Tìm nghiệm nguyên của phương trình x + xy + y = 9

 Ta có \[x + xy + y = 9\]

\[x + xy + y + 1 = 10\]

\[x\left( {y + 1} \right) + \left( {y + 1} \right) = 10\]

\[\left( {x + 1} \right)\left( {y + 1} \right) = 10\].

Mà \[10 = 10 \cdot 1 = 2 \cdot 5 = 5 \cdot 2 = \left( {--10} \right) \cdot \left( {--1} \right) = \left( {--1} \right) \cdot \left( {--10} \right) = \left( {--2} \right) \cdot \left( {--5} \right) = \left( {--5} \right) \cdot \left( {--2} \right).\]

Ta có bảng các trường hợp sau:

\[x + 1\]	1	10	2	5	–10	–1	–2	–5
\[y + 1\]	10	1	5	2	–1	–10	–5	–2
\[x\]	0	9	1	4	–11	–2	–3	–6
\[y\]	9	0	4	1	–2	–11	–6	–3

Vậy \[\left( {x\,;\,\,y} \right)\; \in \;\left\{ {\left( {0;\,\,9} \right),\,\,\left( {9\,;\,\,0} \right),\,\,\left( {1\,;\,\,4} \right),\,\,\left( {4\,;\,\,1} \right),\,\,\left( {--11\,;\,\,--2} \right),\,\,\left( {--2\,;\,\,--11} \right),\,\,\left( {--3\,;\,\,--6} \right),\,\,\left( {--6\,;\,--3} \right)} \right\}.\]