Giải thích các bước giải:
Ta có:
$\begin{array}{l}
2014 \equiv 8\left( {\bmod 17} \right)\\
\Rightarrow {2014^{2015}} \equiv {8^{2015}}\left( {\bmod 17} \right)\left( 1 \right)
\end{array}$
Mà
$\begin{array}{l}
{8^4} \equiv - 1\left( {\bmod 17} \right)\\
\Rightarrow {\left( {{8^4}} \right)^{503}} \equiv {\left( { - 1} \right)^{503}} = - 1\left( {\bmod 17} \right)\\
\Rightarrow {8^{2012}} \equiv - 1\left( {\bmod 17} \right)\\
\Rightarrow {8^{2015}} \equiv - {8^3} \equiv 15\left( {\bmod 17} \right)\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow {2014^{2015}} \equiv 15\left( {\bmod 17} \right)\left( * \right)$
Lại có:
$\begin{array}{l}
2015 \equiv 9\left( {\bmod 17} \right)\\
\Rightarrow {2015^{2016}} \equiv {9^{2016}}\left( {\bmod 17} \right)\left( 3 \right)
\end{array}$
Mà
$\begin{array}{l}
{9^4} \equiv - 1\left( {\bmod 17} \right)\\
\Rightarrow {\left( {{9^4}} \right)^{504}} \equiv {\left( { - 1} \right)^{504}} = 1\left( {\bmod 17} \right)\left( 4 \right)
\end{array}$
Từ $\left( 3 \right),\left( 4 \right) \Rightarrow {2015^{2016}} \equiv 1\left( {\bmod 17} \right)\left( {**} \right)$
Từ $\left( * \right),\left( {**} \right) \Rightarrow {2014^{2015}} + {2015^{2016}} \equiv 15 + 1 = 16\left( {\bmod 17} \right)$
$ \Rightarrow \left( {{{2014}^{2015}} + {{2015}^{2016}}} \right)$ chia $17$ dư $16$
Vậy $\left( {{{2014}^{2015}} + {{2015}^{2016}}} \right)$ chia $17$ dư $16$
