`x/16+1/y=1/32`
`⇒x/16+x/xy=x/32x`
`⇒16+xy=32x`
`⇒32x-xy=16`
`⇒x(32-y)=16`
`⇒x;32-y∈Ư(16)={±1;±2;±4;±8±16}`
Ta có bảng :
$\left[\begin{array}{ccc}32-y&1&2&4&8&16&-1&-2&-4&-8&-16\\x&16&8&4&2&1&-16&-8&-4&-2&-1\\y&31&30&18&24&16&33&34&36&40&48\end{array}\right]$
Vậy `(x;y)=(16;31);(8;30);(4;18);(2;24);(1;16);(-16;33);(-8;34);(-4;36);(-2;40);(-1;48)`