I've been looking at this problem for some time now and just can't seem to get the right idea. The problem is:
Consider the elliptic curve $C:y^2=x^3+bx$ defined over the rational numbers with $b$ a non-zero integer. Show that there are either 2 or 4 points of finite order.
Here's what I know: If $p$ is a prime with $p\equiv 3\mod{4}$ and $p\not| 2b$ then there are $p+1$ points of finite order when considering $C$ over $F_p$. So by the reduction modulo $p$ theorem the number of points with finite order must divide $p+1$.
From this I can deduce that 2 and 4 are options since $p+1\equiv 0\mod{4}$ and that 1 is not an option since I have the point at infinity and $(0,0)$.
But how do I argue that these are the only possibilities i.e. for all $p$ satisfying the above 4 is the only factor that all the $p+1$'s have in common.
You are almost done! Let $q$ be an odd prime. We have to construct a prime $p \equiv 3 \mod 4$ such that $q \nmid p+1$, and such that $p \nmid b$.
First consider the case $q \neq 3$. Consider the arithmetic progression $3+4qk$. By Dirichlet's theorem, it contains infinitely many primes; take one sufficiently large so that it doesn't divide $b$, say $p=3+4qk_0$, which is $\equiv 3 \mod 4$. Then $p+1 = 4(1+qk_0)$ is not divisible by $q$.
For $q=3$, consider instead the progression $7+12k$. Any prime $p$ in this progression is $\equiv 3 \mod 4$ and satisfies $p+1 \equiv -1 \mod 3$. Once again, taking it to be sufficiently large we can make sure that it doesn't divide $b$.
Now, can you show that there exist infinitely many primes $p \equiv 3 \mod 4$ such that $8 \nmid p+1$?