$a_n $ is a positive integer for any $n\in \mathbb {N} $. Let $(a_n)_{n\geq 1}$ be a sequence defined by $a_{n+1}=(2n^2+2n+1)a_n-(n^4+1 )a_{n-1} $. 
$a_1=1$, $a_2=3$.
I have to show that $a_n $ is a positive integer for any $n\in \mathbb {N}, n\geq 1$.
I tried to prove it by induction but it doesn't work.

This answer provides partial results.
For each $n$ put $b_n=\frac {a_{n+1}}{a_n}$. Then $b_1=3$ and 
$$b_n=2n^2+2n+1-\frac{n^4+1}{b_{n-1}}.$$
It suffices to prove that $b_n>0$. Computer evaluation suggests that a sequence $\{c_n=b_n-n^2-2n\}$ decreases and converges to about $-5.78734$. We have $c_1=0$ and 
$$c_n=n^2+1-\frac{n^4+1}{c_{n-1}+n^2-1}= \frac{c_{n-1}(n^2+1)-2}{c_{n-1}+n^2-1}.$$
It seems that we can show that  $c_n<c_{n-1}$ provided the denominator $ c_{n-1}+n^2-1>0$.
