`x/3 + 1/6 = 4/y`
`⇔ ( 2x )/6 + 1/6 = 4/y`
`⇔ ( 2x + 1 )/6 = 4/y`
`⇔ ( 2x + 1 ) . y = 6 . 4`
`⇔ ( 2x + 1 ) . y = 24`
Mà `x , y ∈ Z ⇒ ( 2x + 1 ) . y = 1 . 24 = 2 . 12 = 3 . 8 = 4 . 6 = ( - 1 ) . ( - 24 ) = ( - 2 ) . ( - 12 ) = ( - 3 ) . ( - 8 ) = ( - 4 ) . ( - 6 )`
Ta có bảng sau `:`
\begin{array}{|c|c|c|}\hline \text{2x + 1}&\text{1}&\text{24}&\text{2}&\text{12}&\text{3}&\text{8}&\text{4}&\text{6}&\text{- 1}&\text{- 24}&\text{- 2}&\text{- 12}&\text{- 3}&\text{- 8}&\text{- 4}&\text{- 6}\\\hline \text{y}&\text{24}&\text{1}&\text{12}&\text{2}&\text{8}&\text{3}&\text{6}&\text{4}&\text{- 24}&\text{- 1}&\text{- 12}&\text{- 2}&\text{- 8}&\text{- 3}&\text{- 6}&\text{- 4}\\\hline \text{x}&\text{0}&\text{23/2}&\text{1/2}&\text{11/2}&\text{1}&\text{7/2}&\text{3/2}&\text{5/2}&\text{- 1}&\text{-25/2}&\text{-3/2}&\text{-13/2}&\text{-2}&\text{-9/2}&\text{-5/2}&\text{-7/2}\\\hline\end{array}
Mà `x , y ∈ Z ⇒ ( x ; y ) ∈ { ( 24 ; 0 ) , ( 8 ; 1 ) , ( - 24 ; - 1 ) ; ( - 8 ; - 2 )`
Vậy `, ( x ; y ) ∈ { ( 24 ; 1 ) , ( 8 ; 1 ) , ( - 24 ; - 1 ) ; ( - 8 ; - 2 )`
