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Does anyone know of a proof that every positive integer can be represented as $a^2+b^2+ c^2 + c$ where $a$,$b$ and $c$ are nonnegative integers?

I believe that Goldbach asserted this, along with other ways of representing any number, but I can't find a proof.

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    $\begingroup$ Every positif number ? Did you mean integer ? Otherwise take 3.1 and ... $\endgroup$ – user119228 Feb 4 '14 at 23:18
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    $\begingroup$ Any reasonable reader would conclude that the OP meant "whole number" in that context. $\endgroup$ – Greg Martin Feb 5 '14 at 1:07
  • $\begingroup$ I agree with @GregMartin however I feel the answer should indeed include whole number to avoid ambiguity. $\endgroup$ – Jorge Fernández Hidalgo Feb 5 '14 at 3:28
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Let $N$ be the positive integer that you want to represent as $a^2+b^2+c^2+c$. By the three-squares theorem, $4N+1$ can be written as the sum of three squares, say $4N+1 = x^2+y^2+z^2$. Since all squares are $0$ or $1$ (mod $4$), exactly two of these three squares are even; without loss of generality, suppose $x$ and $y$ are even and $z$ is odd, and write $x=2a$, $y=2b$, and $z=2c+1$. Then $4N+1 = 4a^2 + 4b^2 + 4c^2+4c+1$, which means that $N=a^2+b^2+c^2+c$ as desired.

Of course, the three-squares theorem is not an easy one to prove, but it is standard and classical.

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there exist a and b satisfying $a^2+b^2+c^2+c=k$ if and only if $k-c^2-c$ has all of its factors of the form $4k+3$ an even number of times.

We need to prove for any natural number k there exists a number c such that $k^2-c^2-c$ that has all of its factors of the form $4k+3$ an even number of times.

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  • $\begingroup$ I suspect this idea is not going to lead to a proof, unfortunately. $\endgroup$ – Greg Martin Feb 5 '14 at 1:07
  • $\begingroup$ Well, I'm not sure it's unfortunate, since you found one, I was just offering my two cents, nice answer by the way. $\endgroup$ – Jorge Fernández Hidalgo Feb 5 '14 at 3:26

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