Every number is the sum of three squares with signs 
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:
$$4^k(8n+7).$$
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:
$$31=6^2-2^2-1^2$$
and
$$39=6^2+2^2-1^2.$$
The issue here is that $a$, $b$ and $c$ can be arbitrarily large. For instance:
$$183=14542^2-14541^2-170^2.$$
So I don't really know how to prove or disprove this result, and I think it could go either way.
 A: 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.)
A: 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.
A: $2n+1=(n+1)^2-n^2+0^2$ and $2n=(n+1)^2-n^2-1^2$ cover the odd and even cases respectively.
