GPT 6 Astra Ultra attempt on Math-1034

Inequality of the means

Definitions and mathematical statement

Let $n\ge1$ and let $a_1,\ldots,a_n$ be arbitrary positive real numbers. Set $s=\sum_{i=1}^n a_i$, let $B=\prod_{i=1}^n[0,a_i]$ be a rectangular box, and let $Q=[0,s]^n$. A packing of $m$ congruent copies of $B$ in $Q$ is a family of Euclidean isometries $T_1,\ldots,T_m$ such that $T_j(B)\subseteq Q$ for every $j$ and the interiors of different copies are disjoint. Boundary contacts are allowed; boxes may be moved and rotated but not cut or deformed.

Does such a packing always exist with $m=n^n$? The quantifiers range over every dimension and every choice of side lengths. Permuting the side lengths changes only the presentation of the same box, not the available shape. The definition uses the ordinary unrestricted meaning of geometric packing in the catalogue question. Requiring every edge to be parallel to a coordinate axis would be a stronger version; the source's illustrated constructions use such orientations.

A necessary volume comparison is \[ n^n\prod_{i=1}^n a_i\le\left(\sum_{i=1}^n a_i\right)^n, \] which is exactly the arithmetic--geometric mean inequality after taking $n$th roots. The problem asks for actual disjoint placement, not just this numerical inequality. When all lengths equal $a$, the $n\times\cdots\times n$ grid tiles the target cube. In the unequal case, unused space is permitted and the copies need not cover the entire cube.

Short English statement

Can $n^n$ congruent boxes with side lengths $a_1,\ldots,a_n$ always fit into a cube of side $a_1+\cdots+a_n$?

Research attempt

Scope and literature check.

The constructions on Bar-Natan's page [S3] already establish the two- and three-dimensional cases and explain composition of dimensions. This attempt gives an explicit coordinate certificate for the two-dimensional construction, then extends it to an even-dimensional family of side lengths. These are restricted results; the necessary AM--GM volume inequality is never used as if it supplied a packing. Sources [S2,S3] were checked on 1 October 2026.

A packing with a visible remainder.

Assume $a\ge b>0$ and put $s=a+b$. In $[0,s]^2$ use the four rectangles \[ \begin{split} R_1&=[0,a]\times[0,b],& R_2&=[a,s]\times[0,a],\\ R_3&=[b,s]\times[a,s],& R_4&=[0,b]\times[b,s]. \end{split} \] Each has side lengths $a,b$, in one of the two permitted orientations. Each pair has disjoint interiors: consecutive rectangles are separated by $x=a$, $y=a$, $x=b$, or $y=b$ as appropriate; the two remaining pairs are separated by the horizontal or vertical gap from $b$ to $a$. All endpoints are in $[0,s]$. The uncovered interior is precisely $(b,a)^2$, giving $s^2-4ab=(a-b)^2$. At $a=b$ the gap disappears and the construction becomes a tiling. Swapping the names of the sides handles $b>a$.

An even-dimensional restricted theorem.

Let $n=2k$. Suppose the side lengths can be paired $(a_{2j-1},a_{2j})$, $1\le j\le k$, so that each pair has the same sum $t$. Then $s=kt$. Form the Cartesian product of the four-rectangle packing for each pair. It gives $4^k$ disjoint $2k$-boxes inside $[0,t]^{2k}$, each congruent to the requested box; exchanging coordinates independently within a pair is an orthogonal transformation in the full space. Now partition $[0,kt]^{2k}$ into $k^{2k}$ translates of $[0,t]^{2k}$ and place that product packing in every small cube. Interiors from distinct cubes cannot intersect, so the total is exactly \[ 4^k k^{2k}=(2k)^{2k}=n^n. \] This works for arbitrarily unequal lengths within each pair and arbitrarily large even dimension, provided the pair sums agree. Its occupied volume is $n^n\prod_i a_i$ as required; the geometric construction, rather than the volume calculation, proves feasibility.

Why the most direct generalization fails.

For unequal pair sums $t_j$, the same product has container side lengths $t_1,t_1,\ldots,t_k,t_k$. Repeating it on a grid in a cube of side $s=\sum_jt_j$ guarantees only \[ M=4^k\prod_{j=1}^k\left\lfloor s/t_j\right\rfloor^2 \] copies. There is no universal inequality $M\ge (2k)^{2k}$. For $k=2$, $t_1=3,t_2=2$, one obtains $M=16\cdot1^2\cdot2^2=64$, whereas $4^4=256$. For example the four side lengths $(1,2,1,1)$ realize these pair sums. This does not contradict the known four-dimensional theorem: it disproves only this particular fixed-product grid argument. Mixed orientations of larger blocks, absent from that argument, can change the answer.

Verification and remaining gap.

The accompanying finite checks verify containment and all pairwise interior separations for integer pairs $1\le b\le a\le8$, including the equal-side boundary case, and verify the count identities. The proof itself covers all positive real lengths. To address general $n$ one still needs a placement that accommodates unequal block sums without losing the required number of copies. No exhaustive search in dimension five is claimed.

Reproducibility.

The finite calculations above are implemented in scripts/check_attempt_batch_2026_10_01.py; run it with Python 3 from the repository. It uses exact integer arithmetic and no external packages.

Final status.

UNRESOLVED. There is an explicit packing theorem for equal-sum pairs, and a numerical obstruction to the naive unequal-sum extension. These are not a resolution for arbitrary side lengths and dimensions. Completion estimate: 5%.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3082 (OPG-36901), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Inequality of the means. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/inequality_of_the_means.
• S3 D. Bar-Natan, Inequality of the Means, author’s illustrated formulation and constructions. https://www.math.toronto.edu/~drorbn/projects/ArithGeom/.