Giải hệ phương trình $\left\{ \begin{array}{l}{x^2} = 2x - y\\{y^2} = 2y - z\\{z^2} = 2z - t\\{t^2} = 2t - x\end{array} \right.$

$\left\{ \begin{array}{l}{x^2} = 2x - y\\{y^2} = 2y - z\\{z^2} = 2z - t\\{t^2} = 2t - x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\left( {1 - x} \right)^2} = 1 - y \ge 0\\{\left( {1 - y} \right)^2} = 1 - z \ge 0\\{\left( {1 - z} \right)^2} = 1 - t \ge 0\\{\left( {1 - t} \right)^2} = 1 - x \ge 0\end{array} \right.$

Đặt $\left\{ \begin{array}{l}b = 1 - y \ge 0\\c = 1 - z \ge 0\\d = 1 - t \ge 0\\a = 1 - x \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{a^2} = b\\{b^2} = c\\{c^2} = d\\{d^2} = a\end{array} \right.$

+) Xét $a = 0 \Rightarrow b = c = d = 0 \Rightarrow x = y = z = t = 1$

+) Xét $a \ne 0 \Rightarrow b;c;d \ne 0$.

Ta có $\left\{ \begin{array}{l}{a^2} = b\\{b^2} = c\\{c^2} = d\\{d^2} = a\end{array} \right.$ nhân theo vế ta có ${\left( {abcd} \right)^2} - abcd = 0 \Leftrightarrow abcd = 1$( vì $abcd \ne 0$)

Mặt khác $\left\{ \begin{array}{l}{a^2} = b \ge 0\\{b^2} = c \ge 0\\{c^2} = d \ge 0\\{d^2} = a \ge 0\end{array} \right. \Rightarrow {a^2} + {b^2} + {c^2} + {d^2} = a + b + c + d$

$\begin{array}{l} \Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} + 2{d^2} - 2a + 2b + 2c + 2d = 0\\ \Leftrightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {c - 1} \right)^2} + {\left( {d - 1} \right)^2} + {a^2} + {b^2} + {c^2} + {d^2} - 4 = 0\end{array}$

Ta có ${a^2} + {b^2} + {c^2} + {d^2} \ge 4\sqrt[4]{{{a^2}{b^2}{c^2}{d^2}}} = 4$

$\Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {c - 1} \right)^2} + {\left( {d - 1} \right)^2} + {a^2} + {b^2} + {c^2} + {d^2} - 4 \ge 0$.

Dấu “=” xảy ra $\Leftrightarrow a = b = c = d = 1 \Rightarrow x = y = z = t = 0$.

Vậy hệ phương trình có nghiệm $\left( {x;y;z;t} \right)$là $\left( {0;0;0;0} \right);\left( {1;1;1;1} \right)$.