Question: Why are any two of $k$, $6k+1$, $8k+1$, $24k+1$ for $k\in\{2, 3, 4, 5, \ldots\}$ are mutually coprime? 

I am trying to show that $k$ and $6k+1$ are coprime, I am trying to reason by taking their difference, so $(6k+1)-k=5k+1$ but unable to show why are their coprime?  

Similarly $(8k+1)-(6k+1)=2k$, if their distance is a multiple of $k$, then why they must be coprime, suppose 3 and 27 are both multiple of 3, but 3+1=4 and 27+1=28, their distance 28-4=24 is a multiple of 3 but 4 and 28 are not coprime.

I am on my way to solve a problem but encounter this little result, I hope they are pairwise coprime to proceed, but I need a more convincing reasoning or proof that any two of them are coprime. Many thanks for any helps!

Answer: 
First,
$$GCD(a,b) = GCD(|a-b|,b)=GCD(a,|b-a|).$$

As a corollary:

$$GCD(a,na+l) = GCD(a,l).\qquad (n \in \mathbb{N}, ~~ l\ge 0).$$

Applying this rule, one can get:

$
GCD(k,nk+1) =GCD(k,1) =1; 
$

$\qquad$for any $n\in \mathbb{N}$, our cases: $n=6,8,24$.

Other cases - similar:

$
GCD(6k+1,8k+1)=GCD(6k+1,2k)=GCD(4k+1,2k)=GCD(1,2k)=1.
$

$
GCD(6k+1,24k+1)=GCD(6k+1,6k-2)=GCD(3,6k-2)=GCD(3,1)=1.
$

$
GCD(8k+1,24k+1)=GCD(8k+1,8k-1)=GCD(2,8k-1)=GCD(2,1)=1.
$
Suppose that $p>1$ divides $k$. Then it divides $mk$ for any integer $m$. But then it cannot divide $mk+1$.