<s>[INST] Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$? 
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$

My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and 
$12|(n-2)  \Rightarrow n=12b+2$
so
$\begin{align}k\cdot 2^n+1 &=(13a+3)2^{12b+2}+1 \\
&=4(13a+3)(2^{b})^{12}+1\\
\textrm{ or }
&=(k-3+3)\cdot 2^{n-2+2}+1\\
&=4\cdot 2^{n-2}(k-3+3)+1\\
&=4\cdot 2^{n-2}(k-3)+3\cdot2^n+1\end{align}$
I don't kow how to use the hint :(
 [/INST] ${\rm mod}\ 13\!:\!\! \overbrace{\color{#c00}{ 2^{\large n}}\!\equiv 2^{\large 2+12j}}^{\large\quad\ \ \color{#c00}n\ \equiv\ 2\pmod{\!12}}\!\!\!\!\!\!\!\equiv 2^{\large 2}\color{#0a0}{(2^{\large 12})}^{\large j}\!\equiv 4\color{#0a0}{(1)}^{\large j}\!\equiv \color{#c00}4\,\ $ by $\ \rm\color{#0a0}{ Fermat}$
therefore $\ k\equiv 3\,\Rightarrow\, k\,\color{#c00}{2^{\large n}}\!+1\equiv 3\cdot \color{#c00}4+1\equiv 0\pmod{13}$
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