sequence with a property

by rearranging terms we get $a_n\times a_{n-2}=a_{n}^2 +2$ . replace $n$ by $n-1$ we get $a_{n+1}\times a_{n-1}=a_n^2$ . subtract and trivial transformation yields $\displaystyle \frac{a_{n+1}+a_{n-1}}{a_n}=\frac{a_n + a_{n-2}}{a_{n-1}}=c$ . Substituting the initial conditions $a_1=a_2=1$ we get $c=4$ . therefore we get $a_{n+1}=4a_n-a_{n-1}$ . by induction we can show $a_n$ is always integers.

$a_{1}=1$ and $a_{2}=1$

we calculate $a_{3}$ and $a_{4}$ and find them to be 3,11 respectively

now let us assume $a_{k}$ is an integer

so $\frac{(a_{k-1})^{2}+2}{a_{k-2}}$ is an integer

now $a_{k+1}=\frac{(a_{k})^{2}+2}{a_{k-1}}$

putting value of $a_{k}$ we get

$a_{k+1}=\frac{(\frac{(a_{k-1})^{2}+2}{a_{k-2}})^{2}+2}{a_{k-1}}$

solving it(openig the square and adding) we see that for it to be an integer $a_{k-1}$ must divide $2(a_{k-2})^{2}+4$

so $\frac{(a_{k-2})^{2}+4}{a_{k-1}}$ must be an integer

putting value for $a_{k-1}$ we get

$\frac{(2(a_{k-2})^{2}+4)\times(a_{k-3})}{(a_{k-2})^{2}+2}$

which is an integer if and only if $a_{k-3}$ is an integer

now by induction : if p(k+1) is true only when p(k) and p(k-3) are true and p is true for first four natural numbers then it true for all natural numbers

and we have already done that hence the sequence returns only integer values.