Consider the elliptic curve \begin{equation*} E: y^2 = x^3 + 2015x - 2015~\text{over}~\mathbb{Q}. \end{equation*} I want to prove that $|E(\mathbb{F}_7)| = 12$, that $|E(\mathbb{F}_{19})| = 19$ and that $E(\mathbb{Q})_{\text{tors}}$ is equal to the trivial group.
I'm completely lost. Any suggestions?
The following are some bits and pieces from Section 2.6 of my book. Here I will demonstrate what the OP is trying to do, but with another elliptic curve. Hopefully the OP can replicate this for the elliptic curve above.
Let $E/\mathbb{Q}$ be the elliptic curve $y^2=x^3+3$. Its minimal discriminant is $\Delta_E=-3888=-2^4\cdot 3^5$. Thus, the only primes of bad reduction are $2$ and $3$ and $\widetilde{E}/\mathbb{F}_p$ is smooth for all $p\geq 5$. For $p=5$, there are precisely $6$ points on $\widetilde{E}(\mathbb{F}_5)$, namely $$\widetilde{E}(\mathbb{F}_5)=\{ \widetilde{O}, (1,2), (1,3), (2,1), (2,4),(3,0) \},$$ where all the coordinates should be regarded as congruences modulo $5$. Thus, $N_5=6$, which is in the range given by Hasse's bound: $$ 1.5278\ldots = 5+1-2\sqrt{5}< N_5 < 5+1+2\sqrt{5}=10.4721\ldots .$$ Similarly, one can verify that $N_7 = 13$.
Next, let us quote an important result we will use:
Proposition [Silverman's "The Arithmetic of Elliptic Curves", Ch. VII, Prop. 3.1] Let $E/\mathbb{Q}$ be an elliptic curve, $p$ a prime number and $m$ a natural number not divisible by $p$. Suppose that $E/\mathbb{Q}$ has good reduction at $p$. Then the reduction map modulo $p$, $$E(\mathbb{Q})[m] \longrightarrow \widetilde{E}(\mathbb{F}_p),$$ is an injective homomorphism of abelian groups. In particular, the number of elements of $E(\mathbb{Q})[m]$ divides the number of elements of $\widetilde{E}(\mathbb{F}_p)$.
Let us go back to the example $E/\mathbb{Q} \colon y^2=x^3+3$. Above we have seen that $N_5=6$ and $N_7=13$, and $E/\mathbb{Q}$ has bad reduction only at $2$ and $3$.
If $q\neq 5,7$ is a prime number, then $E(\mathbb{Q})[q]$ is trivial. Indeed, the Proposition implies that $|E(\mathbb{Q})[q]|$ divides $N_5=6$ and also $N_7=13$. Thus, $|E(\mathbb{Q})[q]|$ must divide $\gcd(6,13)=1$.
In the case of $q=5$, we know that $|E(\mathbb{Q})[5]|$ divides $N_7=13$. Moreover, by Lagrange's theorem from group theory, if $E(\mathbb{Q})[p]$ is non-trivial, then $p$ divides $|E(\mathbb{Q})[p]|$ (in fact $E(\mathbb{Q})[p]$ is always a subgroup of $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$). Since $5$ does not divide $13$, it follows that $E(\mathbb{Q})[5]$ must be trivial. Similarly, one can show that $E(\mathbb{Q})[7]$ is trivial, and we conclude that $E(\mathbb{Q})_\text{torsion}$ is trivial.