It's easy let $P(m,n)$ be the assertion $f(m + n) + f(mn -1) = f(m)f(n) + 2$ clearly the constant function doesn't satisfy $P(0,n) \implies f(n)(f(0)-1)=f(-1)-2$ if $f(0) \neq 1$ but since $f$ is not constant then $f(0)=1 , f(-1)=2$ comparing between $P(1,n) , P(-1,n)$ we will have $f(1)=2$ and $f$ is even and $f(n+1)+f(n-1)=2f(n)+2$ then by induction we will have $f(n)=n^2+1$

Ali3085 wrote: It's easy let $P(m,n)$ be the assertion $f(m + n) + f(mn -1) = f(m)f(n) + 2$ clearly the constant function doesn't satisfy $P(0,n) \implies f(n)(f(0)-1)=f(-1)-2$ if $f(0) \neq 1$ but since $f$ is not constant then $f(0)=1 , f(-1)=2$ comparing between $P(1,n) , P(-1,n)$ we will have $f(1)=2$ and $f$ is even and $f(n+1)+f(n-1)=2f(n)+2$ then by induction we will have $f(n)=n^2+1$ How you get $f(1)=2$?

TuZo wrote: How you get $f(1)=2$? $P(-1,-1) \implies f(-2)=5$ $P(1,-1) \implies f(1)=2$

Ok, thank you.