witzar

How about seven points?

Can you put six points on the plane, so that the distance between any two of them is an integer, and no three are collinear?

It use to be one of my favorites. I found it in Mathematical Snapshots, a great book (devoted to recreational math) by Hugo Steinhaus.

It would be true, if all three possibilities were equally probable. Which is not the case.

A problem is similar to (see my solution). You are adding new point on the circle and drawing new chords. For each chord you count intersections with old chords and add 1, this gives you the number of new regions. This is how you arrive at the formula. Last step is to simplify it. But to be honest I just guessed the solution I took the formula from above mentioned problem and made an adjustment for that extra 1s. Few tries and it worked. Edit: Guessing helped me only in the last step (simplification). I needed non-simplified formula to verify guessed one.

Anyone cares to show why the other two solutions I've found (in post #5) are wrong?

Commutativity of multiplication makes it irrelevant, I believe.

It couldn't find 10x10 square.

More squared bishops in attached file. bishop.txt

They do of course. Bushindo made a silly error evaluating your formula.

That's the obvious and natural approach. And that's exactly how I got the formula. When you say "close but", you mean that the formula is wrong?