Is there such an operation that is distributive over addition, and is not multiplication?
Also, please no operations that are defined piece-wise, or that are trivial.
It must apply to all integers, but for all reals for example, the result does not matter.
Best Answer
The only such operations are those of the form $m\otimes n=mna$ for some fixed integer $a$. When $a=1$ this is ordinary multiplication, and when $a=0$ it’s trivial.
Suppose that $\otimes$ is such an operation. Then for any integer $n$ we have $$n\otimes 0=n\otimes(0+0)=(n\otimes 0)+(n\otimes 0)\,;$$ the only integer satisfying $x+x=x$ is $0$, so $n\otimes 0=0$ for every $n\in\mathbb{Z}$. Now let $a=1\otimes 1$. Then $$1\otimes 2=1\otimes(1+1)=(1\otimes 1)+(1\otimes 1)=a+a=2a\,,$$ and an easy induction shows that $1\otimes n=na$ for every positive integer $n$. We already know that $1\otimes 0=0$, so in fact $1\otimes n=na$ for every integer $n\ge 0$.
Now suppose that $n$ is a negative integer. Then $$0=1\otimes 0=1\otimes\big(n+(-n)\big)=(1\otimes n)+\big(1\otimes(-n)\big)=(1\otimes n)-na\,,\tag{1}$$ so $1\otimes n=na$, and we’ve now shown that $1\otimes n=na$ for every $n\in\mathbb{Z}$.
We can now repeat the argument to generalize the operand $1$ to any integer:
$$2\otimes n=(1+1)\otimes n=(1\otimes n)+(1\otimes n)=2an\,,$$ and another easy induction gives us $m\otimes n=mna$ for every non-negative integer $m$ and every integer $n$. Finally, the trick that I used in $(1)$ can be used again to show that $m\otimes n=mna$ for all $m,n\in\mathbb{Z}$.
Added: Because $\otimes$ is just scaled multiplication, it’s certainly both commutative and associative: $$m\otimes n=amn=anm=n\otimes m$$ and $$(k\otimes m)\otimes n=a(akm)n=ak(amn)=k\otimes(m\otimes n)$$ for any $k,m,n\in\mathbb{Z}$.