Q: Show that $(\mathbb{Z},|,+,0,1)$ defines multiplication in $\mathbb{Z}$.
I know how to do this in $\mathbb{N}$, but I'm stuck trying to do this is $\mathbb{Z}$. The idea I have is to define lowest common multiple ($lcm$) and using property of coprimes to define $x^2$. From there $(x+y)^2$ will allow me to define $x\cdot y$.
So I first define $$z = lcm(x,y) \iff x\mid z \land y \mid z \land \forall m(x \mid m \land y\mid m \rightarrow z \mid m).$$ Then I define $$2x^2 = x(x+1) + x(x-1) = lcm(x,x+1) + lcm(x,x-1).$$ However, I realised that my definition of $lcm(x,y)=z$ could give $z$ as either a negative or positive integer. Without the ordering relation ($<$), I am having difficulty finding a way to define $lcm$ such that I only get the positive $z$. How can I overcome this?
It seems I was wrong in my comment and these induction schemes are not really necessary. See this essentially duplication question: Can multiplication be defined in terms of divisibility?
A synopsis: $$ z = x*y \iff \\ \exists u,v,w (\text{Square}(x,u) \land \text{Square}(y,v) \land \text{Square}(x+y, w) \land w = u + v + z + z) $$
However, they don't quite address the problem that the lcm defined in terms of divisibility is not unique. A thought: simply define your LCM as a set, and $$\text{Square(s,t)} \iff \exists u \in \text{LCM}(s,s+1): s+t = u $$
Notice that when $s$ is negative, eg $s=-3$, then $\text{LCM}(-3,-2) = \{6,-6\}$ and $-3 + 9 = 6$. However, this definition also makes $t=-3$ a possibility, which ruins everything.
Thus we still need to inductively define the positive numbers $P(x) \iff x = 0 \lor P(x-1)$ and assert the LCM is positive.