a) Ta có :
$n^3+n^2-n+5$
$=n^2.(n+2) - n.(n+2) + (n+2) + 3$
$=(n+2).(n^2-n+1)+3$
Để : giá trị $n^3+n^2-n+5 \vdots n+2$
$⇔ 3 \vdots n+2$
$⇔n+2 ∈ Ư(3)$
$⇔ n+2 ∈ ${$-1,1,-3,3$}
$⇔n ∈${$-3,-1,-5,1$}
b) Ta có :$n^3+3n+5 = n.(n^2+2) + (n+5)$
Để : giá trị $n^3+3n+5 \vdots n^2+2$
$⇔n+5 \vdots n^2+2$
$⇔n^2-25 \vdots n^2+2$
$⇔ 27 \vdots n^2+2$
$⇔ n^2 + 2 ∈ ${3,9,27$}
Vì $n^2+2 ≥ 2$
$⇔n^2 ∈ ${$1,7,25$}
Mà $n∈Z$
$⇒n^2 ∈ ${$1,25$}
$⇔n ∈ ${$-1,1,-5,5$}
