[proofplan]
We verify the Latin square condition directly by proving that each row and each column is a bijection from $F$ to $F$. For a fixed row label, the row map is translation by a fixed field element, so equality of two outputs forces equality of the two inputs. For a fixed column label, the column map is multiplication by the nonzero element $a$ followed by translation by a fixed field element, and the nonzero factor $a$ can be cancelled in the field. Since $F$ is finite, injectivity of each row and column map implies bijectivity.
[/proofplan]
[step:Prove injective self-maps of finite sets are bijective]
Let $S$ be a finite set and let $f:S \to S$ be an injective map. Since $f$ is injective, distinct elements of $S$ have distinct images, so $|f(S)|=|S|$. Because $f(S) \subseteq S$ and both sets have the same finite cardinality, $f(S)=S$. Therefore $f$ is surjective, and hence $f$ is bijective.
[/step]
[step:Show each row map is injective and hence bijective]
Fix $x_0 \in F$. Define the row map $R_{x_0}: F \to F$ by $R_{x_0}(y)=L_a(x_0,y)=ax_0+y$ for every $y \in F$.
Let $y_1,y_2 \in F$ satisfy $R_{x_0}(y_1)=R_{x_0}(y_2)$. Then $ax_0+y_1=ax_0+y_2$. Adding the additive inverse of $ax_0$ to both sides in the field $F$ gives $y_1=y_2$. Hence $R_{x_0}$ is injective. Since $F$ is finite and $R_{x_0}$ is an injective map from $F$ to itself, the preceding finite-set step implies that $R_{x_0}$ is bijective.
[/step]
[step:Show each column map is injective and hence bijective]
Fix $y_0 \in F$. Define the column map $C_{y_0}: F \to F$ by $C_{y_0}(x)=L_a(x,y_0)=ax+y_0$ for every $x \in F$.
Let $x_1,x_2 \in F$ satisfy $C_{y_0}(x_1)=C_{y_0}(x_2)$. Then $ax_1+y_0=ax_2+y_0$. Adding the additive inverse of $y_0$ to both sides gives $ax_1=ax_2$. Since $a \in F^\times$, the element $a$ has a multiplicative inverse $a^{-1} \in F$. Multiplying both sides by $a^{-1}$ gives
\begin{align*}
x_1 = a^{-1}ax_1 = a^{-1}ax_2 = x_2.
\end{align*}
Thus $C_{y_0}$ is injective. Since $F$ is finite and $C_{y_0}$ is an injective map from $F$ to itself, the preceding finite-set step implies that $C_{y_0}$ is bijective.
[guided]
Fix a column label $y_0 \in F$. The entries in this column are produced by the map $C_{y_0}: F \to F$ defined by $C_{y_0}(x)=L_a(x,y_0)=ax+y_0$ for every $x \in F$.
To prove that this column contains every symbol of $F$ exactly once, it is enough to prove that $C_{y_0}$ is injective: by the preceding finite-set step, every injective map from a finite set to itself is bijective, and here both the domain and codomain are the finite set $F$.
Take two possible row labels $x_1,x_2 \in F$ and suppose they give the same column entry, so $C_{y_0}(x_1)=C_{y_0}(x_2)$. Using the definition of $C_{y_0}$, this means $ax_1+y_0=ax_2+y_0$. The translation by $y_0$ cannot identify two different elements, because we may add the additive inverse of $y_0$ to both sides. This gives $ax_1 = ax_2$.
Now the hypothesis $a \in F^\times$ is used. Since $a$ is nonzero in the field $F$, it has a multiplicative inverse $a^{-1} \in F$. Multiplying both sides by $a^{-1}$ yields
\begin{align*}
x_1 = a^{-1}ax_1 = a^{-1}ax_2 = x_2.
\end{align*}
Therefore different row labels cannot produce the same entry in the fixed column, so $C_{y_0}$ is injective. Because $F$ is finite and $C_{y_0}:F \to F$ is injective, the preceding finite-set step shows that it is also bijective. Hence the column indexed by $y_0$ contains each symbol of $F$ exactly once.
[/guided]
[/step]
[step:Conclude that the affine table is a Latin square]
The preceding two steps prove that for every $x_0 \in F$, the row map $R_{x_0}:F \to F$ is bijective, and for every $y_0 \in F$, the column map $C_{y_0}:F \to F$ is bijective. Thus each row and each column of the table defined by $L_a(x,y)=ax+y$ contains every element of the symbol set $F$ exactly once. Therefore, by the definition of a [Latin square](/page/Latin%20Square), $L_a$ is a Latin square of order $|F|$.
[/step]
