[proofplan]
We convert the given [mutually orthogonal Latin squares](/page/Orthogonal%20Latin%20Squares) into an orthogonal array with $n^2$ rows and $m+2$ columns: two coordinate columns and one symbol column for each [Latin square](/page/Latin%20Square). From each array column we build $n-1$ centered incidence vectors in $\mathbb{R}^{n^2}$. Orthogonality of Latin squares and the Latin property imply that vectors coming from different array columns are orthogonal, while the vectors from a fixed column are linearly independent. Counting these independent vectors inside the hyperplane perpendicular to the all-ones vector gives $(m+2)(n-1)\leq n^2-1$, hence $m\leq n-1$.
[/proofplan]
[step:Encode the Latin squares as an orthogonal array]
Let $S:=\{1,\dots,n\}$ be the common symbol set, after relabelling symbols if necessary. We use the standard definitions of a [Latin square](/page/Latin%20Square) and of [orthogonal Latin squares](/page/Orthogonal%20Latin%20Squares). For each $k\in\{1,\dots,m\}$, view the Latin square $L_k$ as a map
\begin{align*}
L_k:S\times S &\to S.
\end{align*}
Define the column index set
\begin{align*}
C:=\{\rho,\kappa\}\cup\{1,\dots,m\},
\end{align*}
where $\rho$ denotes the row-coordinate column, $\kappa$ denotes the column-coordinate column, and $k\in\{1,\dots,m\}$ denotes the symbol column coming from $L_k$.
Define an array map
\begin{align*}
A:S\times S\times C &\to S
\end{align*}
by declaring the row-coordinate value
\begin{align*}
A(i,j,\rho)&:=i,
\end{align*}
the column-coordinate value
\begin{align*}
A(i,j,\kappa)&:=j,
\end{align*}
and, for each $k\in\{1,\dots,m\}$, the Latin-square value
\begin{align*}
A(i,j,k)&:=L_k(i,j).
\end{align*}
We claim that for any two distinct columns $c,d\in C$ and any ordered pair $(a,b)\in S\times S$, there is exactly one cell $(i,j)\in S\times S$ such that
\begin{align*}
A(i,j,c)=a,\qquad A(i,j,d)=b.
\end{align*}
If $(c,d)=(\rho,\kappa)$, the unique cell is $(a,b)$; if $(c,d)=(\kappa,\rho)$, the unique cell is $(b,a)$. If one column is a coordinate column and the other is $k$, the Latin property gives uniqueness in each ordered placement. When $(c,d)=(\rho,k)$, the conditions are $i=a$ and $L_k(a,j)=b$, so the row $a$ of $L_k$ contains the symbol $b$ at exactly one column $j\in S$. When $(c,d)=(k,\rho)$, the conditions are $L_k(i,j)=a$ and $i=b$, so the row $b$ of $L_k$ contains the symbol $a$ at exactly one column $j\in S$. The two cases involving $\kappa$ are analogous but use columns of $L_k$: for $(c,d)=(\kappa,k)$ one fixes $j=a$ and finds the unique $i\in S$ with $L_k(i,a)=b$, while for $(c,d)=(k,\kappa)$ one fixes $j=b$ and finds the unique $i\in S$ with $L_k(i,b)=a$. If $c=k$ and $d=\ell$ are two distinct Latin-square columns, then orthogonality of $L_k$ and $L_\ell$ says exactly that the map $S\times S\to S\times S$ sending $(i,j)$ to $(L_k(i,j),L_\ell(i,j))$ takes each ordered pair in $S\times S$ exactly once.
[guided]
The point of this encoding is to put the row coordinates, column coordinates, and square entries on equal footing. We use $S=\{1,\dots,n\}$ as the symbol set and regard each Latin square as a function
\begin{align*}
L_k:S\times S &\to S.
\end{align*}
The row coordinate, column coordinate, and each Latin square entry become columns of one array. Thus we define
\begin{align*}
C:=\{\rho,\kappa\}\cup\{1,\dots,m\},
\end{align*}
where $\rho$ records the row coordinate, $\kappa$ records the column coordinate, and $k$ records the value of $L_k$. The array map is $A:S\times S\times C\to S$, with $A(i,j,\rho):=i$, $A(i,j,\kappa):=j$, and $A(i,j,k):=L_k(i,j)$.
We now verify the essential array property: in any two distinct columns, every ordered pair of symbols occurs exactly once. For the two coordinate columns $\rho$ and $\kappa$, the pair $(a,b)$ occurs at the unique cell $(a,b)$. For a coordinate column and a Latin-square column, the Latin property supplies uniqueness. For example, if the row coordinate is fixed as $i=a$ and the symbol in $L_k$ is required to be $b$, then the row $a$ of $L_k$ contains the symbol $b$ exactly once, so there is exactly one $j\in S$ such that $L_k(a,j)=b$. The same argument applies with columns fixed instead of rows. Finally, if the two columns come from distinct Latin squares $L_k$ and $L_\ell$, their orthogonality says precisely that the map $S\times S\to S\times S$ sending $(i,j)$ to $(L_k(i,j),L_\ell(i,j))$ is a bijective listing of all ordered pairs of symbols. Hence every ordered pair occurs exactly once in every pair of distinct array columns.
[/guided]
[/step]
[step:Construct centered incidence vectors for the array columns]
Let $V:=\mathbb{R}^{S\times S}$ be the real [vector space](/page/Vector%20Space) of functions from $S\times S$ to $\mathbb{R}$, equipped with the standard [inner product](/page/Inner%20Product) $(f,g)_V:=\sum_{(i,j)\in S\times S} f(i,j)g(i,j)$. Let $\mathbf{1}:S\times S\to\mathbb{R}$ be the constant function defined by $\mathbf{1}(i,j):=1$.
For each column $c\in C$ and each symbol $a\in S$, define the [incidence vector](/page/Incidence%20Vector) $v_{c,a}:S\times S\to\mathbb{R}$ by $v_{c,a}(i,j):=1$ when $A(i,j,c)=a$ and $v_{c,a}(i,j):=0$ when $A(i,j,c)\neq a$. For each $c\in C$ and each $a\in\{1,\dots,n-1\}$, define the centered incidence vector $w_{c,a}:S\times S\to\mathbb{R}$ by $w_{c,a}(i,j):=v_{c,a}(i,j)-\frac{1}{n}\mathbf{1}(i,j)$.
For every $c\in C$ and $a\in S$, the symbol $a$ occurs exactly $n$ times in column $c$. Therefore
\begin{align*}
(v_{c,a},\mathbf{1})_V=n.
\end{align*}
It follows that, for $a\in\{1,\dots,n-1\}$,
\begin{align*}
(w_{c,a},\mathbf{1})_V=(v_{c,a},\mathbf{1})_V-\frac{1}{n}(\mathbf{1},\mathbf{1})_V=n-\frac{1}{n}n^2=0.
\end{align*}
Thus all vectors $w_{c,a}$ lie in the hyperplane
\begin{align*}
H:=\{f\in V:(f,\mathbf{1})_V=0\}.
\end{align*}
Since $\dim V=n^2$ and $\mathbf{1}\neq 0$, we have $\dim H=n^2-1$.
[/step]
[step:Show vectors from different columns are orthogonal]
Let $c,d\in C$ be distinct columns and let $a,b\in\{1,\dots,n-1\}$. By the array property proved above, exactly one cell has simultaneous values $a$ in column $c$ and $b$ in column $d$. Hence
\begin{align*}
(v_{c,a},v_{d,b})_V=1.
\end{align*}
Also,
\begin{align*}
(v_{c,a},\mathbf{1})_V=(v_{d,b},\mathbf{1})_V=n,
\qquad
(\mathbf{1},\mathbf{1})_V=n^2.
\end{align*}
Expanding the centered inner product by bilinearity gives
\begin{align*}
(w_{c,a},w_{d,b})_V=\left(v_{c,a}-\frac{1}{n}\mathbf{1},v_{d,b}-\frac{1}{n}\mathbf{1}\right)_V.
\end{align*}
Using the already computed values of the four inner products, this becomes
\begin{align*}
(w_{c,a},w_{d,b})_V=(v_{c,a},v_{d,b})_V-\frac{1}{n}(v_{c,a},\mathbf{1})_V-\frac{1}{n}(\mathbf{1},v_{d,b})_V+\frac{1}{n^2}(\mathbf{1},\mathbf{1})_V=1-1-1+1=0.
\end{align*}
Therefore every centered incidence vector from column $c$ is orthogonal to every centered incidence vector from column $d$ whenever $c\neq d$.
[/step]
[step:Prove the centered vectors from one column are linearly independent]
Fix a column $c\in C$. Suppose [real numbers](/page/Real%20Numbers) $\alpha_1,\dots,\alpha_{n-1}\in\mathbb{R}$ satisfy
\begin{align*}
\sum_{a=1}^{n-1}\alpha_a w_{c,a}=0.
\end{align*}
For any symbol $r\in S$, choose a cell $(i_r,j_r)\in S\times S$ such that $A(i_r,j_r,c)=r$; such a cell exists because each symbol occurs in column $c$. Evaluating at a cell with column value $r\in\{1,\dots,n-1\}$ gives
\begin{align*}
0=\sum_{a=1}^{n-1}\alpha_a\left(v_{c,a}(i_r,j_r)-\frac{1}{n}\right)=\alpha_r-\frac{1}{n}\sum_{a=1}^{n-1}\alpha_a.
\end{align*}
Evaluating at a cell with column value $n$ gives
\begin{align*}
0=\sum_{a=1}^{n-1}\alpha_a\left(v_{c,a}(i_n,j_n)-\frac{1}{n}\right)=-\frac{1}{n}\sum_{a=1}^{n-1}\alpha_a.
\end{align*}
Thus $\sum_{a=1}^{n-1}\alpha_a=0$, and then the previous displayed identity gives $\alpha_r=0$ for every $r\in\{1,\dots,n-1\}$. Hence the vectors
\begin{align*}
w_{c,1},\dots,w_{c,n-1}
\end{align*}
are linearly independent.
[guided]
We must check that centering the incidence vectors has not introduced a linear dependence inside a single column. Fix one column $c\in C$ and assume that a real linear combination vanishes:
\begin{align*}
\sum_{a=1}^{n-1}\alpha_a w_{c,a}=0,
\end{align*}
where $\alpha_1,\dots,\alpha_{n-1}\in\mathbb{R}$.
The useful way to test this identity is to evaluate it on cells whose value in column $c$ is prescribed. Since every symbol occurs in column $c$, for each $r\in S$ we may choose a cell $(i_r,j_r)\in S\times S$ with
\begin{align*}
A(i_r,j_r,c)=r.
\end{align*}
If $r\in\{1,\dots,n-1\}$, then $v_{c,r}(i_r,j_r)=1$ and $v_{c,a}(i_r,j_r)=0$ for $a\neq r$. Therefore
\begin{align*}
0=\sum_{a=1}^{n-1}\alpha_a w_{c,a}(i_r,j_r)=\sum_{a=1}^{n-1}\alpha_a\left(v_{c,a}(i_r,j_r)-\frac{1}{n}\right)=\alpha_r-\frac{1}{n}\sum_{a=1}^{n-1}\alpha_a.
\end{align*}
This says that every coefficient $\alpha_r$ equals the same average-like quantity $\frac{1}{n}\sum_{a=1}^{n-1}\alpha_a$.
Now evaluate the same zero vector at a cell $(i_n,j_n)$ whose column value is the omitted symbol $n$. For this cell, all incidence values $v_{c,a}(i_n,j_n)$ with $a\leq n-1$ vanish. Hence
\begin{align*}
0=\sum_{a=1}^{n-1}\alpha_a w_{c,a}(i_n,j_n)=\sum_{a=1}^{n-1}\alpha_a\left(0-\frac{1}{n}\right)=-\frac{1}{n}\sum_{a=1}^{n-1}\alpha_a.
\end{align*}
Thus $\sum_{a=1}^{n-1}\alpha_a=0$. Substituting this into
\begin{align*}
\alpha_r-\frac{1}{n}\sum_{a=1}^{n-1}\alpha_a=0
\end{align*}
gives $\alpha_r=0$ for every $r\in\{1,\dots,n-1\}$. Therefore the $n-1$ centered vectors belonging to the fixed column $c$ are linearly independent.
[/guided]
[/step]
[step:Count dimensions in the hyperplane orthogonal to the all-ones vector]
We now show that the full family
\begin{align*}
\mathcal{W}:=\{w_{c,a}:c\in C,\ 1\leq a\leq n-1\}
\end{align*}
is linearly independent in $H$. Suppose real numbers $\alpha_{c,a}\in\mathbb{R}$ satisfy
\begin{align*}
\sum_{c\in C}\sum_{a=1}^{n-1}\alpha_{c,a}w_{c,a}=0.
\end{align*}
For each $c\in C$, define
\begin{align*}
u_c:=\sum_{a=1}^{n-1}\alpha_{c,a}w_{c,a}\in H.
\end{align*}
Then $\sum_{c\in C}u_c=0$. Taking the inner product with $u_d$ for a fixed $d\in C$ gives
\begin{align*}
0=\left(\sum_{c\in C}u_c,u_d\right)_V=(u_d,u_d)_V+\sum_{\substack{c\in C,\ c\neq d}}(u_c,u_d)_V.
\end{align*}
The cross terms vanish by the orthogonality of vectors from different columns, so
\begin{align*}
(u_d,u_d)_V=0.
\end{align*}
Positive definiteness of the standard inner product on $V$ gives $u_d=0$. Since $d\in C$ was arbitrary, $u_c=0$ for every $c\in C$. The [linear independence](/page/Linear%20Independence) inside each fixed column then implies $\alpha_{c,a}=0$ for every $c\in C$ and every $a\in\{1,\dots,n-1\}$.
Thus $\mathcal{W}$ is a linearly independent subset of $H$. Its cardinality is
\begin{align*}
|\mathcal{W}|=|C|(n-1)=(m+2)(n-1).
\end{align*}
Since $\dim H=n^2-1$, we obtain
\begin{align*}
(m+2)(n-1)\leq n^2-1.
\end{align*}
Because $n\geq 2$, we may divide by $n-1>0$:
\begin{align*}
m+2\leq \frac{n^2-1}{n-1}=n+1.
\end{align*}
Therefore
\begin{align*}
m\leq n-1.
\end{align*}
This proves that there are at most $n-1$ mutually orthogonal Latin squares of order $n$.
[/step]
