Arranging numbers from $1$ to $n^2$ in a $n\times n$ board 
Can the numbers $1,2,...,n^2$ be written in the cells of an $n\times n$ board in such a way that any two consecutive numbers are in adjacent cells (sharing a side), and all perfect squares are in the same column?

Note: The original problem comes from All-Russian Mathematical Olympiad 1995 (fourth round, question 8) for the special case where $n=11$.
By counting the number of cells in the left and right side of that column we know that for an odd number $n$, there's not an arrangement satisfying those conditions. So we only have to consider the case where $n$ is even. 
For $n=4k+2$ we can make explicit construction, but writing it down clearly may be difficult. For $n=4k$, it seems that there's no arrangement. However, I have no idea how to prove it.
Can someone solve the case completely where $n$ is even?
 A: The $n=10$ case can indeed be used to prove that all $n=4k+2$ can be done:

The line that follows the numbers in ascending order just sweeps back and forth in a repeating big letter 'C' pattern, and after the last sweep to the right bottom it 'snakes' up, and then goes left to end at $n^2$.  I realize this picture is not a rigorous mathematical proof, but can easily be made into one by induction.  Oh, and by the way, this pattern works for $n=6$ as well, and I assume that if we express the pattern in a clever way, we might even find $n=2$ an instance of that pattern, though that case is in itself trivial.
As far as the $4k$ goes: I am pretty sure that the above is really the only way to make this work (modulo symmetries of course) for the $4k+2$; before you posted your picture of the $n=10$ I had already come to the conclusion that that was the only way to do it. The $4k$ case is going to be likewise restricted in how to proceed the line, and I am convinced that indeed it's not going to work out there. If I have some more time, I may actually try and generate the argument.
