There exists a function $f:\mathbb N_0\to\mathbb N_0$ such that: $P(x,y):f(x+y,y)=f(x,y)+1\forall x,y\in\mathbb N_0$ $Q(x,y):f(x,y)f(y,x)=0\forall x,y\in\mathbb N_0$ Find $f(2016,121)$ and $f(2016,144)$. $Q(x,x)\Rightarrow f(x,x)=0$ By induction on $P(nx,x)$ we get $f(nx,x)=n-1\forall n\in\mathbb N$ Thus, $f(2016,144)=f(14\cdot144,144)=14-1=\boxed{13}$. $P(1895,121)\Rightarrow f(2016,121)=f(1895,121)+1=f(1774,121)+2=\ldots=f(80,121)+16$ Case 1: $f(121,80)=0$ $P(41,80)\Rightarrow 0=f(41,80)+1$, but $f(x,y)\ge0$, contradiction. Case 2: $f(80,121)=0$ $\boxed{f(2016,121)=16}$