Euler cycle in a $m\times n$ rectangular grid. Let $G=(V,E)$ a graph which consists in an $m\times n$ rectangular grid as the image shows:

I need to find the values of $m,n$ for which this graph has an Euler cycle (or euler circuit, don't repeat edges, and go through them all). $m$ and $n$ refers to the vertices positions as in a matrix ($m$ for rows, $n$ for columns). I could find Euler cycles for $1\times 1$ and $1\times 2$ (and of course $2\times 1$) but i'm not able to find euler cycles for bigger values. I'm trying to think about the parity of $m$ and $n$ but can't conclude anything.
Edit: $m$ and $n$ are edges on the grid, we sould have $m+1,n+1$ vertices as @Daniel Schepler pointed out.
 A: Well, you can't find Eulerian cycle for bigger values because it doesn't exist. Eulerian cycle in your graph exists only if $m = 1$ and $n = 1$, actually.
There is a theorem: Eulerian cycle in a connected graph exists if and only if the degrees of all vertices are even.
If $m>1$ or $n> 1$, you will have vertices of degree 3 (which is odd) on the borders of your grid, i.e. vertices that adjacent to exactly 3 edges. And you will have lots of such vertices as $m$, $n$ grow.
Moving on to the proof, we need just "only if" part in your case, and it is simple. Imagine you have an Eulerian cycle. It means that you can somehow traverse all the edges starting from some vertex and finishing in the same vertex. Let's label each edge with an arrow, pointing accordingly to the direction of this traverse. Now notice that we must have entered each vertex the same number of times that we exited from it. But that means we must have the same amounts of "in" arrows and "out" arrows at each particular vertex. This is not possible if we have at least one vertex with an odd degree (with degree 3, for example).
If you are speaking about Eulerian path (not necessarily cycle), then it exists only in three cases when $mn \leqslant 2$. The difference is that the path, opposing to the cycle, can start and finish at different vertices.
The proof is similar but we need to additionally say that start and finish vertices can be the only two vertices where we have numbers of "in" and "out" arrows not equal. So they are our 2 vertices with an odd degree.
Actually, if a graph has $2k$ vertices of odd degree, you can traverse all edges of this graph using $k$ paths (not cycles). In other words you can "cover" all edges with $k$ edge-non-intersecting paths if you have $2k$ odd vertices. For example, to cover all edges of $4 \times 4$ grid, you'll need 6 paths, because you have 3 odd vertices on each of 4 borders of the grid, totally 12 odd vertices. But this is out of topic of the original question, I believe.
