Proving a solution for the $n$-Queens Puzzle Given an $n$ x $n$ board, assume that $n \geq 5$ and that $n$ is not divisible by $2$ or $3$. Prove that the following positioning of $n$ queens $Q_0, Q_2, ..., Q_{n-1}$ works, i.e no two queens threaten each other:
For $0 \leq i \leq n-1$ we position the queen $Q_i$ on the field $(i, 2i \text{ } \pmod n)$.
Here we're using the ($x$-coordinate, $y$-coordinate) coordinate system, where $x$ describes the horizontal position, and $y$ the vertical. For example, in the formula above, $x$ would be $i$, and $y$ would be $2i \pmod n$.

My idea was to prove by contradiction and break up each case on how the queens are positioned to not threaten each other (horizontally, vertically and diagonally), but I can't see what follows. Can someone offer their thoughts or point me in the right direction?
Consider $i &lt; j$ between 0 and $ n-1$. Of course, the first coordinate can’t be same for any two queens, hence they aren’t in same row. 

Now, suppose, for the sake of contradiction, they lie on the same column. Then, $2(i-j) = nk$  for some natural $k$. Now, as n doesn’t contain 2 as it’s divisor, $k$ must be even. So, either $k$ is 0, which is not possible because that will contradict our assumption that $i$ and $j$ are not equal, or $k\geq 2$. In the latter case, $LHS = 2(i-j) &lt; 2(n-1) &lt; 2n \leq nk$, and hence $LHS \ne RHS$. Hence, we conclude that our assumption must have been incorrect and two queens positioned in this manner can’t lie on same column.

You still need to figure out the diagonal case.