Question

Determinați n€Z pentru care fractia 2n+5.
/n+1 este reductibila. ​

Answer 1

(2n+5)/(n+1)∈Z ⇔ n+1|2n+5, dar n+1|n+1 ⇔ n+1|2(n+1) ⇒ n+1|2n+2.

cum n+1|2n+5 iar n+1|2n+2 ⇒ n+1|(2n+5)-(2n+2) ⇔ n+1|3 ⇒ n+1∈D₃ ⇔

n+1∈{-3,-1,1,3} ⇔ n∈{-4,-2,0,2}.