# Polyomino rectangles by parity: no rectangle for 8, 10 or 16 cells

Partial progress on an open problem. Vamshi Jandhyala.

> One copy of every hole-free n-omino tiles no rectangle for n = 8, 10 or 16, because the pieces' checkerboard imbalances add up to 2 mod 4 (for n = 16, over 11,230,003 pieces). Sizes 9 and 11 to 15 stay open.

Canonical: https://vamshij.com/research/polyomino-rectangle-parity
Code and Lean proofs: https://github.com/jvvk/mathematics/tree/main/polyomino-rectangle-parity
Paper (PDF): https://github.com/jvvk/mathematics/blob/main/polyomino-rectangle-parity/paper/note.pdf

Can one copy of every hole-free `n`-omino tile a rectangle? On [MathOverflow 378923](https://mathoverflow.net/q/378923) (Ralph Morrison, 2020) the answer is yes for `n = 1, 2, 5, 7`, no for `n = 3, 4, 6` by a parity argument, and no for `n ≥ 17` by cavity arguments in the answers (Stanley, Budd, Lehner; `n = 17` by Hoffman's computer count). The cases `8 ≤ n ≤ 16` were open.

**Theorem.** For `n = 8`, `10` and `16` the answer is no.

The reason is the checkerboard count that rules out 4 and 6. A placed piece covers `±c(P)` more black than white squares; for even `n` every `c(P)` is even and the rectangle is balanced, so `∑ c(P)` must be a multiple of 4. Over all hole-free `n`-ominoes the total `S(n)` is

| n | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|
| pieces | 363 | 4,460 | 58,937 | 805,475 | 11,230,003 |
| S(n) | 278 | 4,010 | 59,248 | 886,436 | 13,329,078 |
| S(n) mod 4 | **2** | **2** | 0 | 0 | **2** |

For `n = 12`, `14` and every odd `n` the count gives no obstruction, so `9, 11, 12, 13, 14, 15` remain open.

Preprint v1, 10 October 2026, not peer reviewed and not yet independently reviewed. The parity argument is formally verified in Lean 4 (`lean/`, standard axioms only) for any pieces and any placements; the three totals are its hypotheses, computed by two independent enumerations that both reproduce OEIS A000104, A000105 and A001168 for `n ≤ 16`. The author used AI tools (Claude, Anthropic) in this work, as described in the paper's acknowledgement, and is responsible for its content.