# Prime squares and half-turns

Partial progress on an open problem. Vamshi Jandhyala.

> A p × p square, p prime, cut into p congruent pieces that are only translated or half-turned must be cut into bars; with quarter-turns too, the question is open.

Canonical: https://vamshij.com/research/prime-squares-half-turns
Code and Lean proofs: https://github.com/jvvk/mathematics/tree/main/prime-squares-half-turns
Paper (PDF): https://github.com/jvvk/mathematics/blob/main/prime-squares-half-turns/paper/note.pdf

JetfiRex asked on MathOverflow ([question 487157](https://mathoverflow.net/q/487157)) and Mathematics Stack Exchange ([question 5007342](https://math.stackexchange.com/q/5007342)) whether a `p × p` square, `p` prime, can be cut into `p` congruent `p`-ominoes in any way other than into `p` straight bars. Composite sizes have other tilings (an exhaustive search finds them for every composite `n ≤ 14`), so any proof must use primality. The question is a discrete case of Danzer's open conjecture on cutting a square into congruent pieces.

The note proves two restricted forms, without assuming the piece is connected:

- **Theorem 1.** For every prime `p`, if each piece is a translate of one cell set `P` or of its half-turn, then `P` is a straight bar.
- **Theorem 2.** For odd `p`, the same holds with the half-turn replaced by any one fixed reflection of the square. (For `p = 2` it fails: two diagonal cells and their mirror image tile the `2 × 2` square.)

The proof encodes cells as Laurent monomials, so a tiling becomes `F·U + F*·V = Q_p(x) Q_p(y)` with `F*` the half-turn of the piece. Either `F` and `F*` share a factor, which must be one of the two cyclotomic primes of the board and forces a bar, or they are coprime, and evaluating `F ∣ U* − V` at `(1,1)` shows that only one orientation is used. With translations alone the result is close to known theorems (Szegedy; Horak and Kim) and is not claimed as new; `p = 3` follows from Maltby's classification of trisected rectangles. Quarter-turns, or two different reflections, bring in a third tile polynomial and the argument stops: the full question is open.

Preprint v1, 10 October 2026, not peer reviewed and not yet independently reviewed. The algebraic core is formally verified in Lean 4 (`lean/`, standard axioms only); the unique factorisation of the Laurent ring and the encoding of a tiling are quoted. The author used AI tools (Claude, Anthropic) in this work, as described in the paper's acknowledgement, and is responsible for its content.