$\mathbb{Z}$: Unique operation s.t. $1\star m=m$ and right distributivity hold?

Question

Given $\mathbb{Z}$ and the usual addition $+$ on it, do we have unicity of a binary operation $\star$ such that \begin{align*} \tag{1}1\star m&=m\\ \tag{2}(m+n)\star p&=m\star p+n\star p \end{align*} for all $m,n,p\in\mathbb{Z}$ ?

In an effort to show that the answer is yes, I tried to prove that $$ (a-b)\star(c-d)=ac-ad-bc+bd $$ for all $a,b,c,d\in\mathbb{N}$, where the multiplication on the right-hand side is the usual one in $\mathbb{N}$.

I can show that such a $\star$ satisfies the following:

1. The restriction of $\star$ on $\mathbb{N}$ coincides with the usual multiplication in $\mathbb{N}$.
2. For all $m\in\mathbb{Z}$, $0=0\star m$...

Answer 1

You have proved that for any operation $\star$ satisfying the requirements $n\star m = nm$ for $n,m\in \mathbb{N}$.

Now say that $-n\in -\mathbb{N}$. Then you would need $$ 0 = 0\star m = (n + (-n))\star m = n\star m + (-n)\star m. $$ for all $m\in \mathbb{N}$. So $(-n)\star m = -(n\star m) = -nm$.

Now say $n\in \mathbb{N}$ and $-m\in -\mathbb{N}$ and let's show that $n\star (-m) = -nm$. By induction (you can do this) $$ n\star (-m) = n(1\star (-m))= n(-m) = -nm. $$

So let now $n,m\in \mathbb{Z}$. We want to prove that $n\star m = nm$. If $n$ or $m$ is $0$, then you are done.

1. If $n,m > 0$, then you are done.
2. If ($n > 0$ and $m<0$) or ($n<0$ and $m>0$), then you are done by what we did earlier.
3. If $n < 0$ and $m<0$, then $n\star m = (-(-n))\star -((-m)) = -((-n)\star (-(-m))) = (-n)\star (-m) = (-n)(-m) = nm$.

So in all cases $n\star m = nm$.