Đáp án+Giải thích các bước giải:
Để $n^2012+1$ là số chính phương $(k\in Z)$
$\begin{array}{l}
{n^{2012}} + 1 = {k^2}\\
\Leftrightarrow {k^2} - {n^{2012}} = 1\\
\Leftrightarrow {k^2} - {\left( {{n^{1006}}} \right)^2} = 1\\
\Leftrightarrow \left( {k - {n^{1006}}} \right)\left( {k + {n^{1006}}} \right) = 1\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
k - {n^{1006}} = 1\\
k + {n^{1006}} = - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
k - {n^{1006}} = - 1\\
k - {n^{1006}} = 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2k = 0\\
k - {n^{1006}} = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
2k = 0\\
k - {n^{1006}} = - 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
k = 0\\
{n^{1006}} = - 1\left( l \right)
\end{array} \right.\\
\left\{ \begin{array}{l}
k = 0\\
{n^{1006}} = 1\left( c \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow {n^{1006}} = 1\\
\Leftrightarrow n = \pm 1(tm)
\end{array}$
Vậy $n = \pm 1$
