The sequence $ \{a_n\}$ of integers is defined by
\[ a_1 = 2, a_2 = 7
\]
and
\[ - \frac {1}{2} < a_{n + 1} - \frac {a^2_n}{a_{n - 1}} \leq \frac {}{}, n \geq 2.
\]
Prove that $ a_n$ is odd for all $ n > 1.$
$ -\frac{1}{2}< a_{n+1}-\frac{a^{2}_{n}}{a_{n-1}}\leq\frac{1}{2}$
We are going to use induction to prove $a_{n+2}=3a_{n+1}+2a_{n}$ for $n\ge1$
First, when n=3, $a_{3}=25=3a_{2}+2a_{1}$.
Then assume when $n=k$, $a_{k+2}=3a_{k+1}+2a_{k}$ is true. We consider when $n=k+1$.
$ -\frac{1}{2}< a_{k+3}-\frac{a^{2}_{k+2}}{a_{k+1}}=a_{k+3}-3a_{k+2}-\frac{2a_{k+2}a_{k}}{a_{k+1}}=a_{k+3}-3a_{k+2}-2a_{k+1}-\frac{2a_{k+2}a_{k}-2a^{2}_{k+1}}{a_{k+1}}\leq\frac{1}{2}$
As we have
$ -\frac{1}{2}< a_{k+2}-\frac{a^{2}_{k+1}}{a_{k}}\leq\frac{1}{2}$, which is equivalent to $ -\frac{a_k}{a_{k+1}}< \frac{2a_{k+2}a_{k}-2a^{2}_{k+1}}{a_{k+1}}\leq\frac{a_{k}}{a_{k+1}}$. Note that $ |\frac{a_k}{a_{k+1}}|<\frac{1}{3}$.
Therefore $a_{k+3}-3a_{k+2}-2a_{k+1}$ must be $0$. Induction is complete.
As a result, we have $a_{n+2}=3a_{n+1}+2a_{n}$ for $n\ge1$, which is easily mean that $a_{n}$ is an odd for all $n>1$.
Let $x_{n}=3x_{n-1}+2x_{n-2}$ and $x_1 =2, x_2 =7$
now $x_{n+1}.x_{n-1} -x^2_n=(-2)^{n-2}$ it also satisfy that inequality of $a_n$
and also $x_n>2^n$ so $x_n=a_n$ ,so done.