Family of elliptic curves with trivial torsion I'm wondering, if it is true that the torsion subgroup of $y^2=x^3+p$ (for $p$ some prime, greater than 2), is always trivial?.
I was trying to prove this using Lutz-Nagell, but I can't quite get it.
Thank you
 A: Nagell-Lutz says,as you know, that if a rational point $P=(x,y)$ is torsion, both x and y are integers and either:


*

*$y=0$

*y divides the discriminant of your curve (and hence, so must $y^2$). 
In your case, the discriminant is $-27p^2$. So if $y^2 | -27p^2$ we see that y can either be $\pm 3$, $\pm p$, $\pm 3p$ or $\pm 1$. Suppose first that $p \neq 5$. Then one can show that, since 5 is a place of good reduction and the ramification degree is 1, $E(\mathbb{Q})_{tors}| E(\mathbb{F}_5)$ (here this is the order of the torsion groups of E over the respective field).It is then clear that $E(\mathbb{Q}_{tors})$ must be either 1,2,3 or 6, since  $E_(\mathbb{F}_5)=6$. It is clear that there can be no point of order 2. 
For points of order 3, use the m-division polynomials to find that if P is a 3-torsion point, $\phi_3(x,y) = 3x(x^3+4p)$. Now, $x=0$ is a solution of this equation, and inserting this into your equation you get $y^2=p$, but since p is prime, there can be no points satisfying this equations on your curve. $x^3+4p= 0$ has no rational solutions, so no 3-torsion points exist.
For points of order 4, consider the 4th divisopn polynomial $\phi_4(x,y)/\phi_2(x,y)=2x^6+80px^3-8p^2$. It has no rational roots, so no points of order 4 exist. One knows that point of order 6 exists if points of order 2 and 3 exist. No points of order 2 or 3 exists, so you are done.
So, for the case that $p=5$, you can finish it off easily:
y^3=x^2+5, Nagell-Lutz gives you that $y= \pm 3$, $y= \pm 5$, $y= \pm 1$, $y= \pm 15$. Checking these cases, you are done. 
