Consider sequence of numbers $a_r,\;r\geq0\;$ with $a_0=1\;$ and $a_{r+1}^2=1+a_r\cdot a_{r+2}.\;$ Then which of the following is/are true? Let $\alpha, \beta$ are roots of equation $x^2-a_1 x+1=0$ and consider sequence of numbers $a_r,\;r\geq0\;$ with $a_0=1\;$ and $a_{r+1}^2=1+a_r\cdot a_{r+2}.\;$ Then which of the following is/are true?
(A) $a_r+a_{r+2}=a_1 \cdot a_{r+1}$
(B) $a_r+a_{r+2}=2\cdot a_{r+1}$
(C) $a_n=\dfrac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$
(D) $a_n=\dfrac{\alpha^{n+1}+\beta^{n+1}}{\alpha+\beta}$
My Approach:
Since it is given that  $a_{r+1}^2=1+a_r\cdot a_{r+2}.\; \implies  a_{r+1}^2-1=a_r\cdot a_{r+2}.\;$
Now using option (A)   $\;a_r+a_{r+2}=a_{r}+\dfrac{a_{r+1}^2-1}{a_r}\implies a_r+a_{r+2}=\dfrac{a_{r}^2-1+a_{r+1}^2}{a_r}\implies a_r+a_{r+2}=\dfrac{(a_{r-1}+a_{r+1})\cdot a_{r+1}}{a_r}.$
Now $a_{r-1}+a_{r+1}=a_{r-1}+\dfrac{a_{r}^2-1}{a_{r-1}}\implies a_{r-1}+a_{r+1}=\dfrac{a_{r-1}^2-1+a_{r}^2}{a_{r-1}}\implies a_{r-1}+a_{r+1}=\dfrac{(a_{r-2}+a_{r})\cdot a_{r}}{a_{r-1}}$
When i continue this process i obtain $a_{r}+a_{r+2}=\dfrac{(a_0+a_2)a_2}{a_1}$
But answer given is (A), How do i obtain answer as option (A) which depend upon $\;a_{r+1}?\;$
Also how to obtain option (C) and (D)?
 A: Proof for (A) follows. I've pretty sure induction is the best method; you're already dealing with recurrences, why not add in one more?
I'm going to designate $a_1$ as just $a$ for the sake of simplicity, because it'll show up a lot more than the other $a_r$.
As you already showed, $a_{r+2} = (a_{r+1}^2 - 1)/a_r$. For $r=0$, this is just $a_2 = (a^2-1)/1 = a^2 - 1$.
Since we have an expression in just $a$ for $a_2$, we can see $a_3 = ((a^2-1)^2-1)/a = a^3 - 2a$. We could continue this*, but it's not necessary.
First, we can see quickly that (A) is true for $r=0,1$:
$$a_2 + a_0 = a^2-1+1 = a^2 = a_1 \cdot a_1 \\
a_3 + a_1 = a^3-2a+a = a(a^2-1) = a_1 \cdot a_2$$
Let's do some induction. We want to show that:
$$a_r + a_{r+2} = a \cdot a_{r+1} \implies a_{r+1} + a_{r+3} = a \cdot a_{r+2}$$
Or, having shown $P(0), P(1)$ to be true, we must show that $P(r) \implies P(r+1)$.
I hate big subscripts**, so I'm setting $x = a_r, \ y= a_{r+1}, \ z= a_{r+2}, \ w = a_{r+3}$. This means we want to show:
$$x+z=ay \implies y+w=az$$
Let's begin by ridding ourselves of the $w$. We know $w = (z^2 - 1)/y$, and we can rewrite the RHS of our implication as $y^2+z^2-1 = ayz$. But we also know that $y^2-1=xz$. Making that substitution gives us a RHS of $xz+z^2=ayz$
Now, none of these values can be $0$, as $a_r = 0 \implies a_{r+2} = (a_r^2-1)/0$, and division by zero is a nicely instant contradiction. Since $z \ne 0$, we can divide by it to get $x+z=ay$ on the RHS, which is also the LHS of our implication. Since the LHS and RHS are the same, the implication must be true, and thus $P(r) \implies P(r+1)$.
Because $P(0)$ and $P(1)$ are true, and $P(r) \implies P(r+1)$ is true, $P(r)$ is true for all $r \ge 0$, and statement (A) is proven. $\square$
(Note the importance of not using the LHS as a substitution in the RHS, which would cause logical loops.)

*If we had continued with the resubstitutions, we actually get a very nice pattern:
$$
\begin{array}{l l l l l}
a_0 = & 1 \\
a_1 = & a \\
a_2 = & a^2 & -1 \\
a_3 = & a^3 & -2a \\
a_4 = & a^4 & -3a^2 & +1 \\
a_5 = & a^5 & -4a^3 & +3a \\
a_6 = & a^6 & -5a^4 & +6a^2 & -1 \\
a_7 = & a^7 & -6a^5 & +10a^3 & -4a
\end{array}
$$
The coefficients, if you read going southeast, are the rows of Pascal's Triangle. So column $2$ is the integers, column $3$ is the triangular numbers, etc. It sure looks like each line is:
$$a_r = \binom{r}{0}a^r - \binom{r-1}{1}a^{r-2} + \binom{r-2}{2} a^{r-4} - \binom{r-3}{3}a^{r-6} + \cdots = \sum_{i=0}^{\lfloor r/2 \rfloor} (-1)^i \binom{r-i}{i}a^{r-2i}$$
...but I'm not going to try proving it.
**$\tiny\text{(And I cannot lie)}$
A: Hint: The following are rather hints, as we don't prove claims for each $n$ resp. $r$. The solutions of the quadratic equation $x^2-a_1 x+1=0$ are
\begin{align*}
\alpha=\frac{1}{2}\left(a_1+\sqrt{a_1^2-4}\right)\qquad\qquad
\beta=\frac{1}{2}\left(a_1-\sqrt{a_1^2-4}\right)\tag{1}
\end{align*}
Point (C):
With $n=1$ we obtain
\begin{align*}
a_1=\frac{\alpha^2-\beta^2}{\alpha-\beta}=\alpha+\beta
\end{align*}
which coincides with (1) indicating (C) is correct.
Point (D):
With $n=1$ we obtain
\begin{align*}
a_1&=\frac{\alpha^2+\beta^2}{\alpha+\beta}\\
&=\frac{\frac{1}{4}\left(a_1^2+2a_1\sqrt{a_1^2-4}+a_1^2-4\right)+\frac{1}{4}\left(a_1^2-2a_1\sqrt{a_1^2-4}+a_1^2-4\right)}{a_1}\\
&=\frac{a_1^2-2}{a_1}
\end{align*}
It follows $a_1^2=a_1^2-2$ which is absurd and (D) is not correct.
Point (A):
With $r=0$ we obtain
\begin{align*}
a_0+a_2=a_1^2
\end{align*}
and since $a_0=1$ we have
\begin{align*}
a_1^2=1+a_2
\end{align*}
The recurrence relation $a_{r+1}^2=1+a_r\cdot a_{r+2}$ gives also for $r=0$
\begin{align*}
a_1^2=1+a_0a_2=1+a_2
\end{align*}
which indicates (A) is correct.


Addendum (2022-06-10): Proof of (A)
Since OP has already shown the validity of (C) which is stated in a comment of this post
\begin{align*}
\text{(C)}\qquad a_r=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta}\qquad\qquad r\geq 0\tag{1}
\end{align*}
we will conveniently use it to prove (A). We also recall from the quadratic equation
\begin{align*}
x^2-a_1x+1&=0\\
\alpha+\beta&=a_1\tag{2}\\
\alpha\beta&=1\tag{3}
\end{align*}

We obtain for $r\geq 0$:
\begin{align*}
\color{blue}{a_1a_{r+1}}&=\left(\alpha+\beta\right)\frac{\alpha^{r+2}-\beta^{r+2}}{\alpha-\beta}\tag{$\rightarrow (1),(2)$}\\
&=\frac{\alpha^{r+3}+\beta\alpha^{r+2}-\alpha\beta^{r+2}-\beta^{r+3}}{\alpha-\beta}\\
&=\frac{\left(\alpha\beta\right)\alpha^{r+1}-\left(\alpha\beta\right)\beta^{r+1}}{\alpha-\beta}
+\frac{\alpha^{r+3}-\beta^{r+3}}{\alpha-\beta}\tag{$\rightarrow (3)$}\\
&\,\,\color{blue}{=a_r+a_{r+2}}
\end{align*}
and the claim (A) follows.

