Polynomial ring operations on $\mathbb{Z}$ I have asked this on Math Stack Exchange but without answers:
The usual ring operations on $\mathbb{Z}$ can be defined via polynomials in $\mathbb{Z}[a,b]$ (when viewing $a,b$ as variables): 
Addition: $(a,b) \mapsto a+b \in \mathbb{Z}[a,b]$
Multiplication: $(a,b) \mapsto a \cdot b \in \mathbb{Z}[a,b]$
There are more ring structures on $\mathbb{Z}$ that can be described with polynomials, e.g. $(\mathbb{Z},P_A,P_M)$ with $P_A(a,b) = a+b-1$ and $P_M(a,b)=a+b-ab$. Here the neutral element w.r.t. $P_A$ is $1$ and those w.r.t. $P_M$ is $0$. 
When we don't require the ring to have a neutral element w.r.t. multiplication, we could even take $P_A(a,b)=a+b$ and $P_M(a,b)=0$.
My question is: Can one characterize the pairs $(P_A,P_M)$ of polynomials, such that $(\mathbb{Z},P_A,P_M)$ is a ring? How does this change when we 


*

*dont' require a neutral element w.r.t. $P_M$?

*want to get a commutative ring?

*want to get an integral domain?

 A: Your polynomial $a+b-ab$ is more or less what is used for the formal group associated to $\mathbb G_m$, since the "identity" of a (1-dimensional) formal group is taken to be $0$. I don't know to what extent there's a theory of "formal rings", but my recollection is that the only polynoimals that give a characteristic 0 formal group law (with $0$ as identity) are $X+Y$ (for the additive group) and $X+Y+XY$. Here a formal group of $R$ is a power series $F\in R[\![X,Y,]\!]$ satisfying various axioms:


*

*$F(0,0)=0$

*$F(X+Y) = X+Y+\text{higher order terms}$

*$F(X,F(Y,Z))=F(F(X,Y),Z)$


One could also impose $F(X,Y)=F(Y,X)$, but it turns out that this is automatic unless $R$ has zero divisors and nil-potents. There are lots of interesting power series with these properties, but as I mentioned, only two polynomials. 
A: We can start by looking at $P_A$. The requirement that it be a commutative group implies that for any $a$, $P_A(a, x)$ is a bijective function - so it must be of the form $x + f(a)$ or $-x + f(a)$ for some $f(a)$. By similar consideration on $b$, we can conclude that $P_A(a, b) = a + b + K$ or $P_A(a, b) = -a - b + K$ for some constant $K$. In the latter case, $P_A(P_A(a, b), c) = P_A(-a - b + K, c) = a + b - c + K$, while $P_A(a, P_A(b, c)) = -a + b + c + K$, so this case is eliminated.
We therefore have that $P_A(a, b) = a + b + K$. By translation, as Qiaochu says above, we can take that $P_A(a, b) = a + b$ WLOG. The interesting question then lies in multiplication. 
Distributivity says that for $a$ constant, $P_M(a, b)$ is $\mathbb{Z}$-linear in $b$, so $P_M(a, b) = f(a)b$ for some function $f$. Similarly, $P_M(a, b) = a g(b)$. By dividing, we can see that there must be some constant $L$ with $P_M(a, b) = L a b$. 
Therefore, the answer to your question is that all are of the form Steven Landsburg gave above, and that all of them are equivalent to the usual addition and multiplication on some ideal of $\mathbb{Z}$ (that ideal being the ideal of numbers divisible by $|L|$. 
