\(\left( {2x - 1} \right).\left( {2y + 1} \right) = - 35\)

Ta có: \[ - 35 = \left( { - 1} \right).35 = 1.\left( { - 35} \right) = \left( { - 5} \right).7 = 5.\left( { - 7} \right)\]

Vì \[x,y{\rm{ }} \in \mathbb{Z}\;\] nên \[2x - 1{\rm{ }} \in \mathbb{Z}\;;\,\,2y + 1{\rm{ }} \in \mathbb{Z}\] và \(\left( {2x - 1} \right)\left( {2y + 1} \right) = - 35\)

Suy ra: + \(2x - 1 = \,\, - 1\,\,;2y + 1 = 35\)\( \Leftrightarrow x = \,\,0\,\,;y = 17\)

+ \(2x - 1 = \,\,35\,\,;2y + 1 = - 1\)\( \Leftrightarrow x = \,\,18\,\,;y = \,\, - 1\)

+ \(2x - 1 = \,\,1\,\,;2y + 1 = - 35\)\( \Leftrightarrow x = \,\,1\,\,;y = \,\, - 18\)

+ \(2x - 1 = \,\, - 35\,\,;2y + 1 = 1\)\( \Leftrightarrow x = \,\, - 17\,\,;y = \,\,0\)

+ \(2x - 1 = \,\, - 5\,\,;2y + 1 = 7\)\( \Leftrightarrow x = \,\, - 2\,\,;y = \,\,3\)

+\(2x - 1 = \,\,7\,\,;2y + 1 = - 5\)\( \Leftrightarrow x = \,\,4\,\,;y = \,\, - 3\)

+ \(2x - 1 = \,\,5\,\,;2y + 1 = - 7\)\( \Leftrightarrow x = \,\,3\,\,;y = \,\, - 4\)

+ \(2x - 1 = \,\, - 7\,\,;2y + 1 = \,\,5\)\( \Leftrightarrow x = \,\, - 3\,\,;y = \,\,2\)

Vậy \[\;\left( {x;y} \right) \in \left\{ {\left( {\,0;17\,} \right);\left( {\,18; - 1\,} \right):\left( {\,1; - 18\,} \right);\left( {\, - 17;0\,} \right);\left( {\, - 2;3\,} \right);\left( {\,4; - 3\,} \right):\left( {\,3; - 4\,} \right);\left( {\, - 3;2\,} \right)} \right\}\]

Tìm số nguyên x, biết: x^2 - 5x = 0