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I have two questions.

  1. Does anyone know what the name of this identity is or what I should look up to find out more information about it?

  2. How is this identity used to prove that all integers can be represented as $\sum_{k=1}^n\pm k^2$?

Here and here are the two places where I have seen this identity. I have been unable to find out more about these topics as I don't know what I should be looking for.

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AlgorithmsX
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  • This would be a "derivative of the difference of squares" identity. It follows literally from the difference of squares formula. – Sarvesh Ravichandran Iyer Sep 25 '16 at 04:24
  • Thank you, that answers my first question. What about the second one? – AlgorithmsX Sep 25 '16 at 04:28
  • Do you need an answer via the above identity? Because I think we can answer this question without needing that identity. Besides, you want only the sum or difference of two squares, right? Otherwise, I could just add 1s until I reach the number. – Sarvesh Ravichandran Iyer Sep 25 '16 at 04:39
  • I misread the original post. They were talking about distinct integers. If they weren't distinct, using ones would work. – AlgorithmsX Sep 25 '16 at 04:44
  • Okay, so I have to use distinct integers, but I can add and subtract how many ever squares I want? Then I will tell you, every number that is not an even non-multiple of four can be written as the difference of squares. – Sarvesh Ravichandran Iyer Sep 25 '16 at 04:50
  • I would suggest that you look at the links I provided, as they express what I am looking for much better than I can. – AlgorithmsX Sep 25 '16 at 04:54
  • https://math.stackexchange.com/questions/1079575 – J. Yu Jun 07 '18 at 06:23

2 Answers2

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For 2, you now know you can add $4$ to any number you can represent. If you can represent $0, 1, 2, 3$, you can represent any number by adding in the proper amount of $4$s. You then say $0=0, 1=1^2, 2=6^2-5^2-3^2, 3=6^2-5^2-3^2+1^2$ and declare victory.

Ross Millikan
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  • Thank you. Do you know where I can find out about more ideas like this, such as the similar results for higher powers I cited and Reyley's Theorem? – AlgorithmsX Sep 25 '16 at 15:51
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Actually, the links you provide, also have references where to find more about these topics. Also, why do you want to restrict yourself to the sum of squares? There are so many interesting questions on the sum of cubes. You have given yourself the link here. What do you think about this question, or this one?

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Dietrich Burde
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  • I figured that if I had the name or something to look for to get these general identities and how this proves some result, I could find more things like it. As far as I can tell, there is no way to go from $(n+3)^2-(n+2)^2-(n+1)^2+n^2$ to Reyley's Theorem or any related results. Also, the sources I cited lacked the proof that if they could prove that they could get one through four, they could get all integers. – AlgorithmsX Sep 25 '16 at 15:50