Is there an algebraic formula that gives this weird multiplication?: $(-x)\circ(-x)=(-x)\circ(x)$ I'd like to know whether it's possible to give an equivalent algebraic formula, in terms of normal algebraic operations (i.e. $+, -, ×, ÷, x^y$), if possible avoiding $|x|$, for an operator $\circ$, in the domain ℤ such that:
\begin{array}{|r | r r r r | r r r | r r r r}
\hline
\circ & ... & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & ... \\ \hline
\vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\
-4 & ... & -16 & -12 & -8 & -1 & 0 & -4 & -8 & -12 & -16 & ... \\
-3 & ... & -12 & -9 & -6 & -1 & 0 & -3 & -6 & -9 & -12 & ... \\
-2 & ... & -8 & -6 & -4 & -1 & 0 & -2 & -4 & -6 & -8 & ... \\ \hline
-1 & ... & -4 & -3 & -2 & -1 & 0 & -1 & -2 & -3 & -4 & ... \\
0 & ... & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & ... \\
1 & ... & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & ... \\ \hline
2 & ... & -8 & -6 & -4 & -2 & 0 & 2 & 4 & 6 & 8 & ... \\
3 & ... & -12 & -9 & -6 & -3 & 0 & 3 & 6 & 9 & 12 & ... \\
4 & ... & -16 & -12 & -8 & -4 & 0 & 4 & 8 & 12 & 16 & ... \\
\vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{array}
As you see, this is some sort of weird multiplication where $(-)\cdot(-)=(-)\cdot(+)=(+)\cdot(-)=(-)$ but $(+)\cdot(+)=(+)$. I am mostly interested in the subcase of the square in the middle and, if possible, all the $\circ$ operations where at least one of $-1$, $0$ or $1$ is an argument. If that operation can satisfy at least the square at the middle, that will be enough for me. As a last resort I'm disposed to accept division by zero defined in such a way that for all $x$, $x/0=0$.
But it is important to note that, even if it doesn't matter too much what happens outside the domain $\{-1, 0, 1\}$, the operation must be defined for all integers: no modules, no restricted domains.
If that isn't possible, what strategy shall I use to prove it?
ps: Since I'm not a mathematician, I'd like to apologise for any formal or conceptual error I've made. Corrections, though, are more than encouraged.
 A: You can avoid using the absolute value and sign functions if you change the representation of the domain! Let's use a sign-and-magnitude construction. An integer is a pair $(\sigma, n)$ where $\sigma$ is either the symbol $+$ or the symbol $-$, $n$ is an natural number, and we identify $(+,0)=(-,0)$. For a natural number $n$, we may use the shorthand $n=(+,n)$ and $-n=(-,n)$. But whenever we want to define any operation on integers, we may choose to define the operation on the sign and the magnitude independently; and the operation will be well-defined as long as we respect $(+,0)=(-,0)$.
With that setup, the definition is just:
$$(\sigma,x)\circ(\tau,y)=(\min(\sigma,\tau),xy)$$
Then observe that if either $x=0$ or $y=0$, then $xy=0$, so the sign doesn't matter, and we're done.
If you're unhappy with the $\min$ function, we can do away with that too, by representing the sign symbol as a sign bit. The usual computer representation would be $0=+$ and $1=-$. But it's slightly better for us to choose the opposite, $0=-$ and $1=+$. Then we have:
$$(\sigma,x)\circ(\tau,y)=(\sigma\tau,xy)$$
which is as simple as you could hope for!
