Solution:

Note that we may label half of the cells in our board the number $0$ and the other half $1$, in such a way that squares labeled $0$ are adjacent only to squares labeled $1$ and vice versa. In other words, we make this labeling in a 'checkerboard' pattern. Since cells in an 'X' formation are all labeled with the same number, the number of ways to color the cells labeled $0$ is $\sqrt{N}$, and the same is true of coloring the cells labeled $1$.

Let $a_{2n}$ be the number of ways to color the squares labeled $0$ in a $3$ by $2n$ grid without a monochromatic 'X' formation; we want to find $a_{10}$. Without loss of generality, let the rightmost column of our grid have two cells labeled $0$. Let $b_{2n}$ be the number of such colorings on a $3$ by $2n$ grid which do not have two black squares in the rightmost column and do not contain a monochromatic 'X', which we note is also the number of such colorings which do not have two white squares in the rightmost column.

Now, we will establish a recursion on $a_{2n}$ and $b_{2n}$. We have two cases:

- Case 1: All three squares in the last two columns are the same color. For $a_{2n}$, there are $2$ ways to color these last three squares, and for $b_{2n}$ there is $1$ way to color them. Then, we see that there are $b_{2n-2}$ ways to color the remaining $2n-2$ columns.

- Case 2: The last three squares are not all the same color. For $a_{2n}$, there are $6$ ways to color the last three squares, and for $b_{2n}$ there are $5$ ways to color them. Then, there are $a_{2n-2}$ ways to color the remaining $2n-2$ columns.

Consequently, we get the recursions $a_{2n} = 6 a_{2n-2} + 2 b_{2n-2}$ and $b_{2n} = 5 a_{2n-2} + b_{2n-2}$. From the first equation, we get that $b_{2n} = \frac{1}{2} a_{2n+2} - 3 a_{2n}$. Plugging this into the second equation results in the recursion

$$\frac{1}{2} a_{2n+2} - 3 a_{2n} = 5 a_{2n-2} + \frac{1}{2} a_{2n} - 3 a_{2n-2} \Rightarrow a_{2n+2} = 7 a_{2n} + 4 a_{2n-2}$$

Now, we can easily see that $a_0 = 1$ and $a_2 = 2^3 = 8$, so we compute $a_{10} = 25636$.