Prove that functional equation doesn't have range $\Bbb R.$ Prove that any solution $f: \mathbb{R} \to \mathbb{R}$ of the functional equation
$$ f(x + 1)f(x) + f(x + 1) + 1 = 0 $$
cannot have range $\mathbb{R}$.

I transformed it into 
$$ f(x) = \frac {-1} {f(x + 1)} - 1 = \frac {-1 - f(x + 1)} {f(x + 1)} $$
I tried to evaluate $f(x + 1)$ and $f(x + 2)$ and put them into the transformed equation:
1) after $f(x + 1)$
$$ f(x) = \frac {1} {-1 - f(x + 2)}$$
2) after $f(x + 2)$
$$ f(x) = \frac {1} {\frac {-1}{f(x + 3)}} = -f(x + 3) $$
What am I supposed to do next?
 A: You should always be careful when dividing, because you'll run into trouble if you try and divide by 0.
We have that for all $x$, $$f(x+1)(f(x) + 1) = -1$$
Notice that $f$ can never be zero, then, because that would mean $0$ on the left-hand side was $-1$ on the right-hand side. This justifies your division step, and proves that $0$ is never in the range of $f$; that is all you need.

As an appendix, we prove that there are, in fact, solutions to the functional equation.
If we fix $x=0$, obtain what is effectively the recurrence $$x_{n+1} = \frac{-1}{x_n + 1}$$
where $x_n = f(n)$. This certainly does have solutions (for example, if $x_0 = 1$, then it cycles $1, -\frac{1}{2}, -2, 1, -\frac{1}{2}, -2, \dots$), so the recurrence does have a solution over the reals.
One such solution is therefore $$f(x) =  \begin{cases} 
      1 & \lfloor x \rfloor \equiv 0 \pmod{3} \\
      -\frac{1}{2} & \lfloor x \rfloor \equiv 1 \pmod{3} \\
      -2 & \lfloor x \rfloor \equiv 2 \pmod{3} 
   \end{cases}
$$ 
A: Assume that the range of $f$ is all of $\Bbb R$. Then there are $a,b\in\Bbb R$ with $f(a)=0$ and $f(b)=-1$. But then 
$$ 1=f(a)f(a-1)+f(a)+1=0$$
and 
$$ 1=f(b+1)f(b)+f(b+1)+1=0,$$
both of which are absurd.
