Self-orthogonal Latin squares A Latin square $A$ is called self-orthogonal if $A$ and $A^{T}$ are orthogonal Latin squares.
Use the elements of $\;\mathbb{Z}_v$ as the names of the rows and columns of your Latin square. Let $\boldsymbol{A}=(a_{ij})$ such that $a_{ij}=2i-j\in \mathbb{Z}_v$. Prove that this forms a self-orthogonal Latin square if the $\gcd(v,6)=1$. I have just tried to see that this works with small orders like $v=5$:
$$\boldsymbol{A}=\begin{pmatrix}
0 & 4 & 3 & 2 & 1\\
2 & 1 & 0 & 4 & 3\\
4 & 3 & 2 & 1 & 0\\
1 & 0 & 4 & 3 & 2\\
3 & 2 & 1 & 0 & 4\\
\end{pmatrix}\implies\boldsymbol{A}^{T}=\begin{pmatrix}
0 & 2 & 4 & 1 & 3\\
4 & 1 & 3 & 0 & 2\\
3 & 0 & 2 & 4 & 1\\
2 & 4 & 1 & 3 & 0\\
1 & 3 & 0 & 2 & 4\\
\end{pmatrix}$$
So it seems to work and I could probably sketch a general construction, but I don't know how the assumption $\gcd(v,6)=1$ makes this all work. Can someone explain?
 A: Let's assume $\gcd(v, 6)=1$ and work in the ring $\Bbb{Z}_v$ except where otherwise stated (so I can write $i = i'$ rather than $v \mid i - i'$ or $i \equiv i' \,(\mathrm{mod}\,v)$).
We have $a_{ij} = a_{ij'}$ iff $2i -j = 2i - j'$ iff $j = j'$. Also we have $a_{ij} = a_{i'j}$ iff $2i - j = 2i' - j$ iff $2(i - i') = 0$ iff $i = i'$, since $2$ is invertible in $\Bbb{Z}_v$ as $\gcd(v, 6) \neq  1$. Hence $A = (a_{ij})_{i, j \in \Bbb{Z}{_v}}$ is a Latin square.
$A$ is self-orthogonal if the pairs $(A_{ij}, A^T_{ij}) = (a_{ij}, a_{ji})$ are all distinct. but $(a_{ij}, a_{ji}) = (a_{i'j'}, a_{j'i'})$ iff the following equations hold:
$$
\begin{align*}
2i - j &= 2i' - j' \\
2j - i &= 2j' - i'
\end{align*}
$$
Since $2$ is invertible as $\gcd(v, 6) = 1$, we can rewrite these equations as:
$$
\begin{align*}
i - i' &= \frac{1}{2}(j - j') \\
j - j' &= \frac{1}{2}(i - i')
\end{align*}
$$
whence
$$
\begin{align*}
i - i' &= \frac{1}{4}(i - i') \\
j - j' &= \frac{1}{4}(j - j')
\end{align*}
$$
which gives us
$$
\begin{align*}
4(i - i') &= i - i' \\
4(j - j') &= j - j'
\end{align*}
$$
but if $4x = x$ in $\Bbb{Z}_v$, we must have that $v \mid 3x$ in $\Bbb{Z}$ implying that $x =0$ in $\Bbb{Z}_v$, since $v$ and $3$ are coprime by our assumption that $\gcd(v, 6) = 1$. Hence $i = i'$ and $j = j'$ and $A$ is indeed self-orthogonal.
