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

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The linear Latin squares $L_{a} (i, j) = a i + j$ over $F_{q}$ are pairwise orthogonal

Statement

Let $F$ be a finite field with $q$ elements. For each nonzero $a \in F$ , define $L_{a} : F \times F \to F$ by $L_{a} (i, j) : = a i + j .$ Then each $L_{a}$ is a Latin square of order $q$ , and if $a \neq b$ then $L_{a}$ and $L_{b}$ are orthogonal.

Facts & Assumptions

Given: A finite field $F$ , nonzero elements $a, b \in F$ , and elements $i, j, s, t \in F$ .

[L1]

A Latin square is a function whose row maps and column maps are bijections, and orthogonality means that every ordered pair of symbols occurs exactly once (A Latin square, Orthogonal Latin squares and complete families of them).

Proof

technique · direct

1.1

For fixed $i$ , the map $j \mapsto a i + j$ is a translation of $F$ , so it is a bijection. For fixed $j$ , the map $i \mapsto a i + j$ is the composition of multiplication by the nonzero scalar $a$ and a translation, so it is also a bijection. Thus $L_{a}$ is a Latin square of order $q$ .

L1algebra

2.1

Assume $a \neq b$ . Given symbols $s, t \in F$ , a cell $(i, j)$ satisfies $L_{a} (i, j) = s$ and $L_{b} (i, j) = t$ exactly when $a i + j = s$ and $b i + j = t$ . Subtracting gives $(a - b) i = s - t$ , and since $a - b \neq 0$ there is a unique solution $i$ . Then $j = s - a i$ is also unique. Therefore every ordered pair $(s, t)$ occurs exactly once, so $L_{a}$ and $L_{b}$ are orthogonal.

L1algebra∎

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Sources

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

The linear Latin squares La(i,j)=ai+jLa​(i,j)=ai+j over FqFq​ are pairwise orthogonal