<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] Note: $p,q,r,s,t, a ,b,  n \in \mathbb{N},$
$f(n):= (8^n-3^n) - (6^n-1^n)= (8-3)p-(6-1)q=5(p-q).$
On the other hand:
$f(n)= (8^n-6^n) -(3^n-1^n)=(8-6)r -(3-1)s=2(r-s)$
Hence: 
$f(n)=5(p-q)=2(r-s).$
Euclid's Lemma: 
$2$ divides $(p-q)$, i.e.
$p-q=2t$.
Combining:
$f(n)= 5\cdot 2 t.$
Used:
$(a^n-b^n)=$
$(a-b)(a^{n-1}+a^{n-2}b+.....+b^{n-1}).$
$p,q,r$ and $s$ were used for the above second factor.
Euclid's Lemma:
If a prime $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.
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