Prove two squares and a cube equal an integer

In summary, Max tried 8 and 15 as counter examples, and they both didn't work. Petek tried 1, 2, 4, 9, 16 and found that 15 won't work.
  • #1
maxsthekat
55
0

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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  • #2
If I was trying to solve this problem, I would start by working it out for small positive integers:

[itex]1 = 1^2[/itex]

[itex]2 = 1^2 + 1^2[/itex]

[itex]3 = 1^2 + 1^2 +1^3[/itex]

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
 
  • #3
Is 8 the only number you tried as a counter example??
 
  • #4
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 :)
 
  • #5
7 doesn't work either. You must have missed it.
 
  • #6
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 :)
 

Related to Prove two squares and a cube equal an integer

1. How do you prove that two squares and a cube equal an integer?

To prove that two squares and a cube equal an integer, we must show that there exist integers a, b, and c such that a^2 + b^2 + c^3 = n, where n is the integer we want to prove. This can be done through various mathematical techniques such as substitution, induction, or proof by contradiction.

2. What is the significance of proving that two squares and a cube equal an integer?

The significance of this proof lies in its application to various mathematical problems, particularly in number theory and algebra. It can also be used to show the relationship between squares and cubes in a larger context.

3. Can you provide an example of how this concept is used in real life?

Yes, one example is in the field of cryptography, where the RSA encryption algorithm uses the concept of two squares and a cube to generate large prime numbers for secure encryption.

4. What are the common misconceptions about this concept?

One common misconception is that the two squares and a cube must be consecutive. In reality, they can be any two squares and one cube, as long as the sum equals an integer.

5. Is there a general formula for finding the integers a, b, and c in this proof?

Yes, the formula is a = m^2 - n^2, b = 2mn, and c = m^3 - n^3, where m and n are any positive integers. This is known as the Pythagorean triple formula.

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