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Statement

Hypothesis (*) (p. 1). $f:\mathbb N\to\mathbb R$ satisfies, for every positive integer $m$ and all $a\le b$,

$$a f(m)\le a^2-b^2+b\,f(b^2-a^2+m)\qquad\text{and}\qquad f(m^2+1)\ge m.$$

Write $\epsilon(k):=f(k^2+1)-k$, which (*) makes nonnegative.

Theorem 1 (p. 1). Let $f$ satisfy (*).

1. If $\epsilon(n)=0$ for some $n$, then $\epsilon(k)=0$ for every $k\le n$. The print's statement reads "then $\epsilon(k)$ for all $k\leqslant n$" [sic], dropping "$=0$"; the proof's opening sentence and its conclusion, $f(k^2+1)=k$ for all $k\le n$, give the reading above.
2. If $\epsilon(n)>0$ for some $n$, then $\epsilon(k)=\Omega(1/k)$: there are a constant $c>0$ and an integer $k_0$ with $\epsilon(k)\ge c/k$ for all $k\ge k_0$.

The theorem concerns an abstract $f$; the paper applies it to packing functions in Sections 3 to 5, see Theorem 2 and the square case.

Proof pointer

P. 1. For part 1, put $m=k^2+1$, $a=k$, $b=n$ in the first inequality of (); with $f(n^2+1)=n$ it gives $k f(k^2+1)\le k^2$, and the second inequality of () supplies the reverse bound. For part 2, write $\epsilon(n)=\alpha>0$ and put $a=n$, $m=n^2+1$ and any $b\ge n$; this gives $\epsilon(b)\ge n\alpha/b$ for all $b\ge n$, so $c=n\alpha$ and $k_0=n$ work.

Read depth

Claims checked: hypothesis (*), both parts of the theorem and the proof on p. 1 were read clause by clause on the page images of arXiv v1. Nothing here is independently reviewed.

Dependencies

None.

Source. Anshul Raj Singh, On a square packing conjecture of Erdős, arXiv:2601.22163 (2026); the edition read is named on the source card.

Bears on

• Problem 106: the theorem is stated for any $f$ obeying (); the paper asserts in Section 4 (p. 3) that the square-packing function of the problem obeys (), and with that part 1 says that $f(n^2+1)=n$ at one $n$ gives $f(k^2+1)=k$ for every $k\le n$. The theorem does not decide the problem.