Răspuns:
2n-1|3n+8|*2
2n-1|2n-1|*3 (inmultim in asa fel incat n sa aiba acelasi coeficient)
(orice numar se divide cu el insusi)
=> 2n-1|6n+16
2n-1|6n-3 => (scadem)=> 2n-1|6n+16-(6n-3)=> 2n-1|19 =>2n-1∈D19
2n-1∈{-1, 1, -19, 19} |+1
2n∈{0, 2, -18, 20} |:2
n∈{0, 1, -9, 10}
