<s>[INST] $1^n-3^n-6^n+8^n$ 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? 
 [/INST] Perhaps another method to notice is that, by Euler's Generalization of Fermat's Little Theorem, we have that:
$$1^n-3^n-6^n+8^n \equiv 1^{n \mod \phi(10)} - 3^{n \mod \varphi(10)} - 6^{n \mod \varphi(10)} + 8^{n \mod \varphi(10)} \mod 10$$
Since $\varphi(10) = 4$, you really only need to check the cases of $n = 0, 1, 2$ and $3$. For $n = 0$, you have:
$$1^0-3^0-6^0+8^0 \equiv 1 - 1 - 1 + 1 \equiv 0 \mod 10$$
For $n = 1$:
$$1^1-3^1-6^1+8^1 \equiv 1 - 3 - 6 + 8 \equiv 0 \mod 10$$
For $n = 2$:
$$1^2-3^2-6^2+8^2 \equiv 1 - 9 - 36 + 64 \equiv 20 \equiv 0 \mod 10$$
For $n = 3$:
$$1^3-3^3-6^3+8^3 \equiv 1 - 27 - 216 + 512 \equiv 270 \equiv 0 \mod 10$$
Since all the other exponents leading forward will be one of these cases, you are done. It is not the most elegant way to prove this, however, it is fairly direct.
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