Given a complex elliptic K3 surface $\pi\colon X\rightarrow \mathbb P^1$, its *discriminant locus* is the divisor $$D = \sum\_{i = 1}^s n\_i P\_i$$ on $\mathbb P^1$ such that $n\_i$ is equal to the Euler-Poincaré characteristic of the fiber $\pi^{-1}(P\_i)$, where the sum runs over the points $P\_i \in \mathbb P^1$ such that $\pi^{-1}(P\_i)$ is singular. It is well known that $\deg D = 24$. 



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> Conversely, given an effective divisor $D$ of degree $24$ on $\mathbb P^1$, when is it the discriminant locus of a complex elliptic K3 surface?
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I am particularly curious about the minimal possible $s$. The maximal Euler-Poincaré characteristic of a singular fiber is $20$, so $s \geq 2$. But in case, say, $n\_1 = 20$, then the fibration is of type $I\_{14}^\*,I\_1,I\_1,I\_1,I\_1$ (see [Schütt-Schweizer](https://arxiv.org/abs/0712.3873)), so indeed $s = 5$. Are smaller $s$ possible?



Discriminant locus of elliptic K3 surfaces
The minimal $s$ is $3$.


It is attained by several elliptic K3's,
including $y^2 = x^3 + (t^2-t)^4$ which has IV\* fibers at
$t = 0, 1, \infty$ and no other singular fibers. 
The comment by **Ariyan Javanpeykar** gives one argument that 
$s$ can be no smaller. (*See postscript.*
This uses characteristic zero; in small positive characteristic
$s$ can be as small as $1$, e.g. in characteristic 2 the elliptic K3 surface
$y^2 + y = x^3 + t^9$ has only one reducible fiber, at $t = \infty$.)


P.S. There are other ways to prove $s>2$; for example it follows from
Szpiro's inequality, which has an elementary proof via the 
Mason-Stothers theorem (polynomial ABC). See MO 190530,
[Are there nonisotrivial elliptic curves over $\mathbb{G}\_m$?](https://mathoverflow.net/questions/190530), for this and
some related ideas. (That question is related because
${\bf G}\_m({\bf C}) = {\bf CP}^1 - \{0, \infty\}$ and
if an elliptic surface $\pi: X \to {\bf P}^1$ has $s \leq 2$
then one can choose the coordinate on ${\bf P}^1$ so that
each bad fiber maps to $0$ or $\infty$.)


