# Prove two squares and a cube equal an integer

maxsthekat

## Homework Statement

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?

Thanks!

-Max

Gold Member
If I was trying to solve this problem, I would start by working it out for small positive integers:

$1 = 1^2$

$2 = 1^2 + 1^2$

$3 = 1^2 + 1^2 +1^3$

and so on. If you're lucky, you'll find a relatively small counter example. Otherwise, you might see a pattern that suggests a proof. Try that and see what you find.

Petek

Homework Helper
Is 8 the only number you tried as a counter example??

maxsthekat
Ah. By brute force, it appears 15 won't work. Somehow, prior to eating dinner, I must've skipped over that one :P I had thought I checked up until 50.

So, I guess the moral of the story for these kinds of problems is never let elegance get in the way of plain ol' plug and chug.

Thanks for the insight, guys :)