How many ring can $(\mathbb{Z},+)$ upgrade to? (Can we define $(-1)\times(-1)=-1$?) We know that $(\mathbb{Z},+)$ is an abelian group. And under usual multiplication, $(\mathbb{Z},+,\times)$ become a ring. Is there another binary operation $\square$ on $\mathbb{Z}$ that can make $(\mathbb{Z},+,\square)$ to be a ring?

PS: The question comes with a more philosophical argument: junior high students often ask why we should define $(-3)\times(-5)=15$, why not define $(-3)\times(-5)=-15$ or just another number, say $(-3)\times(-5)=155555$? I think a good motivation and reason for operations with negative numbers defined like so, is because only when we defined like that, the integers $\mathbb{Z}$ with $+$ and $\times$ can have associativity, distribution law and so on(i.e. forms a ring). If we didn't choose to define so, then we don't have such good properties!
Yes, there is a different ring structure on $(\mathbb{Z}, +)$: letting $\cdot$ denote the usual product, define $*$ as $$a*b=-(a\cdot b).$$

Why does this work? Well, let $(G, \oplus, \otimes)$ be any ring, and $\alpha$ any automorphism of $(G, \oplus)$. Then we can "push" $\otimes$ through $\alpha$ to get a new operation, $\otimes_\alpha$, such that $(G, \oplus,\otimes_\alpha)$ is also a ring. Namely, $\otimes_\alpha$ is defined as $$a\otimes_\alpha b=\alpha^{-1}(\alpha(a)\otimes\alpha(b)).$$ In our case, we have the automorphism $\alpha:x\mapsto -x$, and $\alpha(a)\cdot \alpha(b)=a\cdot b$, so this yields the operation described above.



In fact, $\cdot$ and $*$ are the only operations which yield a ring structure: suppose $k$ is a unit of $\odot$ with $k\not\in\{1, -1\}$. WLOG, $k$ is positive - but then $1\odot 1+1\odot 1+...$ ($k^2$ times) equals $k$, but no integer has this property if $k&gt;1$.

It's now enough to show that any ring expansion with unit $1$ is the usual one, and with unit $-1$ is the "twisted" version described above; and this isn't too hard.
I think you can answer this to your students : 

Let $\mathbb{Z}[1/2] = \{ \frac{a}{2^b}, a \in \mathbb{Z},b \in \mathbb{N}\}$.  With the usual operation it is a (commutative) ring. 

But you can define another multiplication $\ast$ on it :
$$a \ast b = 2 ab$$
This way $1/2$ is the multiplicative identity, and $(\mathbb{Z}[1/2],+,\ast)$ is a  (commutative) ring where

$$(-1) \ast (-1) = 2$$