Beautiful problem , solution : assume that $ h(0) = c$ put $ x = y = 0$ so we have : $ 2c = c^{2}+1$ which gives $ c = 1$ , assume that $ h(1) = a$ and $ h(-1) = b$ , put $ x = 1 , y =-1$ we get : $ ab = b$ which follows $ h(1) = a = 1$ or $ h(-1) = b = 0$ so we have 2 case : 1)$ h(1) = 1$ then put $ y = 1$ then we have $ h(x+1) = 1$ so we have $ h(x) = 1$ for all $ x$ . 2)$ h(-1) = 0$ then put $ y =-1$ then we have $ h(x-1)+h(-x) = 1$ so we have $ h(1)+h(-2) = 1\longrightarrow a+h(-2) = 1$ now put $ y = 1$ we have : $ h(x+1)+h(x) = ah(x)+1$ now put $ x =-2$ we have $ h(-1)+h(-2) = ah(-2)+1$ and we have $ h(-2) = ah(-2)+1$ now put $ h(-2) = 1-a$ so we have $ 1-a = a(1-a)+1\longrightarrow a^{2}-2a = 0$ so $ a = 0$ or $ a = 2$ so we have 2 case : $ 1)a = 0$ then put $ y = 1\longrightarrow h(x+1)+h(x) = 1$ but $ h(0) = 1$ and $ h(1) = 0$ so we $ h(x) = 1$ if $ x = 2k$ and $ h(x) = 0$ if $ x = 2k+1$ . u can check this solution ! $ 2)a = 2$ , now put $ y = 1$ so we have : $ h(x+1) = h(x)+1$ and u can prove that $ h(x) = x+1$ very easily . so there are 3 solutions !

Let $h(x)=f(x)$ and let $P(x,y)$ denote the given assertion. $P(0,0)\implies f(0)=1.$ Let $f(1)=c+1,$ where $c$ is any constant. $P(x,1)\implies f(x+1)=cf(x)+1.$ Making some basic substitutions one gets that $c=-1,0,1.$ a. $c=-1$: then $f(x)=1$ for even and $f(x)=0$ for odd, which works. b. $c=0$: Then $f$ is identically 1, which works. c. $c=1$: Then $P(1,-1)\implies f(-1)=0.$ Comparing $P(x,1)$ and $P(x,-1)$ we get $f(x)=x+1,$ which works.

Let $f(x) = Ax + B$ and substitute this into the functional equation to yield: $A(x+y) + B + Axy + B = A^{2}xy + AB(x+y) + B^{2} + 1$ and coefficient matching further produces: $A = A^{2}, A = AB, 2B = B^{2} + 1$ which yield the ordered pairs: $(A, B) = (0, 1); (1, 1)$. Hence, the solutions include: $\fbox{f(x) = 1, x + 1}$.