This question is about elliptic curves defined over the rational numbers $\mathbb{Q}$, though the question will make sense for any number field with class number one (due to the existence of unique minimal models). 


Let 


$$\displaystyle E\_{A,B} : y^2 = x^3 + Ax + B, A, B \in \mathbb{Z}$$


be a minimal Weierstrass model of an elliptic curve defined over $\mathbb{Q}$ (in particular, we demand that for all primes $p$, if $p^4 | A$ then $p^6 \nmid B$). The discriminant $\Delta(E\_{A,B})$ of the curve is given by 


$$\displaystyle \Delta(E\_{A,B}) = -16(4A^3 + 27B^2).$$


Let $S = \{2, p\_1, \cdots, p\_s\}$ be a set of primes. A rational $S$-unit is a rational number $n = 2^{k\_0} p\_1^{k\_1} \cdots p\_s^{k\_s}$, with $k\_i \in \mathbb{Z}$ for $i = 0, \cdots, s$. We say that $n$ is an *exact* $S$-unit if $k\_i \ne 0$ for $i = 0, \cdots, s$.


Is the set 


$$\displaystyle M(S) = \{D \text{ is an exact }S\text{-unit}: \exists E\_{A,B} \text{ s.t. } \Delta(E\_{A,B}) = D\}$$


finite for all $S$? Is it *uniformly* finite? That is, does there exist a number $N$ such that for any set $S$ of primes containing $2$ we have $|M(S)| \leq N$?



$S$-units which can be discriminants of elliptic curves over $\mathbb{Q}$
The finiteness follows from Falting's theorem on finiteness of abelian varieties with good reduction outside a given set of primes (or some easier stuff too, probably).


The uniformity is false. Take an elliptic curve $E$ with bad reduction at primes $p\_1,\dots, p\_n$. Assume that at none of these primes does a quadratic twist of $E$ have good reduction. Then there are $\approx 2^n$ quadratic extensions ramified only at $p\_1,\dots p\_n$. If we twist $E$ by any of them, it will still have bad reduction at those primes and good reduction at all other primes. Moreover, at every prime where the quadratic extension is ramified, the discriminant will change by a factor of $p^6$, so these elliptic curves all have distinct discriminants.


