Proving $x$ is divisible by $20$ I need to prove that $x$ divisible by $20$ if and only if $x=0\pmod4$ and $x=0 \pmod 5$
proving that if $x=0 \pmod 4$ and $x=0 \pmod 5$ than $x$ divisible by $20$ is by the Chinese theorem (am I right??)
But the other way - I dont understand why its not enough to be divisible only by one of them ($5$ OR $4$)?
P.S. How does it help me prove $(7^n+4*2^n+8^n-3^n)|20$?? How do I open this?

Hint $\rm\ \ 4,5\mid n\:\Rightarrow\: \dfrac{n}4,\dfrac{n}5\in\Bbb Z\:\Rightarrow\: \dfrac{n}4-\dfrac{n}5 = \dfrac{n}{20}\in\Bbb Z\:\Rightarrow\:20\mid n$
Edit $\ $ Regarding the new question in your edit, the above applies as follows
$\rm mod\ 4\!:\ f(n) = \color{#C00}{7^n}\!+4\cdot2^n\!+8^n\!-3^n\equiv \color{#C00}{3^n}+\,0\,+\,0\,-\,3^n\equiv 0$
$\rm mod\ 5\!:\ f(n) = \color{#C00}{7^n}\!+4\cdot2^n\!+\color{#0A0}{8^n}\!-3^n\equiv \color{#C00}{2^n}\!-2^n+ \color{#0A0}{3^n}-3^n\equiv 0$
Thus $\rm\:4,5\mid f(n)\:\Rightarrow\:20\mid f(n).$
