I am having trouble with the following question any help would be great!
Suppose that $n$ is an odd positive integer with $n\geq3$. Let $A$ be the $n\times n$ Latin square whose rows and columns are indexed by the elements of $\mathbb{Z}_n= {0,1,2,...,n-1}$ with $(i,j)$-entry $a_{i,j}=i+j$ and let $B$ be the $n\times n$ array with $(i,j)$-entry $b_{i,j}=i+2j$ (where the addition and multiplication is that of $\mathbb{Z}_n$). Show that $B$ is a Latin square and the Latin square $A$ and $B$ are orthogonal.
I understand what it mean for two Latin square to be orthogonal but I am unsure how to go about this.
A Latin square is a square in which each number appears exactly once in each row and column. It is clear that both $A$ and $B$ are Latin squares.
Proof: By contradiction. Suppose there are 2 entries in $A$ in the same row that have the same value, then $i_1 + j_1 = 1_2 + j_2$. But since $i_1 = i_2$, this implies that $j_1 = j_2$. The same proof works for 2 entires being the same in a column.
The proof works for $B$, since $2$ has a multiplicative inverse.
2 Latin squares are orthogonal if the corresponding pairs of entries $(a_{ij}, b_{ij})$ are distinct. We will show that $A$ and $B$ are orthogonal.
Proof: By contradiction. Suppose not. Then there are 2 distinct entries such that $( a_{i_1 j_1}, b_{i_1 j_1} ) = ( a_{i_2 j_2}, b_{i_2 j_2} ) $ Let's write out what this means. In the first coordinate, we have $i_1 + j_1 = i_2 + j_2$. In the second coordinate, we have $i_1 + 2j_1 = i_2 + 2j_2$. Taking the difference of these equations, we get that $j_1=j_2$, and hence that $i_1=i_2$. This contradicts the assumption that the entries were distinct.
It's easy to see how for a prime $p$, we can thus form $p-1$ Latin squares which are pairwise orthogonal. This is the maximum possible.