Proving that $1^n+2^n+3^n+4^n$ $(n\in \Bbb N)$ is divisible by 10 when $n$ is not divisible by 4 I was solving some math problems to prepare for math contests and came across this one:
Prove that $1^n+2^n+3^n+4^n$ $(n\in N)$ is divisible by 10 if and only if $n$ is not divisible by 4.
So, from what I understand, we have to prove that:
$1+16^n+81^n+256^n$ $(n\in \Bbb N)$ is never divisible by 10;
$1+2 \cdot 16^n+3 \cdot 81^n+4 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10;
$1+4 \cdot 16^n+9 \cdot 81^n+16 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10;
$1+8 \cdot 16^n+27 \cdot 81^n+64 \cdot 256^n$ $(n\in \Bbb N)$ is always divisible by 10.
The problem is, these equations seem even more complex than the starting one. How would you prove these? Am I on the right track or is there an easier and more elegant way to do this? Thanks.

$1^n+2^n+3^n+4^n$ is divisible by $10$ if and only if it's divisible by $2$ and $5$.
Clearly $1^n+2^n+3^n+4^n$ is even for all $n\in\mathbb Z^+$, so let's work with mod $5$.
If $n=4k$ for some $k\in\mathbb Z^+$, then:
$$1^k+16^k+81^k+256^k\equiv 1^k+1^k+1^k+1^k\equiv 4\not\equiv 0\pmod{5}$$
If $n$ is odd, then:
$$1^n+2^n+3^n+4^n\equiv 1^n+2^n+(-2)^n+(-1)^n\pmod{5}$$
$$\equiv 1^n+2^n-2^n-1^n\equiv 0\pmod{5}$$
If $n=4k+2$ for some $k\in\mathbb Z_{\ge 0}$, then:
$$1^n+4\cdot 16^k+9\cdot 81^k+16\cdot 256^k\pmod{5}$$
$$\equiv 1+(-1)\cdot 1^k+(-1)\cdot 1^k+1\cdot 1^k\equiv 1-1-1+1\equiv 0\pmod{5}$$
