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?
 A: You could use the Chinease remainder theorem, but you don't really need to.  You just need to know that $x = 0 \pmod a$ if and only if $a$ divides $x$.  So $x = 0$ modulo $4$ and $5$ if and only if both of them divide $x$.  As $4$ and $5$ don't share any prime factors they both divide $x$ if and only if their product, $20$, divides $x$.
p.s. That line about not sharing prime factors is a simplified version of the Chinease remainder theorem in disguise, but I think calling it the Chinease remainder theorem makes it seem like a much more complicated fact than it really is.
Edit: If you want to know that $20$ divides $7^n = 4\cdot2^n + 8^n - 3^n$ then reduce it modulo $4$ and modulo $5$.  If you get $0$ both times (you will) then you are done.
A: $\Longrightarrow$
Assume $\;20\mid x.\;$ Clearly then, $x = 20k$ for some integer k. 
$20k = 4\cdot 5k\;$ and $\;20k = 5 \cdot 4k.\;\;\implies \;\;4\mid 20k\;$ and $5\mid 20k\;$ $$\iff 4\mid x\;\;\text{and}\;\;5 \mid x$$
$$\iff x \equiv 0 \pmod 4\;\;\text{and}\;\;x \equiv 0 \pmod 5$$ 
$\Longleftarrow$
$\quad x \equiv 0\pmod 4\;$ AND $\;x \equiv 0 \pmod 5\;$ means $4 \mid x \iff x = 4m$ and $\;5\mid x \iff x = 5n\;$ for some integers $m,n.\;$ Since the $\;\gcd(4, 5) = 1$, the least common multiple $\operatorname{lcm}(4, 5) = 20.$ So any integer $x$ satisfying both $\;x = 4m\;$ AND $\;x = 5n,\;$ must be such that $x = 4\cdot 5 \cdot k = 20k$ for some integer $k$. That is, we have that $\;20 \mid x$.
A: 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 $\rm\:4,5\mid f(n)\:\Rightarrow\:20\mid f(n).$
A: $20 = 4*5$ so if $x$ is divisible by $20$ then it is divisible by $4$ and $5$. And hence by definition $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$ 
Conversely, if $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$  then $x = 4n$ and $x= 5m$ for some integers $n$ and $m$.
So $4n$ = $5m$. 
Now since $5$ does not divide $4$, it must divide $n$. Hence $n = 5l$ for some integer $l$.
Hence $x = 4n = 4*5*l$.
Hence $x$ is divisible by $20$.
