$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.
 A: It can be proven that for $n\ge 4$, $$n^2+\frac 32n\lt \frac{a_{n+1}}{a_n}\lt n^2+3n\tag1$$
From $(1)$, we have, for $n\ge 4$,  $$\frac{a_{n+1}}{a_n}\gt 0$$
Since we have
$$a_1=1,\quad a_2=3,\quad a_3=22,\quad a_4=304,\quad a_5=6810$$
it follows that $a_n$ is a positive integer for all $n\ge 1$.

Now, let us prove that $(1)$ holds for $n\ge 4$.
Proof :
Let us prove $(1)$ by induction on $n$.
We see that $(1)$ holds for $n=4$ since
$$n^2+\frac 32n=22,\quad \frac{a_{n+1}}{a_n}=22+\frac{61}{152},\quad n^2+3n=28$$
Suppose that $(1)$ holds for some $n\ (\ge 4)$.
Then, we have
$$\begin{align}&\frac{a_{n+2}}{a_{n+1}}-\left((n+1)^2+\frac 32(n+1)\right)
\\\\&=2(n+1)^2+2(n+1)+1-\frac{a_n}{a_{n+1}}\left((n+1)^4+1\right)-\left((n+1)^2+\frac 32(n+1)\right)
\\\\&\gt 2(n+1)^2+2(n+1)+1-\frac{(n+1)^4+1}{n^2+\frac 32n}-\left((n+1)^2+\frac 32(n+1)\right)
\\\\&=\frac{n^2 - n - 8}{2 n (2 n + 3)}
\\\\&\gt 0\end{align}$$
and
$$\begin{align}&(n+1)^2+3(n+1)-\frac{a_{n+2}}{a_{n+1}}
\\\\&=(n+1)^2+3(n+1)-\left((2(n+1)^2+2(n+1)+1)-\frac{a_n}{a_{n+1}}\left((n+1)^4+1\right)\right)
\\\\&\gt (n+1)^2+3(n+1)-(2(n+1)^2+2(n+1)+1)+\frac{(n+1)^4+1}{n^2+3n}
\\\\&=\frac{2 n^2 + n + 2}{n (n + 3)}
\\\\&\gt 0\qquad\blacksquare\end{align}$$
A: 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$.
