n²+n+4/n+1 ∈Z ⇒n+1 divide n²+n+4
n²+n+4=n²+4n+4-3n=(n+2)²-3n
(n+2)²-3n/n+1∈Z ⇒n+1 divide (n+2)²
n+1/n+1
n+1/(n+2)² -3n⇒din cele doua relatii va rezulta
n+1/(n+2)²-3n-n-1
n+1/(n+2)²-4n-1
n+1/n²+4n+4-4n-1
n+1/n²+4-1
n+1/n²+3 ⇔
n/n²+2
n∈Z ⇒de unde obtinem n∈{-2,-1,1,2};
