Every cell of a $m \times n$ chess board, $m\ge 2,n\ge 2$, is colored with one of four possible colors, e.g white, green, red, blue. We call such coloring good if the four cells of any $2\times 2$ square of the chessboard are colored with pairwise different colors. Determine the number of all good colorings of the chess board.

[i]Proposed by N. Beluhov[/i]
1. **Define the colors and initial conditions:**
   Let the four colors be denoted by \(1, 2, 3, 4\). We are given an \(m \times n\) chessboard where \(m \geq 2\) and \(n \geq 2\). We need to determine the number of good colorings such that any \(2 \times 2\) sub-square has all four cells colored differently.

2. **Fix the colors in the first two rows:**
   Without loss of generality (WLOG), assume that the colors in cells \((1,1)\) and \((2,1)\) are \(1\) and \(3\) respectively. This implies that the colors in cells \((1,2)\) and \((2,2)\) must be \(2\) and \(4\) respectively to satisfy the condition of having all four colors in the \(2 \times 2\) sub-square.

3. **Determine the number of valid configurations for the first two rows:**
   For the first two rows, there are \(2^n\) possible arrangements. However, not all of these arrangements will predetermine the rest of the table. Specifically, there are \(2^n - 4\) configurations that will predetermine the rest of the table, and 4 configurations that do not. The 4 configurations that do not predetermine the table are:
   - \((1,3), (2,4)\)
   - \((3,1), (2,4)\)
   - \((1,3), (4,2)\)
   - \((3,1), (4,2)\)

4. **Count the number of good colorings for the remaining rows:**
   For the 4 configurations that do not predetermine the table, the first number in every row determines that row. We can only choose a number that does not appear in the previous row, giving us 2 choices for each row. Since there are \(m-2\) rows left, for each of the 4 configurations, there are \(2^{m-2}\) good colorings.

5. **Calculate the total number of good colorings:**
   The total number of good colorings for the pair \((1,3)\) in cells \((1,1)\) and \((2,1)\) is:
   \[
   2^n - 4 + 4 \cdot 2^{m-2}
   \]
   Since there are 6 possible pairs of colors that can be placed in cells \((1,1)\) and \((2,1)\), the final answer is:
   \[
   6 \left(2^n - 4 + 4 \cdot 2^{m-2}\right)
   \]

The final answer is \(\boxed{6 \left(2^n - 4 + 4 \cdot 2^{m-2}\right)}\)