dam valori
a=1 2+3b=12, b∉N
a=2, 4+3b=12, b∉N
a=3, 6+3b=12, b=2, solutie
a=4, 8+3b=12, b∉N
a=5, 10+3b=12, b∉N
a=6, 12+3b=12, b=0, dar a,b∈N*
pentru a>6, 2a+3b>12 ∀b∈N
b=1, 2a+3=12, a∉N
b=2, 2a+6=12, a=3 solutie
b=3, 2a+9=12, a∉N
b=4, 2a+12=12, a=0, dar a, b∈N
Am gasit o singura solutie
a=3, b=2
