(1) Consider the recurrence relation modulo $n^2+1$: This yields $-a_n\equiv -n \pmod{n^2+1}$. Hence there exist integers $b_1,b_2,\ldots$ with $a_n=(n^2+1)b_n+n$. (2) Plugging the expressions $a_n=(n^2+1)b_n+n$ into the original recurrence relation yields $(n^2+1)( (n^2+2n+2)b_{n+1}+n+1) - (n^2+1)b_n-n = n^3+n^2+1$, which simplifies to $(n^2+2n+2)b_{n+1}=b_n$ for every $n\ge1$. $~~~(\ast)$ (3) If one of the numbers $b_n$ is zero, then all of them are zero. This yields the solution $a_n=n$ for $n\ge1$. (4) If none of the numbers $b_n$ is zero, we multiply the first $k$ equations $(\ast)$ and divide by the product $b_2b_3\cdots b_{k-1}$: $(k^2+2k+2)\cdot(k^2+1)\cdot(k^2-2k+2)\cdots5\cdot b_{k+1}=b_1$. This shows that $b_1$ must be divisible by every number $(k^2+2k+2)\cdot(k^2+1)\cdot(k^2-2k+2)\cdots5$ with $k\ge1$. As these divisors become arbitrarily large, $b_1$ must be zero; contradiction.

MarkBcc168 wrote: Determine all sequences $a_1,a_2,a_3,\dots$ of positive integers that satisfy the equation $$(n^2+1)a_{n+1} - a_n = n^3+n^2+1$$for all positive integers $n$. See https://artofproblemsolving.com/community/c6h2727217p23739966

The answer is $a_n=n$ for all $n=1,2,\dots$ By taking $\pmod{n^2+1}$, we get $a_n\equiv n\pmod{n^2+1}$, motivated by this we consider a sequence $\{b_n\}$ such that $a_n=n+b_n(n^2+1)$, Assume for the contradiction that $b_1\neq0$. replacing it into the first equality yields \[\frac{b_{n+1}}{b_n}=\frac{1}{n^2+2n+2}\]it turns out that $b_i\neq 0$ for all $i\in\mathbb{N}$. So\[\frac{b_n}{b_1}=\frac{b_n}{b_{n-1}}\times\frac{b_{n-1}}{{b_{n-2}}}\times\dots\times\frac{b_2}{b_1}=\frac{1}{A(n)}\]for some increasing integer function $A$. (assume to be $A$ since I am too lazy to write all of the fractions.) Take a large enough $n$ that make $A(n)>b_1$, it turns out that $b_n<1$ is not a natural number, which is a contradiction. So $b_1=0$ and $b_{n+1}(n^2+2n+2)=b_n$ gives $b_n=n$ for all $n\in\mathbb{N}$, Thus $a_n=n$ for all $n=1,2,\dots$