Possible proof that $1 = 2$? No it can't be! So why do I keep concluding to that?

So my teacher gives me a problem to work with:

Let $n = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{\cdots}}}} \ $ then how must you solve for $n$?

Solution:

$$n = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{\cdots}}}} \Rightarrow n = \sqrt{1 + n} \Rightarrow n^2 - 1 = n \Rightarrow n^2 - n - 1 = 0$$ $$\therefore n = \frac{1\pm \sqrt{5}}{2}$$ However, $n > 0$ $$\therefore n = \frac{1 + \sqrt{5}}{2} = 1.6180339887\ldots = φ \neq \frac{1 - \sqrt{5}}{2}$$ Now we can substitute $n = φ$ into the following equation to double-check: $$\begin{align} n^2 - 1 &= n \Rightarrow φ^2 - 1 = φ \Rightarrow \bigg(\frac{1 + \sqrt{5}}{2}\bigg)^2 - 1 = \frac{(1 + \sqrt{5})^2}{2^2} - 1 = \frac{6 + 2\sqrt{5}}{4} - 1 \\ &= \frac{6 + 2\sqrt{5}}{4} - \frac{4}{4} = \frac{2 + 2\sqrt{5}}{4} = \frac{2}{2}\times \frac{1 + \sqrt{5}}{2} = 1\times \frac{1 + \sqrt{5}}{2} = \boxed{\frac{1 + \sqrt{5}}{2}} \quad \color{green}{\checkmark} \end{align}$$

My teacher said, "Very good... Now try this problem."

Let $n = \frac{2}{3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\cdots}}}} \ $ then how do you solve for $n$?

No matter what I do, I always get that $n = 1 = 2 \Rightarrow \boxed{1 = 2}$ !!! I used the same method as I did for the previous question, so why isn't it working? And of course my teacher just said in response, "work on it as part of your homework". Could somebody please help me?

Thank you in advance.

Answer

If you were to use the same method as in your first example, (substituting n back into your infinite continuous fraction)

$$ 3-n=\frac{2}{n}$$
You can simplify to obtain: $ 3n-n^2=2 $, which works for any n
, as $n \neq 0$.
By factorising, you get the solutions $n = 1, $ OR $2$, not $1=2$.

Hope this helps!