Prove two squares and a cube equal an integer

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Homework Help Overview

The discussion centers around the statement that every positive integer can be expressed as the sum of at most two squares and a cube of non-negative integers. Participants are exploring the validity of this claim through examples and counterexamples.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to disprove the statement by identifying counterexamples, starting with small integers. They discuss specific numbers like 8, 15, and 7, questioning their validity as sums of two squares and a cube.

Discussion Status

The discussion is active, with participants sharing their findings and counterexamples. Some guidance has been offered regarding the approach of testing small integers, and there is an acknowledgment of the need for careful checking of calculations.

Contextual Notes

There is a focus on the definitions of non-negative integers versus positive integers, which influences the interpretation of the problem. Participants also note the importance of thoroughness in checking potential counterexamples.

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 :)
 

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