Finding continued fractions of a root of polynomial (Solution verification) I was trying to solve this problem from my number theory class. I think the solution had a typo, so I wanted to verify my understanding. (Just to make sure I'm not missing something and mistaking a typo where there isn't)

Let $a$ be a positive integer, and let $\alpha$ be the positive root of the polynomial $x^{2}-a x+1$. Find the continued fraction expansion of $\alpha$.

The official solution: We have $\alpha=a+\frac{1}{\alpha}$, so the continued fraction is $[a ; a, a, a, \ldots]$

*

*But surely, the above is the case of $\alpha$ being the positive root of the polynomial $x^{2}-a x-1$?


*Is it correct that for the original question, the solution is given $[a ; -a, a, -a, \ldots]$?
 A: I'm going ahead and summing up what was said in the comments.
To your 1. point, we all agree that you are completely correct that $[a ; a, a, a, \ldots]$ is a root of the $x^{2}-a x-1$ polynomial, and not $x^{2}-a x+1$. So indeed, there must be a typo in the source document.
About your 2. point, and assuming $a>1$, it is easy to verify that if some $\alpha$ verifies
$$ \require{cancel}
\alpha-1=a-2+\cfrac{1}{1+\cfrac{1}{\alpha-1}} \tag{1}
$$
then it follows that
$$ \begin{aligned}
\alpha-1 &= a-2+\frac{\alpha-1}{\alpha} \\
\alpha^2-\cancel{\alpha} &= a\alpha-\cancel{2\alpha}+\cancel{\alpha}-1 \\
\alpha^2 - a\alpha + 1 &= 0
\end{aligned} $$
$\alpha$ is a root of $x^{2}-a x+1$. Consequently it can be written in continued fraction form as $\alpha = \left[a-1;\overline{1,a-2}\right]$. Also, mon cher ami François Viète assures me that the other root of the polynomial verifies $\alpha\beta=1$, and is therefore $\beta = \left[0; a-1,\overline{1,a-2}\right]$.
As for the $[a ; -a, a, -a, \ldots]$ continued fraction suggested as a possible root of $x^{2}-a x+1$, I can't be sure, but if it is one, then it must be equivalent to either $\alpha$ or $\beta$.

In answer to OP's query about how I arrived at $(1)$: it was essentially a form of reverse-engineering. As I mentioned in the comments, I had found a pattern while playing with such polynomials on WolframAlpha, suggesting the continued fraction form $\left[a-1;\overline{1,a-2}\right]$, so I tried to build the expression from there. Notice that you can add $-1$ on both sides of $\alpha = \left[a-1;\overline{1,a-2}\right]$ and obtain $\alpha-1 = \left[a-2;\overline{1,a-2}\right]$, which allows to write the recursive fraction $(1)$. I verified algebraically that the resulting $\alpha$ was indeed a root of the considered polynomial, and voila.
