Disprove or prove the statement that every positive integer is the sum of at most two squares and a cube of non-negative integers.
2. The attempt at a solution
I'll call the numbers that can be squares a and b. C will be the cube.
The easiest way to disprove something is to find a counter example. I figured 8 would work if one doesn't include 0 in the allowed values of a, b, and c. However, since the problem says "non-negative integers" instead of "positive" integers, I figure this is in err. (Edit: also, now thinking about it, it says "at most"-- so, 0 must be an allowed state for a, b, and/or c)
Just from glancing at it, I see that we have a situation where, since a can equal b, we can have 2a + 0 or 2a + 1 (depending on c)... So, that would seem to cover all even/odd integers. The only thing that's bugging me is the spacing. Since a and b are perfect squares, I can't simply say that they will cover all of the possible integers.... I'm not sure how to go about proving/disproving this. Does anyone have any thoughts, or any other thoughts in general as to how to approach this problem?