\mathcal O\right\} \cong \mathbb Z/ 11 \mathbb Z.$
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? (See the edit in paragraph 2 please).
 A: Yes, your answer to part (a) is correct. It's probably good to be slightly more careful in the following way: Let $p$ be an odd prime at which $E$ has good reduction. We know that $E(\mathbb{Q})[m]$ injects into $E(\mathbb{F}_p)$ for each $m$ coprime to $p$. In particular we have that $\#E(\mathbb{Q})_{tors}$ divides $ p^r \# E(\mathbb{F}_p)$ for some $r$. Then applying this with $p = 3, 7$ gives the result.
For part (b) you have showed (via the same observation above) that $\#E(\mathbb{Q})_{tors}$ divides $4$. It remains to find a point of order $4$ to show this is the entrie thing. Notice that $P = (-1, 0)$ is an obvious (unique) point of order $2$. Now, when is $2Q$ a $2$-torsion point? Precisely when the tangent to $E$ at $Q$ meets $E$ at the $x$-axis.
All you need to do at this point is to solve the arising equation to find that $(1,2)$ is a point of order $4$.

Aside: A (generalisation of a) similar approach to what I mention for classifying $4$-torsion points at the end is used in Silverman VIII Ex8.13 to provide a cute way of constructing $X_1(N)$ for $N \geq 4$.
