[proofplan] We first show that each array $L_a$ is a [Latin square](/page/Latin%20Square) by checking rows and columns separately. Rows are translations in the second variable, while columns are affine maps with nonzero slope $a$. To prove mutual orthogonality, we fix distinct $a,b \in F^\times$ and show that the map sending a cell $(x,y)$ to the ordered pair $(L_a(x,y),L_b(x,y))$ is injective; since its domain and codomain are finite sets with the same cardinality $q^2$, injectivity then forces surjectivity. [/proofplan]

[step:Show that every row of $L_a$ is a bijection of $F$] Fix $a \in F^\times$ and fix a row index $x \in F$. Define the row map $R_x: F \to F$ by sending each $y \in F$ to $R_x(y) = L_a(x,y) = ax + y$. If $R_x(y_1) = R_x(y_2)$ for $y_1,y_2 \in F$, then $ax + y_1 = ax + y_2$, so cancellation in the additive group of the [finite field](/page/Finite%20Field) $F$ gives $y_1 = y_2$. Thus $R_x$ is injective. Since $F$ is finite and $R_x: F \to F$ is an injective map from a finite set to itself, $R_x$ is bijective. [/step]

[step:Show that every column of $L_a$ is a bijection of $F$] Fix $a \in F^\times$ and fix a column index $y \in F$. Define the column map $C_y: F \to F$ by sending each $x \in F$ to $C_y(x) = L_a(x,y) = ax + y$. If $C_y(x_1) = C_y(x_2)$ for $x_1,x_2 \in F$, then $ax_1 + y = ax_2 + y$, so $a(x_1 - x_2) = 0$. Since $a \in F^\times$, multiplication by $a$ is cancellative in the field $F$, hence $x_1 - x_2 = 0$ and $x_1 = x_2$. Thus $C_y$ is injective. Since $F$ is finite and $C_y: F \to F$ is an injective map from a finite set to itself, $C_y$ is bijective. Therefore $L_a$ is a Latin square of order $q$. [/step]

[step:Prove that two distinct squares determine each cell uniquely]Fix distinct elements $a,b \in F^\times$. Define the paired-symbol map $\Phi_{a,b}: F \times F \to F \times F$ by \begin{align*} \Phi_{a,b}(x,y) = (L_a(x,y), L_b(x,y)). \end{align*} We prove that $\Phi_{a,b}$ is injective. Let $(x,y),(x',y') \in F \times F$ satisfy \begin{align*} \Phi_{a,b}(x,y) = \Phi_{a,b}(x',y'). \end{align*} By equality of ordered pairs in $F \times F$, this means that both identities $ax + y = ax' + y'$ and $bx + y = bx' + y'$ hold. Subtracting the right-hand side of each equation from the left-hand side gives $a(x-x') + (y-y') = 0$ and $b(x-x') + (y-y') = 0$. Subtracting the second displayed equation from the first gives \begin{align*} (a-b)(x-x') = 0. \end{align*} Since $a \neq b$, we have $a-b \in F^\times$. Therefore multiplication by $a-b$ is cancellative in $F$, so $x-x' = 0$ and hence $x=x'$. Substituting $x=x'$ into $ax+y=ax'+y'$ gives $y=y'$. Thus $(x,y)=(x',y')$, proving that $\Phi_{a,b}$ is injective.[/step]

[guided]The orthogonality condition says that the two symbols \begin{align*} L_a(x,y) \quad \text{and} \quad L_b(x,y) \end{align*} together identify the cell $(x,y)$ uniquely. We encode this by the map $\Phi_{a,b}: F \times F \to F \times F$ defined by \begin{align*} \Phi_{a,b}(x,y) = (L_a(x,y), L_b(x,y)). \end{align*} To prove uniqueness of the cell, suppose two cells $(x,y),(x',y') \in F \times F$ produce the same ordered pair: \begin{align*} \Phi_{a,b}(x,y) = \Phi_{a,b}(x',y'). \end{align*} By the definition of equality in the Cartesian product $F \times F$, the first coordinates are equal and the second coordinates are equal. Therefore both identities $ax + y = ax' + y'$ and $bx + y = bx' + y'$ hold. The point of using two distinct parameters $a$ and $b$ is that subtracting these equations eliminates the common term involving $y-y'$. First rewrite both equations by moving all terms to the left: $a(x-x') + (y-y') = 0$ and $b(x-x') + (y-y') = 0$. Now subtract the second equation from the first. The terms $y-y'$ cancel, giving \begin{align*} (a-b)(x-x') = 0. \end{align*} Because $a$ and $b$ are distinct elements of the field $F$, the difference $a-b$ is nonzero, so $a-b \in F^\times$. A nonzero field element has a multiplicative inverse, hence multiplication by $a-b$ is cancellative. It follows that \begin{align*} x-x' = 0, \end{align*} so $x=x'$. Finally substitute $x=x'$ into the first coordinate equality: \begin{align*} ax + y = ax' + y'. \end{align*} Since $x=x'$, the terms $ax$ and $ax'$ are equal, and additive cancellation gives $y=y'$. Therefore $(x,y)=(x',y')$. This proves that $\Phi_{a,b}$ is injective.[/guided]

[step:Conclude that the family is mutually orthogonal] For distinct $a,b \in F^\times$, the map $\Phi_{a,b}: F \times F \to F \times F$ is injective by the previous step. Since $|F|=q$, both the domain and codomain have cardinality \begin{align*} |F \times F| = q^2. \end{align*} Since $\Phi_{a,b}$ is injective, the restricted map from $F \times F$ onto the subset $\Phi_{a,b}(F \times F) \subset F \times F$ is a bijection, so this image subset has cardinality $q^2$. Because $F \times F$ itself has cardinality $q^2$, the image subset is all of $F \times F$. Hence $\Phi_{a,b}$ is bijective for every distinct pair $a,b \in F^\times$, so $L_a$ and $L_b$ are orthogonal. Since each $L_a$ is a [Latin square](/page/Latin%20Square) and every distinct pair in the family is orthogonal, the $q-1$ arrays $(L_a)_{a \in F^\times}$ are mutually orthogonal Latin squares of order $q$. [/step]