Question 1208189
<pre>
Here is my approach. In the diagram below, I have placed the number of
forbidden places for the second queen in the event that we choose to put the
first queen in that space. I will explain below, where I got them. 

<font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font>
<font color="red"><b>21</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="red"><b>21</b></font>
<font color="red"><b>21</b></font> <font color="green"><b>23</b></font> <font color="red"><b>25</b></font> <font color="red"><b>25</b></font> <font color="red"><b>25</b></font> <font color="red"><b>25</b></font> <font color="green"><b>23</b></font> <font color="red"><b>21</b></font>
<font color="red"><b>21</b></font> <font color="green"><b>23</b></font> <font color="red"><b>25</b></font> <font color="green"><b>27</b></font> <font color="green"><b>27</b></font> <font color="red"><b>25</b></font> <font color="green"><b>23</b></font> <font color="red"><b>21</b></font>
<font color="red"><b>21</b></font> <font color="green"><b>23</b></font> <font color="red"><b>25</b></font> <font color="green"><b>27</b></font> <font color="green"><b>27</b></font> <font color="red"><b>25</b></font> <font color="green"><b>23</b></font> <font color="red"><b>21</b></font>
<font color="red"><b>21</b></font> <font color="green"><b>23</b></font> <font color="red"><b>25</b></font> <font color="red"><b>25</b></font> <font color="red"><b>25</b></font> <font color="red"><b>25</b></font> <font color="green"><b>23</b></font> <font color="red"><b>21</b></font>
<font color="red"><b>21</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="green"><b>23</b></font> <font color="red"><b>21</b></font>
<font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font> <font color="red"><b>21</b></font>

Case 1. We put a queen on the edge of the outer 8x8 square.

If we put a queen on the edge of the outer 8x8 square, there are 21 
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 7 on the same diagonal.  That's 21 forbidden places.  That's
why I put 21's around the outer edge of the 8x8 square and colored them
red. So for each of the 28 places we choose for one queen, there are 
63-21=42 safe places to put the other queen. That's (28)(42)=1176.

Case 2. We put a queen on the edge of the inner 6x6 square.

If we put a queen on the edge of the inner 6x6 square, there are 23 
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 9 on the same diagonal, since we pick up 2 more forbidden
places on the diagonals. That's 23 forbidden places.  That's
why I put 23's around the edge of the 6x6 square and colored them green.
So for each of the 20 places we choose for one queen, there are 63-23=40 
safe places to put the other queen. That's (20)(40)=800. 

Case 3. We put a queen on the edge of the inner 4x4 square.

If we put a queen on the edge of the inner 4x4 square, there are 25 
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 11 on the same diagonal, since we pick up 2 more forbidden
places on the diagonals. That's 25 forbidden places.  That's
why I put 25's around the edge of the 4x4 square and colored them red.
So for each of the 12 places we choose for one queen, there are 63-25=38 
safe places to put the other queen. That's (12)(38)=456. 

Case 4. We put a queen on the center 2x2 square.

If we put a queen on the edge of the center 2x2 square, there are 27 
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 13 on the same diagonal, since we pick up 2 more forbidden
places on the diagonals. That's 27 forbidden places.  That's
why I put 27's for the center 2x2 square and colored them green.
So for each of the 4 places we choose for one queen, there are 63-27=36 
safe places to put the other queen. That's (4)(36)=144.

Total for the 4 cases = 1176 + 800 + 456 + 144 = 2576.

Since we can swap the queens, we take half, and get 1288.

Edwin</pre>