Half of solution: If $n=2k$, we have ${{(4-1)}^{n}}+47=4k+47+{{(-1)}^{n}}\vdots 4$ , $20{{(4+1)}^{n}}-2=4m+18$ and this is not divisible by $4$. So the denominatot is divisible by $4$, and the numerator not! Done!

Here is my solution: Assume the opposite. If \(n\) is even then 4 divides denominator but does not divide numerator. If \( n \equiv 1\) (\(mod\) 4) 5 divides denominator but does not divide numerator. Thus \(n \equiv 3\)(\(mod\) 4) and \(\frac{3^n+47}{2} \equiv 5\) (\(mod\) 8). Let \(p\) be a prime number. It is well known that if 2 is a quadratic residue (\(mod\) \(p\)) then we have \(p \equiv 1\), \(-1\) (\(mod\) 8). Because the numerator is of the form \(x^2-2\), \(\frac{3^n+47}{2}\) must be 1 or -1 (\(mod\) \(8\)). Contradiction.

dame dame

Turkey 2021 IMO TST Problem 1 wrote: Prove that for any positive integer $n$, $$\frac{20\cdot 5^n-2}{3^n+47}\not \in\mathbb{Z}.$$ Firstly, we have $20\cdot 5^n-2\equiv 2\pmod 4$, but if $n$ is even, then $3^n+47\equiv 0\pmod 4$, therefore in fact $n$ is odd, thus let $n=2k-1$, $k\in \mathbb{N}$, now $$3^{2k-1}+47\mid (2\cdot 5^k)^{2}-2,$$now we must have $x^{2}\equiv 2\pmod{p}$, where $p\mid 3^{2k-1}+47$. In fact $2$ is quadratic residue modulo $p$, iff $p\equiv 1,7\mod{8}$. By modulo $5$ we obtain $3^n\not \equiv 3\pmod{5}\implies n\not\equiv 1\pmod{4}$, thus $n\equiv 3\pmod{4}$ as $n$ is odd, write $n=4r+3$. $$2,14\equiv 2\cdot \pm 1\equiv 3^{4r+3}+47\equiv 27\cdot 81^r+47\equiv 11-1\equiv 10\pmod{16},$$impossible.

by mod4 $n$ is odd. by mod5 $n\equiv -1 (mod4)$ $$n=4k+3$$$$3^n+47=3^{4k+3}+47 \equiv 10(mod16) (*)$$exist $p$ prime number.$p|3^n+47$ $p^2$ is not 1 mod 16.by (*). $20\cdot 5^n=A^2$ $$A^2 \equiv 2 (modp)$$$$\frac{2}{p}=(-1)^(\frac{p^2-1}{8})=-1$$contradiction.