Even more fun facts: if you change the problem instead to 4 queens and one knight, there is exactly one solution up to symmetries (rotations and flipping). Here it is:

    ..N.....
    ........
    ........
    ....Q...
    .....Q..
    ...Q....
    ......Q.
    ........

Edit: even more fun facts: if we take the standard piece values of Q=9, R=5, B/N=3, then we can ask for the smallest piece budget that attacks every square. The cheapest possible configuration is 24 points, you can see one with 8 bishops:

    ........
    ....B...
    ....B...
    .B......
    ...B.B..
    .....B..
    ..B.....
    ....B...

Which has pleasing symmetry when you view it as a composition of light-square bishops and dark-square bishops.