Tìm nghiệm nguyên của phương trình: y² – 5y + 62 = (y – 2)x² + (y² – 6y + 8)x.

y² – 5y + 62 = (y – 2)x² + (y² – 6y + 8)x

\( \Leftrightarrow \) y² – 5y + 6 + 56 = (y – 2)x² + (y² – 6y + 8)x

\( \Leftrightarrow \) (y – 2)(y – 3) + 56 = (y – 2)x² + (y – 2)(y – 4)x

\( \Leftrightarrow \) (y – 2)[x² + (y – 4)x – (y − 3)] = 56

\( \Leftrightarrow \) (y – 2)(x² + xy – 4x − y + 3) = 56

\( \Leftrightarrow \) (y – 2)(x² – 4x + 3+ xy − y) = 56

\( \Leftrightarrow \) (y – 2)[(x – 1)(x – 3)+ y(x − 1)] = 56

\( \Leftrightarrow \) (y – 2)(x – 1)(x + y – 3) = 56 = 1.2.28 = 1.4.14 = 1.8.7 = 2.2.14

Nhận thấy x – 1 + y – 2 = x + y – 3 nên ta xét các trường hợp sau:

TH1: \(\left\{ \begin{array}{l}x - 1 = 1\\y - 2 = 7\\x + y - 3 = 8\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = 9\end{array} \right.\);

TH2: \(\left\{ \begin{array}{l}x - 1 = 7\\y - 2 = 1\\x + y - 3 = 8\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 8\\y = 3\end{array} \right.\);

TH3: \(\left\{ \begin{array}{l}x - 1 = - 8\\y - 2 = 7\\x + y - 3 = - 1\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = - 7\\y = 9\end{array} \right.\);

TH4: \(\left\{ \begin{array}{l}x - 1 = 7\\y - 2 = - 8\\x + y - 3 = - 1\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 8\\y = - 6\end{array} \right.\);

TH5: \(\left\{ \begin{array}{l}x - 1 = - 8\\y - 2 = 1\\x + y - 3 = - 7\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = - 7\\y = 3\end{array} \right.\);

TH6: \(\left\{ \begin{array}{l}x - 1 = 1\\y - 2 = - 8\\x + y - 3 = - 7\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = - 6\end{array} \right.\).

Vậy các cặp giá trị nguyên là nghiệm của phương trình là: (2; 9), (8; 3), (−7; 9), (8; −6), (−7; 3), (2; −6).