A number theory problem by Wildan Bagus Wicaksono

$\begin{aligned} \large a & = \large 3 + \frac 4{3+\frac 4{3+\frac 4{\ddots}}} \\ \implies a & = 3+\frac 4a \\ a^2 -3a - 4 & = 0 \\ (a-4)(a+1) & = 0 \\ \implies a & = 4 & \small \color{#3D99F6} \text{Note that }a > 0 \\ \implies b & = \frac a2 = 2 \\ \implies c & = \frac b2 = 1 \end{aligned}$

Therefore,

$\begin{aligned} S & = \sum_{k=0}^{2^a3^b7^c} 3^k {2^a3^b7^c \choose k} = \sum_{k=0}^{2^43^27^1} 3^k {2^43^27^1 \choose k} = \sum_{k=0}^{1008} 3^k {1008 \choose k} & \small \color{#3D99F6} \text{Note that }(1+x)^n = \sum_{k=0}^n {n \choose k} \\ & = (1+3)^{1008} = 4^{1008} = \left(\sqrt 2 \right)^{4 \times 1008} = \left(\sqrt 2 \right)^{4032} \end{aligned}$

$\implies m = \boxed{4032}$

$\large a=3+\frac { 4 }{ 3+\frac { 4 }{ 3+\frac { 4 }{ \ddots } } } \\ \large a=3+\frac { 4 }{ a } \\ { a }^{ 2 }=3a+4\\ \large 0={ a }^{ 2 }-3a-4\\ 0=(a-4)(a+1)\\ \large \Rightarrow a=4,a=-1$

Because $\large 3+\frac { 4 }{ 3+\frac { 4 }{ 3+\frac { 4 }{ \ddots } } } >0\\$ , so the value of $\large 3+\frac { 4 }{ 3+\frac { 4 }{ 3+\frac { 4 }{ \ddots } } } =4\\$ .

$b=\frac { a }{ 2 } \\ b=\frac { 4 }{ 2 } \\ b=2\\ c=\frac { b }{ 2 } \\ c=\frac { 2 }{ 2 } \\ c=1$

$\large =\sum _{ k=0 }^{ { 2 }^{ a }{ 3 }^{ b }{ 7 }^{ c } }{ { 3 }^{ k }\left( \begin{matrix} { 2 }^{ a }{ 3 }^{ b }{ 7 }^{ c } \\ \large k \end{matrix} \right) } \\ \large =\sum _{ k=0 }^{ { 2 }^{ 4 }{ 3 }^{ 2 }{ 7 }^{ 1 } }{ { 3 }^{ k }\left( \begin{matrix} 2^{ 4 }{ 3 }^{ 2 }{ 7 }^{ 1 } \\ \large k \end{matrix} \right) } \\ \large =\sum _{ k=0 }^{ 16\cdot 9\cdot 7 }{ { 3 }^{ k }\left( \begin{matrix} 16\cdot 9\cdot 7 \\ \large k \end{matrix} \right) } \\ \large =\sum _{ k=0 }^{ 1008 }{ { 3 }^{ k }\left( \begin{matrix} 1008 \\ k \end{matrix} \right) } \\ \large ={ 3 }^{ 0 }\left( \begin{matrix} 1008 \\ \large 0 \end{matrix} \right) +{ 3 }^{ 1 }\left( \begin{matrix} 1008 \\ \large 1 \end{matrix} \right) +{ 3 }^{ 2 }\left( \begin{matrix} 1008 \\ 2 \end{matrix} \right) +...+{ 3 }^{ 1007 }\left( \begin{matrix} 1008 \\ \large 1 \end{matrix} \right) +{ 3 }^{ 1008 }\left( \begin{matrix} 1008 \\ \large 1008 \end{matrix} \right)$

Based on ${ \left( a+b \right) }^{ n }=\left( \begin{matrix} n \\ 0 \end{matrix} \right) { a }^{ 0 }{ b }^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) { a }^{ 1 }{ b }^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { a }^{ 2 }{ b }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) { { a } }^{ n-1 }{ b }^{ 1 }+\left( \begin{matrix} n \\ n \end{matrix} \right) { a }^{ n }{ b }^{ 0 }$

So we get

$\large ={ 3 }^{ 0 }\left( \begin{matrix} 1008 \\ \large 0 \end{matrix} \right) +{ 3 }^{ 1 }\left( \begin{matrix} 1008 \\ \large 1 \end{matrix} \right) +{ 3 }^{ 2 }\left( \begin{matrix} 1008 \\ 2 \end{matrix} \right) +...+{ 3 }^{ 1007 }\left( \begin{matrix} 1008 \\ \large 1 \end{matrix} \right) +{ 3 }^{ 1008 }\left( \begin{matrix} 1008 \\ 1008 \end{matrix} \right) \\ \large =\left( \begin{matrix} 1008 \\ \large 0 \end{matrix} \right) { 1 }^{ 1008 }{ 3 }^{ 0 }+\left( \begin{matrix} 1008 \\ \large 1 \end{matrix} \right) { 1 }^{ 1007 }{ 3 }^{ 1 }+\left( \begin{matrix} 1008 \\ \large 2 \end{matrix} \right) { 1 }^{ 1006 }{ 3 }^{ 2 }+...+\left( \begin{matrix} 1008 \\ \large 1 \end{matrix} \right) { 1 }^{ 1 }{ 3 }^{ 1007 }+\left( \begin{matrix} 1008 \\ \large 1 \end{matrix} \right) { 1 }^{ 0 }{ 3 }^{ 1008 }\\ \large ={ (1+3) }^{ 1008 }\\ ={ 4 }^{ 1008 }\\ \large ={ 2 }^{ 2016 }$

$\large { 2 }^{ 2016 }={ \sqrt { 2 } }^{ m }\\ \large { 2 }^{ 2016 }={ 2 }^{ \frac { 1 }{ 2 } m }\\ \large \Rightarrow 2016=\frac { 1 }{ 2 } m\\ \large \therefore m=4032$