Vì ƯCLN(a; b) = 16⇒ $\left \{ {{a=16×x} \atop {b=16×y}} \right.$
(a; b; x; y ∈ N) ; (m; n) = 1
Ta có: a + b = 128
⇒ 16 × m + 16 × n = 128
⇒ 16 × ( m + n ) = 128
⇒ m + n = $\frac{128}{16}$
⇒ m + n = 8
Mà ( m; n ) = 1
⇒$\left \{ {{m=7} \atop {n=1}} \right.$ ⇒$\left \{ {{a=112} \atop {b=16}} \right.$
$\left \{ {{m=5} \atop {n=3}} \right.$⇒$\left \{ {{a=80} \atop {b=48}} \right.$
$\left \{ {{n=7} \atop {m=1}} \right.$ ⇒$\left \{ {{a=16} \atop {b=112}} \right.$
$\left \{ {{n=5} \atop {m=3}} \right.$⇒$\left \{ {{a=48} \atop {b=80}} \right.$
Vậy ( a; b ) = ( 112; 16 ); ( 16; 112 ); ( 80; 48 ); ( 48; 80 )
