In [Er94b] Erdős dates this conjecture to 'more than 60 years ago'. Erdős proved that $f(2)=1$ in an early mathematical paper for high school students in Hungary. Newman proved (in personal communication to Erdős) that $f(5)=2$.

It is trivial from the Cauchy-Schwarz inequality that $f(k^2)=k$. Erdős also asks for which $n$ is it true that $f(n+1)=f(n)$.

It is easy to see that $f(k^2+1)\geq k$, by first dividing the unit square into $k^2$ smaller squares of side-length $1/k$, and then replacing one square by two smaller squares of side-length $1/2k$. Halász [Ha84] gives a construction that shows $f(k^2+2)\geq k+\frac{1}{k+1}$, and in general, for any $c\geq 1$, \[f(k^2+2c+1)\geq k+\frac{c}{k}\] and \[f(k^2+2c)\geq k+\frac{c}{k+1}.\] Halász also considers the variants where we replace a square by a parallelogram or triangle.

Erdős and Soifer [ErSo95] and Campbell and Staton [CaSt05] have conjectured that, in general, for any integer $-k<c<k$, $f(k^2+2c+1)=k+\frac{c}{k}$, and proved the corresponding lower bound. Praton [Pr08] has proved that this general conjecture is equivalent to $f(k^2+1)=k$.

Baek, Koizumi, and Ueoro [BKU24] have proved $g(k^2+1)=k$, where $g(\cdot)$ is defined identically to $f(\cdot)$ with the additional assumption that all squares have sides parallel to the sides of the unit square. More generally, they prove that $g(k^2+2c+1)=k+c/k$ for any $-k<c<k$, which determines all values of $g(\cdot)$.