[proofplan]
Write $n=p^r$ with $p$ prime and construct a [finite field](/page/Finite%20Field) $F$ with $|F|=n$; write $F^\times:=F\setminus\{0\}$ for its [multiplicative group](/page/Multiplicative%20Group). For each $a \in F^\times$, define an affine square $L_a(x,y)=ax+y$ on the symbol set $F$. The field axioms make each $L_a$ a [Latin square](/page/Latin%20Square), and solving a two-by-two linear system over $F$ shows that $L_a$ and $L_b$ are [orthogonal Latin squares](/page/Orthogonal%20Latin%20Squares) whenever $a \ne b$. Since $F^\times$ has exactly $n-1$ elements, these squares form the desired family.
[/proofplan]
[step:Construct a finite field with $n$ elements]
By the theorem statement, there exist a prime number $p$ and an integer $r \ge 1$ such that $n=p^r$.
[claim:There exists a field with exactly $p^r$ elements]
[/claim]
[proof]
Let $\mathbb{F}_p$ denote the field with $p$ elements, and define the polynomial $f \in \mathbb{F}_p[X]$ by
\begin{align*}
f(X):=X^{p^r}-X.
\end{align*} We first construct a [splitting field](/page/Splitting%20Field) for $f$ over $\mathbb{F}_p$. Starting from $E_0:=\mathbb{F}_p$, repeatedly choose an irreducible factor $h \in E_j[X]$ of the part of $f$ that has not yet split over $E_j$, and set $E_{j+1}:=E_j[T]/(h)$; the residue class of $T$ is a root of $h$ in $E_{j+1}$. Since the degree of the remaining unsplit part strictly decreases at each adjoining step, this process terminates after finitely many steps. Let $K$ denote the resulting [field extension](/page/Field%20Extension) of $\mathbb{F}_p$ over which $f$ splits into linear factors, and define
\begin{align*}
F := \{\alpha \in K : \alpha^{p^r}=\alpha\}.
\end{align*}
The derivative of $f$ is
\begin{align*}
f'(X)=p^r X^{p^r-1}-1=-1
\end{align*}
in characteristic $p$, since $r \ge 1$ implies that the integer coefficient $p^r$ is zero in $\mathbb{F}_p$. Hence $f$ has no repeated roots in $K$. Therefore $F$ contains exactly $\deg f=p^r$ elements.
It remains to check that $F$ is a field under the operations inherited from $K$. Let $\alpha,\beta \in F$. Since $K$ has characteristic $p$, the [Frobenius endomorphism](/page/Frobenius%20Endomorphism) $\varphi:K\to K$, $\gamma\mapsto \gamma^p$, is a field homomorphism. Iterating $r$ times gives an additive map $\varphi^r:K\to K$, $\gamma\mapsto \gamma^{p^r}$, and hence
\begin{align*}
(\alpha+\beta)^{p^r}=\alpha^{p^r}+\beta^{p^r}=\alpha+\beta,
\end{align*}
so $\alpha+\beta \in F$. Also
\begin{align*}
(-\alpha)^{p^r}=-\alpha^{p^r}=-\alpha,
\end{align*}
so $-\alpha \in F$, and
\begin{align*}
(\alpha\beta)^{p^r}=\alpha^{p^r}\beta^{p^r}=\alpha\beta,
\end{align*}
so $\alpha\beta \in F$. Finally, if $\alpha \ne 0$, then $\alpha^{-1} \in K$ and
\begin{align*}
(\alpha^{-1})^{p^r}=(\alpha^{p^r})^{-1}=\alpha^{-1},
\end{align*}
so $\alpha^{-1} \in F$. Thus $F$ is a subfield of $K$ with exactly $p^r=n$ elements.
[/proof]
Fix such a field $F$, and define $F^\times:=F\setminus\{0\}$, the multiplicative group of nonzero elements of $F$. We shall use $F$ itself as the symbol set for the Latin squares.
[/step]
[step:Define one affine Latin square for each nonzero field element]
For each $a \in F^\times$, define the map $L_a: F \times F \to F$ by
\begin{align*}
L_a(x,y):=ax+y.
\end{align*}
We prove that $L_a$ is a Latin square. Fix $x \in F$. The row map $\rho_{a,x}: F \to F$ defined by
\begin{align*}
\rho_{a,x}(y):=L_a(x,y)=ax+y
\end{align*}
is bijective, because for every $z \in F$ the unique solution of $\rho_{a,x}(y)=z$ is $y=z-ax$.
Fix $y \in F$. The column map $\kappa_{a,y}: F \to F$ defined by
\begin{align*}
\kappa_{a,y}(x):=L_a(x,y)=ax+y
\end{align*}
is bijective, because $a \in F^\times$ and for every $z \in F$ the unique solution of $\kappa_{a,y}(x)=z$ is $x=a^{-1}(z-y)$. Hence every symbol of $F$ appears exactly once in each row and exactly once in each column, so $L_a$ is a Latin square of order $|F|=n$.
[guided]
Fix $a \in F^\times$. We define the map $L_a: F \times F \to F$ by
\begin{align*}
L_a(x,y):=ax+y.
\end{align*}
For this square indexed by $F \times F$ to be Latin, each row and each column must contain every field element exactly once. We therefore verify the two bijectivity conditions directly.
Fix $x \in F$. In row $x$, the entry as a function of the column variable is the map $\rho_{a,x}: F \to F$ defined by
\begin{align*}
\rho_{a,x}(y):=ax+y.
\end{align*}
Given a desired symbol $z \in F$, the equation $\rho_{a,x}(y)=z$ is
\begin{align*}
ax+y=z.
\end{align*}
Solving in the field $F$ gives
\begin{align*}
y=z-ax.
\end{align*}
This solution exists and is unique because addition in a field is cancellative. Hence $\rho_{a,x}$ is bijective.
Now fix $y \in F$. In column $y$, the entry as a function of the row variable is the map $\kappa_{a,y}: F \to F$ defined by
\begin{align*}
\kappa_{a,y}(x):=ax+y.
\end{align*}
Given $z \in F$, the equation $\kappa_{a,y}(x)=z$ is
\begin{align*}
ax+y=z.
\end{align*}
Because $a \in F^\times$, the inverse $a^{-1}$ exists, and the unique solution is
\begin{align*}
x=a^{-1}(z-y).
\end{align*}
Thus each column also contains every symbol exactly once. Therefore $L_a$ is a Latin square of order $n$.
[/guided]
[/step]
[step:Show distinct affine squares are orthogonal]
Let $a,b \in F^\times$ with $a \ne b$. Define the superposition 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))=(ax+y,bx+y).
\end{align*}
We prove that $\Phi_{a,b}$ is bijective. Let $(u,v) \in F \times F$. Solving the two equations
\begin{align*}
ax+y=u
\end{align*}
and
\begin{align*}
bx+y=v
\end{align*}
and subtracting the first equation from the second gives
\begin{align*}
(b-a)x=v-u.
\end{align*}
Since $a \ne b$, the element $b-a$ is nonzero and hence invertible in $F$. Therefore the unique possible value of $x$ is
\begin{align*}
x=(b-a)^{-1}(v-u),
\end{align*}
and then the unique possible value of $y$ is
\begin{align*}
y=u-a(b-a)^{-1}(v-u).
\end{align*}
Substituting these values into the two equations verifies that $\Phi_{a,b}(x,y)=(u,v)$. Hence $\Phi_{a,b}$ is bijective, so $L_a$ and $L_b$ are orthogonal.
[/step]
[step:Count the constructed squares]
The family
\begin{align*}
\mathcal{L}:=\{L_a : a \in F^\times\}
\end{align*}
has one Latin square for each nonzero element of $F$. Since $F$ has $n$ elements, its multiplicative group $F^\times=F\setminus\{0\}$ has $n-1$ elements. The preceding step shows that any two distinct members of $\mathcal{L}$ are orthogonal. Therefore $\mathcal{L}$ is a family of $n-1$ mutually orthogonal Latin squares of order $n$.
[/step]
