Galloping Queens #2

We can simplify things by looking at the "difference in position" and seeing if it fits a queen or knight move:

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def attacking_squares(x,y):
    #difference in position
    a,b = x[0]-y[0], x[1]-y[1]
    #queen move (same axis or diagonal)
    if a*b == 0 or abs(a) == abs(b):
        return True
    #knight move
    if abs(a*b) == 2:
        return True
    return False

c = 0
for x0 in range(1,9):
    for x1 in range(1,9):
        for y0 in range(1,9):
            for y1 in range(1,9):
                    if attacking_squares((x0,x1),(y0,y1)) == False:
                        c += 1
print c/2

This gives 1120, where we divided by 2 in the last step because each pair of points gets counted twice.

For a more mathematical approach, we can work with values of $(a,b)$ which satisfy $ab=0,$ $|a|=|b|,$ or $|ab|=2,$ and then back out how many pairs of $(x,y)$ this corresponds to. This would get a little messy (don't overcount!), and is a great example of where I'd rather just use a computer rather than working it out...