Every number is the sum of three squares with signs

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.

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.)