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I've got to reduce it to this

$n+1|n^3+3n^2+n+2$

$n+1|n^3+3n^2+1$

$n+1|2n^2+1$

$n+1|2n^2+2n-2n+1$

$n+1|2n(n+1)-2n+1$

$n+1|-2n+3-2$

$n+1|-2(n+1)+3$

$n+1|3$

All this using propertie $c|a \land c|b \implies c|a \pm b$, however, I don't know how to find the $n$ integers solutions.

Bill Dubuque
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not_aphysicsdr
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    $\dfrac{n^3+3n^2+n+2}{n+1}=n^2+2n-1+\dfrac{3}{n+1}$ so $n+1$ should be equal to $\pm3$ and $\pm1$and the solutions are $\pm2,\space -4$ and $0$. – Piquito Sep 27 '23 at 16:47
  • To solve $,n+1\mid 3,$ simply list all divisors $d$ of $3$ then equate them to $n+1,,$ i.e. $,n+1 = d\iff n = d-1,,$ i.e. subtracting $1$ from all divisors of $d$ yields all solutions $,n.\ \ $ – Bill Dubuque Sep 29 '23 at 00:58

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