Prove that matrix A so that $a_{ij}$ = (m+1)*(i+j) mod n is a symmetric Latin Square with different elements on its diagonal (m is a freely chosen Z) It's obvious that it's symmetric because $a_{\left(i+1\right)j}=\left(m+1\right)\left(i+1+j\right) = a_{i\left(j+1\right)}=\left(m+1\right)\left(i+j+1\right)$, but how can I prove that it's a Latin square and that it's diagonal consists of different elements?
I thought about showing that the sum of the elements in any row/column is equal to 0+1+...+n-1, but it's not working out
$\left(m+1\right)\left[\left(i+j\right)+\left(i+j+1\right)+...+\left(i+j+n-1\right)\right] mod(n)$
$\left(m+1\right)\left[n\left(i+j\right)+\left(1\right)+...+\left(n-1\right)\right] mod(n)$
$\left(m+1\right)\left[n\left(i+j\right)+\frac{n\left(n-1\right)}{2}\right] mod(n)$
$\left(m+1\right) \left[0\right]mod(n)$
Which will get me 0 or k*n where k is an integer, so we only know that if summing up any row/column (I get the same for other rows also)
 A: It's true if and only if $m+1$ and $n$ are coprime.  In particular, the cells $(i,j)$ and $(i,j+n/\gcd(m+1,n))$ are in the same row and contain the same symbol.  However, they are the same cell if and only if $m+1$ and $n$ are coprime, i.e., provided $\gcd(m+1,n)=1$).
As you say, it's obvious that it's symmetric, since $a_{ij} = a_{ji}$ by commutativity of + (i.e., we know $i+j = j+i$).  However, checking $a_{(i+1)j} = a_{i(j+1)}$ does not show it's symmetric (it shows some other property, not symmetry).  We need only show the rows are Latin, i.e., there are no repeated symbols in any row (the columns are the same but transposed, so we don't need to re-check them).  I don't think summing the rows is going to lead to a proof: it won't show there are no duplicate symbols in each row.
To prove the rows are Latin, suppose $a_{ij} = a_{ij'}$ for columns $j$ and $j'$.  Then we deduce from the definition of $a_{ij}$ and $a_{ij'}$ that $$j(m+1) \equiv j'(m+1) \pmod n.$$
If $m+1$ and $n$ are coprime, then we can "cancel out" the $m+1$ (or more formally, multiply by the multiplicative inverse of $m+1$ on both sides) to obtain
$j \equiv j' \pmod n$ which implies $a_{ij}$ and $a_{ij'}$ refer to the same cell, and thus there are no duplicate symbols in the arbitrary row $i$.
A: $(m+1)$ is any number $\in$ Z. I think , there is relation between $m$ and $n$ to get the require . Also, you mean the operation * is usual multiplication.
