$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough 
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$. 
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it is divisible by $2\cdot5=10$.
Is it correct? 

Another way to see the solution is by using 
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})\tag{1}$$
leading to
$$1^n-3^n-6^n+8^n=(8^n-3^n)-(6^n-1^n)=\\
(8-3)(8^{n-1}+8^{n-2}\cdot3+...+8\cdot3^{n-2}+3^{n-1})-\\
(6-1)(6^{n-1}+6^{n-2}\cdot1+...+6\cdot1^{n-2}+1^{n-1})=...$$
we can see that $8^{n-1}+8^{n-2}\cdot3+...+8\cdot3^{n-2}$ is always even and so it $6^{n-1}+6^{n-2}\cdot1+...+6\cdot1^{n-2}$ then
$$...=5(2m+3^{n-1})-5(2n+1)=5\left(2(m-n)+3^{n-1}-1\right)$$
and $3^{n-1}-1$ is even from the same application of $(1)$, thus we can "extract" a 2 from $\left(2(m-n)+3^{n-1}-1\right)$ and the result follows.
