CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Every prime power order $q$ admits $q - 1$ mutually orthogonal Latin squares

Statement

For every prime power $q$ , there exists a complete family of $q - 1$ mutually orthogonal Latin squares of order $q$ .

Facts & Assumptions

Given: A prime power $q$ .

[L1]

There exists a field $F$ with $q$ elements (For every prime $p$ and $n \geq 1$ , a field with $p^{n}$ elements exists).

[L2]

For each nonzero $a \in F$ , the square $L_{a} (i, j) = a i + j$ is Latin, and distinct nonzero $a$ give orthogonal squares (The linear Latin squares $L_{a} (i, j) = a i + j$ over $F_{q}$ are pairwise orthogonal).

[L3]

A complete family of order $q$ consists of $q - 1$ pairwise orthogonal Latin squares (Orthogonal Latin squares and complete families of them).

Proof

technique · direct

1.1

Choose a field $F$ with $q$ elements by [L1]. It has exactly $q - 1$ nonzero elements.

L1choose

2.1

By [L2], the squares $L_{a}$ for $a \in F^{\times}$ are pairwise orthogonal Latin squares of order $q$ . Since there are exactly $q - 1$ of them, [L3] makes this family complete.

step 1.1L2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Deductive Press, Section 16.2: Latin Squares and MOLS (standard reference, not scraped)

Every prime power order qq admits q−1q−1 mutually orthogonal Latin squares