Đáp án đúng:
Giải chi tiết:Gọi \(d = \left( {{m^2} + {n^2};m + n} \right) \Rightarrow \left\{ \begin{array}{l}{m^2} + {n^2}\,\, \vdots \,\,d\\m + n\,\, \vdots \,\,d\end{array} \right.\).
\(\eqalign{ & \Rightarrow \left\{ \matrix{ {\left( {m + n} \right)^2} - 2mn\,\, \vdots \,\,d \hfill \cr m + n\,\,\, \vdots \,\,d \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 2mn\,\, \vdots \,\,d \hfill \cr m + n\,\, \vdots \,\,d \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 2n\left( {m + n} \right) - 2mn\,\, \vdots \,\,d \hfill \cr 2m\left( {m + n} \right) - 2mn\,\, \vdots \,\,d \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ 2{n^2}\,\, \vdots \,\,d \hfill \cr 2{m^2}\,\, \vdots \,\,d \hfill \cr} \right. \Leftrightarrow d|\left( {2{m^2};2{n^2}} \right) = 2\left( {{m^2} + {n^2}} \right) = 2 \Rightarrow d \in \left\{ {1;2} \right\} \cr} \)
Nếu \(m+n\,\,\vdots \,\,2\Rightarrow d=2\)
Nếu \(\left( m+n \right)\) lẻ \( \Rightarrow d = 1\).
