Wrong reasoning yields $1=0$ I was thinking about this sequence
\begin{equation}\tag{1}
n=\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{\ldots}}}}}
\end{equation}
It is well known that this converges to n. For example:
$$3=\sqrt{9}=\sqrt{6+3}=\sqrt{6+\sqrt{9}}=\sqrt{6+\sqrt{6+3}}=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$$
The problem is when $n=0$ and $n=1$, because when we apply formula $(1)$, in both cases we get 
$$\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{\ldots}}}}}$$
This yields
$$0=1$$
Can you tell me where the mistake is in the reasoning?

If we look at $(1)$, say we set
$$x=\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{\dots}}}}$$
Because this goes on forever, we clearly have that
$$x^2-(n^2-n)=x$$
Solving this using the quadratic formula yields $x=1-n$ or $x=n$. At this point, we simply choose whatever makes since (i.e. if $n>1$, then $x$ is clearly positive so we take $x=n$). In our case of $n=1$, it makes sense to have $x=1-n$.
