A different ring on the integers such that addition and multiplication are polynomials. One can define rings other than the standard one on the set of integers so that both of the operations are polynomials, for example $(\mathbb Z, +', \times')$ with $a+'b=(a+1)+(b+1)-1$ and $a\times'b=(a+1)(b+1)-1$. I think I saw a less silly example a few years ago. However both examples turn out to be isomorphic to $(\mathbb Z, +, \times)$. Is there such a ring that isn't isomorphic to $(\mathbb Z, +, \times)$?
The question is loosely motivated by computing. Computation models are often defined to have the addition and multiplication operations as basic operations. Some techniques use fields (in this case I was thinking about polynomial/rolling hashes) but can run into issues with the standard integers, so it would be nice to have a ring that can be easily "implemented" with integers.
 A: Define a commutative ring structure on $\mathbb Z$ as follows: $a+'b=p(a,b)$ and $a\times'b=q(a,b)$, where $p(X,Y),q(X,Y)\in\mathbb Z[X,Y]$. Note that since $p$ and $q$ are symmetric, we have that $p,q\in \mathbb Z[X+Y,XY]$.
For each $a\in\mathbb Z$ we must have $\mathbb Z\to\mathbb Z:x\mapsto x+'a=p(x,a)$ is bijective, hence $p(X,a)$ must be a degree-$1$ polynomial of $X$. Thus $p(X,Y)$ must be of the form $\alpha+\beta(X+Y)+\gamma XY$ for $\alpha,\beta\in\mathbb Z$. Now we need the map $\mathbb Z\to\mathbb Z:x\mapsto x+'a=\alpha+\beta a+(\beta+\gamma a)x$ to be bijectice for each $a\in\mathbb Z$, so that $\gamma=0$ and $\beta=\pm1$. However, if $\beta=-1$ we must have that
$$p(p(X,Y),Z)=p(\alpha-X-Y,Z)=X+Y-Z$$
is symmetric with respect to permutation of $X,Y,Z$, which is clearly not true. Thus we conclude that $\beta=1$ and $p(X,Y)=\alpha+X+Y$. Thus, $0'=-\alpha$, and by shifting everything by $\alpha$ (that is, applying the bijection $\mathbb Z\to\mathbb Z:x\mapsto x+\alpha$) we may simply assume $\alpha=0$.
Now, $q$ must be a polynomial such that $q(X+Y,Z)=q(X,Z)+q(Y,Z)$, hence it must be of the form $q(X,Y)=f(X)Y$ for some polynomial $f(X)\in\mathbb Z[X]$. Since $q$ must be symmetric, it must be the case that $f(X)=aX$ for $a\in\mathbb Z\setminus\{0\}$. The existence of a multiplicative unit tells us that $a=\pm1$. Either way, by considering the isomorphism $\mathbb Z\to\mathbb Z:x\to ax$ we may assume $a=1$, so that $(\mathbb Z,+',\times')\cong(\mathbb Z,+,\times)$.
P.S.
We may arrive at the same conclusion without assuming commutativity, since distributivity on the left tells us that $q(X,Y)$ is of the form $Xg(Y)$ for some $g(Y)\in\mathbb Z[Y]$.
