The modular curve $X_0(15)$ has a canonical model over $\mathbf{Q}$, and it has genus $1$. As the cusp $\infty$ is rational, it is an elliptic curve. Roughly, my question is whether we can find all the rational points, and properties of the elliptic curves corresponding to the non-cuspidal ones, *after a reasonable computation*.

Let me elaborate: To find the rank, one could compute the unique newform of weight $2$ and level $\Gamma_0(15)$ using Sage (or by hand if one is keen) to find $$q - q^2 - q^3 - q^4 + q^5 + q^6 + 3q^8 + q^9 - q^{10} - 4q^{11} + \ldots$$

Now this information (maybe compute some more terms) can be used to approximate the value at $1$ of its $L$-function and observe it is non-zero. A result by Kolyvagin shows the rank is $0$. Moreover, we notice from inspection of the above Fourier coefficients that the order of the torsion group has to be a divisor of $8$. The cusps are $0,\infty, \frac{1}{3}$ and $\frac{1}{5}$, and hence they correspond to rational points on the canonical model as they all have different denominators.

In conclusion: To prove that the rational torsion is of order $8$, it would suffice to find a single rational elliptic curve with a rational subgroup of order $15$. It turns out there are four such curves, an they form an isogeny class of conductor $50$.

1) Could we somehow have expected that there are indeed $4$ non-cuspidal rational points? How could one have guessed that the corresponding curves have conductor $50$?

Note that these points are not Heegner as their associated curves are not CM, so you can't "cheat". Actually, in this particular case I am mainly concerned about the curves being semi-stable, so finding an actual model for them is quite a lot stronger.

2) Is there an argument that shows they are not semi-stable, without computing the Néron model? Could we have predicted some of their properties/invariants?

**Remark:** Note that I consider the computation of the $q$-expansions of a certain space of newforms to be 'reasonable', but not the computation of the canonical model followed by a $2$-descent (say).

**Remark:** This question originated as follows. In one of the arguments leading up to a proof of modularity of semi-stable elliptic curves over $\mathbb{Q}$, one argues that the canonical model $X_0(15)_{/\mathbf{Q}}$ has exactly $8$ rational points, and the non-cuspidal points are not semi-stable. This led me to wonder how much of this we could actually deduce without a large scale search, and how much of this could be anticipated without computing anything.

Modular Forms in One Variable=LNM476). The curve $X_0(15)$ can be computed via $q$-expansions or otherwise, and it happens to have a $2$-torsion point so a $2$-descent is easy and turns out to yield rank $0$. Finding the full torsion group is then routine, and $q$-expansions (or other methods) can be used to find the $j$-invariants etc. of the 15-isogenous curves parametrized by the non-cusp rational points. – Noam D. Elkies Jul 2 '13 at 16:41