While I wouldn't do the check using root of n but instead the square of the running index since that is faster, he is generally right to just check until the squareroot since every divisor you find already has another factor on the other side of the root you need to multiply with to get n. Think of this as an area of same sice but different side length were you shorten one side so the other get's longer. When you arrive at the perfect square, you don't have to go further since all other sidelengths were already checked "the other way round" (hence add 2 to the output for every divisor except for the root - there the other factor is identical).

why is that?

Elegant solution.

Amazing, first I didn't even checked what's going on in your code, but after I checked it it's an amazing result

Mindblowing solution, good job

I think this is a great kata, it gives clear instructions and the kata itself doesn't seem to have any problems. It is very creative and simple to think through. I think it could do very well if it leaves beta (though the developers don't look to have updated this in a while).