(3n+9)/(2n+1) ∈N
n∈N =>
=>(2n+1)|(3n+9) <=> (2n+1)|2(3n+9)
(2n+1)|(2n+1) (2n+1)|3(2n+1) <=>
<=> (2n+1)|(6n+18)
(2n+1)|(6n+3) <=> (2n+1)| 6n+18-6n-3 <=> (2n+1)|15 =>
=>(2n+1)={1,3,5,15}
1) 2n+1= 1 <=> 2n=0 =>n=0
2) 2n+1=3 <=> 2n=2 =>n=1
3) 2n+1=5 <=> 2n=4 =>n=2
4) 2n+1=15 <=> 2n=14 =>n=7
n={0,1,2,7}
