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The question. Can every $n\in \mathbb N$ can be written:

$$n=a^2\pm b^2\pm c^2$$

where $\pm$ are signs of your choice?

We know with Lagrange's four-square theorem that every integer can be written as the sum of four squares.

Plus, with have Legendre's three-square theorem stated that an integer can not be written as the sum of three squares if, and only if, it is of the form:


So we just have to prove (or disprove) it for every number of this form.

I have checked it until $55$, and it seems to work so far. So the number we have to check are these ones.

For instance:




The issue here is that $a$, $b$ and $c$ can be arbitrarily large. For instance:


So I don't really know how to prove or disprove this result, and I think it could go either way.


marked as duplicate by Dietrich Burde, user91500, Claude Leibovici, Rohan, TastyRomeo Feb 15 '17 at 14:25

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Hang on, it's actually quite simple!

So suppose that we have a number $l$. Suppose that $l=pq$, with $p,q$ having the same parity. That is, both $p$ and $q$ are even, or both $p$ and $q$ are odd.

If this is the case, consider $a= \frac{p+q}{2}, b= \frac{p-q}{2}$. Then, note that $a^2 - b^2 = pq = l$!

For example, $183 = 61 \times 3$, so $a=32$ and $b = 29$, and $32^2-29^2 = 1024 - 841 = 183$.

Now, when can $l$ be written in this form? At least when $l$ is odd, because then you can split it into two odd factors (even if one of those factors is $1$ : for example $7=7 \times 1 = 4^2-3^2$) and carry out the above procedure.

Finally, given an even number, just subtract (or add!) $1^2=1$ to make it an odd number,which can be expressed as a difference of squares.

For example: given $39$, we can write $39=13 \times 3 = 8^2 - 5^2$. Given $78$, we can write $78 = 77 + 1 = 11 \times 7 +1 = 9^2-2^2+1^2$.

What is the reason for so much flexibility? Simple : $(a^2-b^2)$ has a non-trivial factorization, while $a^2+b^2$ does not. This is what makes the whole additive theory of squares (and the Waring problem) so interesting and difficult.


$2n+1=(n+1)^2-n^2+0^2$ and $2n=(n+1)^2-n^2-1^2$ cover the odd and even cases respectively.

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    $\begingroup$ Or, to avoid use of zeroes, $2n+1=(n+3)^2-(n+2)^2-2^2$. $\endgroup$ – Adam Bailey Feb 15 '17 at 11:32

Hint: show that every $n$ which is not of the form $4k+2$ can be written in the form $a^2-b^2+0^2$ for some $a$ and $b$. Then $4k+2$ can be written in the form $a^2-b^2-1^2$ for some $a$ and $b$.

(Thanks to John Bentin for pointing out the silly error in my original post.)

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    $\begingroup$ Thanks, I made a mistake. It is actually anything not 2 mod 4, will edit. $\endgroup$ – Especially Lime Feb 15 '17 at 9:42
  • $\begingroup$ @JohnBentin I do not see the connection either... I want to say odd number since $(n+1)^2-n^2=2n+1$ which is odd. $\endgroup$ – E. Joseph Feb 15 '17 at 9:42
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    $\begingroup$ @E.Joseph you can also get all multiples of $4$ from $(n+1)^2-(n-1)^2$. $\endgroup$ – Especially Lime Feb 15 '17 at 9:47

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