a) Tìm tất cả các bộ số nguyên \[\left( {x\,\,;\,\,y} \right)\] thỏa mãn phương trình:

\[{x^2} - 2x + 2{y^2} = 2\left( {xy + 1} \right)\]

b) Cho \[p\] là số nguyên tố sao cho tồn tại các số nguyên dương \[x\,\,;\,\,y\] thỏa mãn \[{x^3} + {y^3} - p = 6xy - 8.\] Tìm giá trị lớn nhất của \[p\].

a) Ta có:

\[\begin{array}{l}{x^2} - 2x + 2{y^2} = 2\left( {xy + 1} \right)\,\,\\ \Leftrightarrow \,\,{x^2} - 2x + 2{y^2} = 2xy + 2\,\,\\ \Leftrightarrow \,\,{x^2} - 2xy + {y^2} + {y^2} - 2x = 2\,\,\end{array}\]

\[\begin{array}{l} \Leftrightarrow {\left( {x - y} \right)^2} - 2x + {y^2} - 2y + 1 + 2y = 3\,\,\\ \Leftrightarrow \,\,{\left( {x - y} \right)^2} - 2\left( {x - y} \right) + {\left( {y - 1} \right)^2} = 3\,\,\\ \Leftrightarrow \,\,{\left( {x - y} \right)^2} - 2\left( {x - y} \right) + 1 + {\left( {y - 1} \right)^2} = 4\end{array}\]

\[ \Leftrightarrow {\left( {x - y - 1} \right)^2} + {\left( {y - 1} \right)^2} = 4\,\,\left( { = {0^2} + {2^2}} \right)\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}x - y - 1 = 0\\y - 1 = 2\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x - y - 1 = 0\\y - 1 = - 2\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}y - 1 = 0\\x - y - 1 = - 2\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}y - 1 = 0\\x - y - 1 = 2\end{array} \right.\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}x = 4\\y = 3\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 0\\y = - 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}y = 1\\x = 0\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}y = 1\\x = 4\end{array} \right.\]

Vậy \[\left( {x\,\,;\,\,y} \right)\,\, = \,\,\left( {4\,\,;\,\,3} \right)\,\,;\,\,\left( {0; - 1} \right)\,\,;\,\,\left( {0\,\,;\,\,1} \right)\,\,;\,\,\left( {4\,\,;\,\,1} \right).\]

b) Ta có:

\[\begin{array}{l}{x^3} + {y^3} - p = 6xy - 8\,\,\\ \Leftrightarrow \,\,p\, = \,{x^3} + {y^3} - 6xy + 8\,\,\\ \Leftrightarrow \,\,p\, = \,{\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) - 6xy + 8\end{array}\]

\[\begin{array}{l} \Leftrightarrow \,\,p\, = \,\left[ {{{\left( {x + y} \right)}^3} + 8} \right] - 3xy\left( {x + y + 2} \right)\,\\\, \Leftrightarrow \,p\, = \,\left( {x + y + 2} \right)\,\left[ {{{\left( {x + y} \right)}^2} - 2\left( {x + y} \right) + 4 - 3xy} \right]\end{array}\]

Do \[p\] là số nguyên tố nên:

\[\left[ \begin{array}{l}\left( {x + y + 2} \right) = 1\,\,\\{\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 4 - 3xy\, = 1\end{array} \right.\,\,\, \Rightarrow \,\,{\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 4 - 3xy\, = 1\,\,\,\]

(Vì: \[x\,;\,y\,\, \in \,\,{\mathbb{Z}^ + } \Rightarrow x + y + 2\,\, \ge \,\,4\])

\[\begin{array}{l} \Rightarrow \,\,{\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 4 - 3xy\, = 1\,\,\,\\ \Leftrightarrow {x^2}\, + 2xy + {y^2} - 2x - 2y - 3xy = - 3\,\,\\ \Leftrightarrow \,{x^2}\, - xy + {y^2} - 2x - 2y = - 3\end{array}\]

\[\begin{array}{l}\, \Leftrightarrow \,\,4{x^2}\, - 4xy + 4{y^2} - 8x - 8y = - 12\,\,\\ \Leftrightarrow \,\,{\left( {2x - y} \right)^2}\, + 3{y^2} - 4\left( {2x - y} \right) + 4 - 12y + 12 = 4\end{array}\]

\[\, \Leftrightarrow \,\,{\left( {2x - y - 2} \right)^2}\, + 3{\left( {y - 2} \right)^2} = 4\,\left( { = {1^2} + {{3.1}^2}} \right)\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}2x - y - 2 = 1\\y - 2 = 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}2x - y - 2 = 1\\y - 2 = - 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}2x - y - 2 = - 1\\y - 2 = 1\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}2x - y - 2 = - 1\\y - 2 = - 1\end{array} \right.\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}x = 3\\y = 3\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 2\\y = 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 2\\y = 3\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 1\\y = 1\end{array} \right.\]

TH1: \[\left\{ \begin{array}{l}x = 3\\y = 3\end{array} \right.\,\,\, \Rightarrow \,\,p = 8\,\,\left( {KTM} \right)\]

TH2: \[\left\{ \begin{array}{l}x = 2\\y = 1\end{array} \right.\,\,\, \Rightarrow \,\,p = 5\,\,\left( {TM} \right)\,\]

TH3: \[\left\{ \begin{array}{l}x = 2\\y = 3\end{array} \right.\,\,\, \Rightarrow \,\,p = 7\,\,\left( {TM} \right)\]

TH4: \[\left\{ \begin{array}{l}x = 1\\y = 1\end{array} \right.\,\,\, \Rightarrow \,\,p = 4\,\,\left( {KTM} \right)\]

Vì: \[p\] là số nguyên tố lớn nhất \[ \Rightarrow \,\,p = 7\]

Vậy \[p = 7\] thỏa mãn yêu cầu bài toán.