Prove for every $1 \leq n$, that $47\mid3^n4^{2n}-1$ I tried to solve with induction, base $n=1$ is correct such that, $\frac{3*16-1}{47}=1$.
Then assuming that for $n$ the statement is correct such that,  $3^n4^{2n} \equiv 0 \pmod {47}$.
We need prove that for $n+1$ is true such that, $\frac{(3*16*48^n)-1}{47}$ .
Then I tried to simplify it to $\frac{48}{47}*48^n-\frac{1}{47}= 1\frac{1}{47}*48^n-\frac{1}{47}=\frac{(3^n4^{2n})+47(3^n4^{2n})}{47}-\frac{1}{47}=\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$
We get that from $\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$, the expression  $\frac{(3^n4^{2n})} {47}-\frac{1}{47}$ is from the assumption so we are left with $3^n4^{2n}$ which is a natural number.
Is that proof enough to show that $47\mid3^n4^{2n}-1$?

Transform ($3^n4^{2n}-1) \to$ ($3^n16^n-1) \to (48^n-1)$
For sum of geometric series
$1+x+x^2+x^3+...+x^{n-1}=\frac{x^n-1}{x-1}$
or
$x^n-1=(x-1)(1+x+x^2+x^3+...+x^{n-1})$
Set x=48,
$48^n-1=(48-1)(1+48+48^2+48^3+...+48^{n-1})$
=
$47 (1+48+48^2+48^3+...+48^{n-1}) \to  47\mid3^n4^{2n}-1$
