gpt pro 5.2 attempt on Math-5429

Draw $n$ squares inside the unit square with no common interior point. Let $f(n)$ be the maximum possible sum of the side-lengths of the squares. Is $f(k^2+1)=k$? 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 $g(\cdot)$. References [BKU24] Baek, J. and Koizumi, J. and Ueoro, T., A note on the Erdős conjecture about square packing . arXiv:2411.07274 $2024$. [CaSt05] Campbell, Connie and Staton, William, A square-packing problem of Erdős . Amer. Math. Monthly $2005$, 165--167. [Er94b] Erdős, Paul, Some problems in number theory, combinatorics and combinatorial geometry . Math. Pannon. $1994$, 261-269. [ErSo95] Erdős, Paul and Soifer, Alexander, Squares in a square . Geombinatorics $1995$, 110--114. [Ha84] Halász, Sylvia, Packing a convex domain with similar convex domains . J. Combin. Theory Ser. A $1984$, 85--90. [Pr08] Praton, I., Packing squares in a square . Math. Mag. $2008$, 358-361.

OUTPUT 1) FORMAL RESTATEMENT Let $U=[0,1]^2$ be the unit square.

A (closed) square $S\subseteq U$ has a side-length $s(S)>0$ (no restriction on orientation). A family of squares $S_1,\dots,S_n\subseteq U$ is admissible if their interiors are pairwise disjoint: \[\operatorname{int}(S_i)\cap\operatorname{int}(S_j)=\emptyset\quad\text{for all }i\neq j.\]

Define \[f(n):=\sup\left\{\sum_{i=1}^n s(S_i):\; S_1,\dots,S_n\subseteq U\text{ admissible}\right\}. \] (The problem statement phrases this as a maximum; using $\sup$ avoids an attainment issue.)

Main conjecture asked in the problem block: \[\forall k\in\mathbb N:\quad f(k^2+1)=k\,?\]

Secondary question asked in the problem block: Determine for which $n$ one has $f(n+1)=f(n)$.

2) QUICK LITERATURE/CONTEXT CHECK Web browsing available: YES, but in this write-up I do not use external sources. I only rely on statements already present in the PROBLEM block.

3) ATTACK PLAN The trivial global bound is the area/Cauchy--Schwarz inequality: \[\sum s(S_i)^2\le 1\implies \sum s(S_i)\le \sqrt n.\] This immediately yields $f(k^2)=k$ (sharp via the $k\times k$ grid construction).

To prove $f(k^2+1)=k$ one needs a strict improvement over $\sqrt{k^2+1}$, i.e. show that the geometric constraints beyond area force $\sum s(S_i)\le k$. I do not complete such an argument here.

4) WORK FAST REALITY CHECK (small cases / sanity checks) - $n=1$: One square of side $1$ fills $U$, so $f(1)=1$. - $n=4$: Partition $U$ into a $2\times 2$ grid of side $1/2$ squares. Sum of side-lengths $=4\cdot(1/2)=2$. The general upper bound (Lemma 4.2 below) gives $f(4)\le\sqrt 4=2$, hence $f(4)=2$. - $n=5$: Taking $k=2$ in Lemma 4.5 gives an explicit packing with total side-length $2$, so $f(5)\ge 2$, consistent with the problem text claim $f(5)=2$.

Lemma 4.1 (Area bound). For any admissible squares $S_1,\dots,S_n\subseteq U$, \[\sum_{i=1}^n s(S_i)^2\le 1.\]

Proof. Each square $S_i$ has area $\operatorname{area}(S_i)=s(S_i)^2$. Because the interiors are pairwise disjoint, the squares overlap (if at all) only on boundaries, which have area $0$. Therefore the area of the union is the sum of the areas: \[\operatorname{area}\left(\bigcup_{i=1}^n S_i\right)=\sum_{i=1}^n \operatorname{area}(S_i)=\sum_{i=1}^n s(S_i)^2.\] Since $\bigcup_i S_i\subseteq U$ and $\operatorname{area}(U)=1$, we get $\sum_i s(S_i)^2\le 1$. $\square$

Lemma 4.2 (Cauchy--Schwarz upper bound). For any admissible squares $S_1,\dots,S_n\subseteq U$, \[\sum_{i=1}^n s(S_i)\le \sqrt n.\] Consequently $f(n)\le \sqrt n$ for all $n$.

Proof. Apply Cauchy--Schwarz to the vectors $(s(S_1),\dots,s(S_n))$ and $(1,\dots,1)$: \[(\sum_{i=1}^n s(S_i))^2\le (\sum_{i=1}^n 1^2)(\sum_{i=1}^n s(S_i)^2)=n\sum_{i=1}^n s(S_i)^2.\] By Lemma 4.1, $\sum_i s(S_i)^2\le 1$, hence \[(\sum_i s(S_i))^2\le n\quad\Rightarrow\quad \sum_i s(S_i)\le\sqrt n.\] Taking the supremum over admissible families gives $f(n)\le\sqrt n$. $\square$

Corollary 4.3 (Exact value at perfect squares). For every integer $k\ge 1$, \[f(k^2)=k.\]

Proof. Upper bound: Lemma 4.2 gives $f(k^2)\le\sqrt{k^2}=k$. Lower bound: Lemma 4.4 constructs an admissible family with sum $k$, so $f(k^2)\ge k$. Hence equality. $\square$

Lemma 4.4 ($k\times k$ grid construction). For every integer $k\ge 1$ there is an admissible family of $k^2$ squares in $U$ with total side-length exactly $k$.

Proof. Partition $U$ into a $k\times k$ grid of congruent axis-parallel squares, each of side $1/k$. These $k^2$ squares have disjoint interiors and cover $U$. Their total side-length is \[k^2\cdot\frac{1}{k}=k.\] $\square$

Lemma 4.5 (Elementary lower bound at $k^2+1$). For every integer $k\ge 1$, \[f(k^2+1)\ge k.\]

Proof. Start from the $k\times k$ grid in Lemma 4.4 (total side-length $k$). Pick one of the $k^2$ grid squares of side $1/k$. Replace it by two smaller axis-parallel squares of side $1/(2k)$ placed side-by-side inside that grid cell. This keeps interiors disjoint and contained in $U$.

The new family has $(k^2-1)+2=k^2+1$ squares and total side-length \[(k^2-1)\cdot\frac{1}{k} + 2\cdot\frac{1}{2k} = \frac{k^2-1}{k}+\frac{1}{k}=k.\] So $f(k^2+1)\ge k$. $\square$

5) VERIFICATION - Lemma 4.1 uses only additivity of area up to measure-zero boundaries. - Lemma 4.2 uses Cauchy--Schwarz and Lemma 4.1. - Lemma 4.5: the replacement increases the count by $+1$ while preserving total side-length exactly; the two $1/(2k)$ squares fit inside the removed $1/k$-square because $2\cdot(1/(2k))=1/k$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result obtained here. - For all $n$, $f(n)\le \sqrt n$ (Lemma 4.2). - For all $k\ge 1$, $f(k^2)=k$ (Corollary 4.3). - For all $k\ge 1$, $f(k^2+1)\ge k$ (Lemma 4.5).

(ii) Exact first gap. Show the matching upper bound $f(k^2+1)\le k$ for all integers $k\ge 1$. Equivalently: rule out any admissible packing of $k^2+1$ squares in $U$ with \[\sum_{i=1}^{k^2+1} s(S_i) > k.\]

(iii) Top 3 next moves (concrete targets). 1. Prove a stability refinement of Lemma 4.2 at $n=k^2+1$: if $\sum s(S_i)$ is close to $k$, then many squares must have side close to $1/k$. Try to make this quantitative. 2. Use that $k^2+1$ squares force at least two squares to fall in the same cell of a $k\times k$ coarse partition; attempt to convert this to a strict upper bound on total side-length via a local packing inequality inside one cell. 3. Attempt a linear-programming / interval-projection bound: choose a direction $\theta$ and bound total projected lengths of the squares along $\theta$ by the projection length of $U$; optimize over $\theta$ and use the fact that a square's projection length depends on its orientation.

(iv) Minimal counterexample structure (if the conjecture is false). A smallest $k\ge 1$ for which there exists an admissible family of $k^2+1$ (non-degenerate) squares inside $U$ whose total side-length exceeds $k$. Such a counterexample would necessarily beat the trivial area bound by a $\Theta(1/k)$ margin, i.e. have $\sum s(S_i)\in (k,\sqrt{k^2+1}]$.