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Claim. Let $\mathfrak c=\lambda$ with $\operatorname{cf}(\lambda)> \omega_1$, and add $\omega_1$ Cohen reals. In the extension, every family of $\mathfrak c$ pairwise distinct entire functions has a point $z$ at which it takes $\mathfrak c$ distinct values. In particular, starting from $\mathfrak c=\aleph_2$, the extension has $\mathfrak c=\aleph_2$ and every family of entire functions taking at most $\aleph_1$ values at each point has at most $\aleph_1$ members, so the question of Problem 1119 has a positive answer there for its only admissible cardinal, $\mathfrak m=\aleph_1$. This is Theorem 2.1 of the preprint Sh:1078 in Shelah's archive (version of 2017-01-05), which dates the page; the paper was published as Ashutosh Kumar and Saharon Shelah, On a question about families of entire functions, Fund. Math. 239 (2017), no. 3, 279--288, and the theorem numbering of the published version was not compared. The source card records the statements.

Covers. The case $\mathfrak m^+=\mathfrak c$ with the continuum hypothesis failing: in the model of Theorem 2.1 built from $\mathfrak c=\aleph_2$, the cardinal $\mathfrak m=\aleph_1$ has $\mathfrak m^+=\mathfrak c$ and the answer is yes. For the literal statement, non-refutability needs no such model: under the continuum hypothesis no cardinal $\mathfrak m$ satisfies $\aleph_0<\mathfrak m<\mathfrak c$, so the statement holds vacuously in every model of CH, such as Gödel's constructible universe, and whenever $\mathfrak m^+<\mathfrak c$ the answer is yes by Erdős's counting argument. What this theorem adds is a positive answer in a model of the negation of CH in the hard case $\mathfrak m^+=\mathfrak c$. Independence of that case needs both this model and Schilhan and Weinert's result, a model with $\mathfrak c=\aleph_2$ in which the answer for $\mathfrak m=\aleph_1$ is no. The paper's Theorem 3.1 also gives a model of the negation of the continuum hypothesis with a family of $\mathfrak c$ entire functions taking fewer than $\mathfrak c$ values at every point, but there $\mathfrak c=\aleph_{\omega_1}$ is singular and the value sets have no uniform bound below $\mathfrak c$, so that theorem does not answer the problem's question for a single $\mathfrak m$.

Argument, in outline. A Cohen-generic point $z$ avoids every meager set coded before it, and two distinct entire functions agree on at most a countable set; the authors use both to show that $\mathfrak c$ distinct functions cannot all take few values at $z$. The proof is not reconstructed on this page.

Context. Erdős showed in 1964 (source card) that the countable case depends on the continuum hypothesis, and his counting argument gives a positive answer whenever $\mathfrak m^+<\mathfrak c$, which Hayman's 1974 problem list calls easy; the open case is $\mathfrak m^+=\mathfrak c$, which this model realizes with $\mathfrak m=\aleph_1$ and $\mathfrak c=\aleph_2$.

Acceptance. The result appeared in a refereed journal, Fundamenta Mathematicae, in 2017, the refereed evidence; the preprint was not compared with the published version. The site's curator, Thomas Bloom, marks the problem INDEPENDENT and records in the commentary that Kumar and Shelah gave a model with $\mathfrak c=\aleph_2$ in which the answer is yes for $\mathfrak m=\aleph_1$: that curator credit is the reviewed evidence. No formalization of this result is known.