Prove two squares and a cube equal an integer

maxsthekat
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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
 
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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
 
Is 8 the only number you tried as a counter example??
 
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 :)
 
7 doesn't work either. You must have missed it.
 
Ah. You're entirely right! For some reason, I had written down the squares as 1, 2, 4, 9, 16...

Better moral of the story: be sure to have full stomach when doing discrete math :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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