Proof $\frac{a_n^2+a_{n+1}^2+1}{a_{n}a_{n+1}} $ is constant I would appreciate if somebody could help me with the following problem:
Question:   Defined by $a_{1} =1,a_{2}=2$   and    $a_n a_{n+2}=a_{n+1}^2+1(n\geq 1)$ 
Proof. $\frac{a_n^2+a_{n+1}^2+1}{a_{n}a_{n+1}} $ is constant for  $n\geq 1$
I tried (Mathematical Induction)  but couldn’t get it that way.

The formula $$\frac{a_n^2+a^2_{n+1}+1}{a_na_{n+2}}=3$$ holds for $n=1,2$. Assume it is true for $k$ and show it is true for $k+1$. Here is another approach:
$$
\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k+1}a_{k+2}}\cdot\frac{a_{k}}{a_{k}}
=\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k}a_{k+2}}\cdot\frac{a_{k}}{a_{k+1}}
=\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k+1}^2+1}\cdot\frac{a_{k}}{a_{k+1}}
$$
$$
=\frac{a_k}{a_{k+1}}\left(\frac{a_{k+1}^2+1}{a_{k+1}^2+1}+\frac{a_{k+2}^2}{a_{k+1}^2+1}\right)
=\frac{a_k}{a_{k+1}}\left(1+\frac{a_{k+2}^2}{a_k a_{k+2}}\right)
=\frac{a_k+a_{k+2}}{a_{k+1}}.
$$
So now we need to show that is constant in $k$. See solution above by NikolajK.
