Is there a function over which multiplication is distributive which isn't addition? Does there exist a binary operation on $\mathbb{Z}$ over which multiplication is distributive which isn't addition? Formally does there exist $f:\mathbb{Z^2}\to\mathbb{Z}$ s.t. $af(b,c)=f(ab,ac) \forall a, b, c \in \mathbb{Z}$ and $f$ isn't just the usual addition? It would also be nice if this made $(\mathbb{Z},f)$ into an Abelian group.
Edit 1. Ok, there are lots of examples, but are there any which make an Abelian group?
Edit 2. Ok, Andreas Blass has given a rather good example, but I believe his example is a group which is isomorphic to the usual additon. Are there any which make an Abelian group which is not isomorphic to $(\mathbb{Z},+)$?
 A: Let $\pi$ be any permutation of the set of prime natural numbers, and extend it to map $\mathbb Z$ to itself by sending any integer, written as a signed product of primes $\pm p_1^{e_1}\cdots p_k^{e_k}$, to $\pm \pi(p_1)^{e_1}\cdots \pi(p_k)^{e_k}$ and sending $0$ to $0$.  This extended $\pi$ commutes with multiplication, but it transforms addition into a new function $f(a,b)=\pi(\pi^{-1}(a)+\pi^{-1}(b))$, and multiplication distributes over $f$.  Indeed, $\pi$ is an isomorphism from $\mathbb Z$ with ordinary addition and multiplication to $\mathbb Z$ with $f$ and ordinary multiplication.
A: I thought of $\max(x,y)$, but this fails for negative $a$.
Maybe only linear functions work.
I'll play around a bit and see what happens.
Set $c = 0$ and $g(x) = f(x, 0)$.
Then $ag(b) = g(ab)$.
Let $b = 1$, so $a g(1) = g(a)$,
so $g$ is multiplication by $g(1)$.
Similarly, if $h(x) = f(0, x)$,
$a h(1) = h(a)$, so $h$ is also linear.
Setting $a = 0$, $f(0, 0) = 0$.
Sure looks like only linear might work.
The problem seems to me to be that
the two arguments to $f$
are completely independent,
so there is no way to get them to interact.
I'll leave it at this for now.
