a.) Let $n$ be even. Find a permutation $x_1, x_2,\dots, x_n$ of the elements of $\mathbb{Z}_n$ such that the differences $x_{i+1} - x_i, 1 \leq i < n$, are all different.
Say if I take $n=6$, then one possible permutation would be $6\,1\,5\,2\,4\,3$. How would I generalize this? I'm suspecting ring subtraction at this point.
b.) Show that this is not possible if $n$ is odd.
I know that I have add up the differences, but not sure how to.
c.) Consider the permutation of (a). Define $a_{ij} := x_i+x_j$. Show that the square with entries $a_{ij}, 1 \leq i, j \leq n$, is a Latin square that has the following property: The $n(n-1)$ adjacent pairs $(a_{ij}, a_{i,j+1})$ are different. Such a square is called row-complete. This square is also column-complete.
Any help, tips, or a fully worked out solution would be appreciated!
(a) Use the same pattern as their given example. Interleave the higher half of the numbers in descending order with the lower half of the numbers in ascending order. For example, with $n=10$, $$10,1,9,2,8,3,7,4,6,5$$ You just need to prove this works in general.
(b) On the one hand, the sum of all the differences is telescoping:$$(x_2-x_1)+(x_3-x_2)+\dots+(x_n-x_{n-1})=x_{n}-x_1.$$ On the other hand, $$(x_2-x_1)+(x_3-x_2)+\dots+(x_n-x_{n-1})=1+2+\dots+(n-1),$$ since all differences are distinct and nonzero. What is $1+2+\dots+(n-1)$ modulo $n$, when $n$ is odd? If you cannot guess, look at small values of odd $n$ to find a pattern, then prove it. Using this, use the equation $x_n-x_1=1+2+\dots+(n-1)$ to get a contradiction.
(c) Suppose that $(a_{i,j},a_{i,j+1})=(a_{h,k},a_{h,k+1})$. In particular, this means $$ a_{i,j+1}-a_{i,j}=a_{h,k+1}-a_{h,k},\\ (x_i+x_{j+1})-(x_i+x_j)=(x_h+x_{k+1})-(x_h+x_k)\\ x_{j+1}-x_j=x_{k+1}-x_k $$ Since the differences $x_{i+1}-x_i$ are distinct, this implies $j=k$. Since $a_{i,j}=a_{j,k}$, we have $x_i+x_j=x_h+x_k$, so subtracting $x_j=x_k$ from both sides, we get $x_{i}=x_h$. Since the values of $x_i$ are distinct, this implies $i=h$. Therefore, the ordered pairs $(a_{i,j},a_{i,j+1})$ are distinct.