Question

$\frac{n^{2}+n+4}{n+1}$∈Z. Determina n. Va rog, dau coroana

Answer 1

(n²+n+4)/(n+1)=[n*(n+1)+4]/n+1)=n*(n+1)/(n+1)+4/(n+1)=n+4/(n+1)
n=numar intreg  Pui  conditia  ca  si  fractia  sa  fie  nr  intreg adica n+1  l4
n+1=D4
n+1={-4  .-2.-1.1,2,4}
n1=-4-1=-5
n2=-2-1=-3
n3=-1 -1=-2
......................
n6=4-1=3
n∈{-5 ,-3, -2,0,1.3}