I am looking for the solutions of the following functional equation: $$f(x+1)+f(x)=(f(x))^2+1$$
For all non-negative integers $n$, we get a recurrence $f(n+1)+f(n)=(f(n))^2+1$. Suppose $f(0)=r$, then we have $f(1)=r^2-r+1, f(2)=r^4 - 2 r^3 + 2 r^2 - r + 1..$. etc. We may choose $r$ such that the sequence converges $\{f(n)\}_n$.
What to do next?
Choose an arbitrary function $$t:[0,1) \mapsto \mathbb{R}$$ then there is a unique function $$f:[0,\infty)\mapsto \mathbb{R}$$ such that $$f|_{[0,1)}=t \tag1$$ and $$f(x+1)=(f(x))^2+1-f(x)\tag 2$$ Because if $x \in [0,\infty)$ then $x=n+\xi$, where $\xi \in [0,1)$ and $n \in \mathbb{{Z_0}^+}$. So $$f(\xi)=t(\xi)\\f(1+\xi)=f(\xi)^2+1-f(\xi)\\ f(2+\xi)=f(1+\xi)^2+1-f(1+\xi)\\ \ldots\\ f(n+\xi)=f((n-1)+\xi)^2+1+f((n-1)+\xi)$$ So now we have calculated $f(x)=f((n-1)+\xi)$ Of course we can use an arbitrary half open interval of length $1$ to start with our definition of $f$.
But given such an half open interval of length, can we extend $f$ in a similar way towards $-\infty$? From the functional equation $(2)$ we have
$$f(x-1)=\frac{1+\sqrt{4x-3}}2\tag 3$$ or $$f(x-1)=\frac{1-\sqrt{4x-3}}2\tag 4$$ If $x>1$ then for $f(x-1)$ calculated by $(3)$ holds $$f(x-1)\ge 1,$$ too, so if $\xi \in [0,1)$ and $t(\xi)>1$ then $f(x)$ is defined for all $$x=\xi -n, n \in \mathbb{{Z_0}^{+}}$$
The investigation of other values of $\xi$ and the formula $(4)$ is up to you.