WakeUp wrote: Let $a\circ b=a+b-ab$. Find all triples $(x,y,z)$ of integers such that \[(x\circ y)\circ z +(y\circ z)\circ x +(z\circ x)\circ y=0\] By expanding, the $LHS=3(x-1)(y-1)(z-1)+3$.

$a\circ b=a+b-ab$ and so $1-a\circ b=1-a-b+ab=(1-a)(1-b)$. So $\circ$ multiplication is isomorphic to regular multiplication under the transformation $x\rightarrow1-x$. In other words, $\phi(ab)=\phi(a)\circ\phi(b)$ where $\phi(a)=1-a$. Also note $\phi(\phi(a))=a$. So $(a\circ b)\circ c=\left(\phi\left(\phi\left(a\right)\right)\circ \phi\left(\phi\left(b\right)\right)\right)\circ\phi\left(\phi\left(c\right)\right)=\phi\left(\phi\left(a\right)\phi\left(b\right)\right)\circ\phi\left(\phi\left(c\right)\right)$ $=\phi\left(\phi\left(a\right)\phi\left(b\right)\phi\left(c\right)\right)=1-(1-a)(1-b)(1-c)=(a-1)(b-1)(c-1)+1$ So $(x\circ y)\circ z=(y\circ z)\circ x=(z\circ x)\circ y=(x-1)(y-1)(z-1)+1$ and their sum is just $3(x-1)(y-1)(z-1)+3$. So this is 0, and hence $(x-1)(y-1)(z-1)=-1$. So we need 3 integers whose product is $-1$. They all must be either $1$ or $-1$, and the number of $-1$'s must be odd. There are at most 3, so either there's 1 or 3 $-1$'s. The first case gives $(x, y, z)=(2, 2, 0); (2, 0, 2); (0, 2, 2)$ and the second gives $(x, y, z)=(0, 0, 0)$. So the only possibilities are $(x, y, z)=(0, 0, 0); (0, 2, 2); (2, 0, 2); (2, 2, 0)$.