Pairwise relatively prime proof (number theory) Show that if $k \in \mathbb{Z}$, then the integers $6k-1$, $6k+1$, $6k+2$, $6k+3$, and $6k+5$ are pairwise relatively prime. 
I am still new and uncomfortable with proofs. Any help would be great. 
 A: Hint: Look at these numbers as numbers $\pmod 6$. They are $-1, 1, 2, 3, 5 \pmod 6$. If you take the difference of any pair you will notice both numbers are not divisible by their difference. For example numbers in the form $1 \pmod 6$ and $5 \pmod 6$ aren't divisible by 4. Also, none of the numbers are more than 6 apart, so they can't have a common factor of more than 6. If two numbers share a factor their difference will be a factor of both numbers, and none of the numbers do, so all the numbers are relatively prime. 
A: When you're not sure what to do, look at examples.  (This can be a good idea even if you are sure what to do!)  In this case,
$$\begin{align}
k=1&\Rightarrow 5,7,8,9,11\\
k=2&\Rightarrow 11,13,14,15,17\\
k=3&\Rightarrow 17,19,20,21,23\\
k=4&\Rightarrow 23,25,26,27,29\\
k=5&\Rightarrow 29,31,32,33,35\\
&\vdots
\end{align}$$
You might notice that only the middle number in each list is even, and only the one next to it is divisible by $3$.  Is it obvious that this should be the case?  You might also notice that there's always a multiple of $5$, which moves around.  Can you see why that happens?  Finally, you might notice that the difference between the largest and smallest numbers in each list is always $6$.  So how large can a number $d$ possibly be if it divides two numbers in one of the lists?
A: A common prime factor $\,p\,$ divides their difference $\,\le 6 = (5k\!+\!5)-(5k\!-\!1),\,$
so  $\,p\in\{2,3,5\}$
By Euclid $\,\gcd(6,6k\!+\!i,6k\!+\!j) = \gcd(6,i,j)$ is $1$ if $\,i$ or $\,j\in\{\pm1,5\}$ else $\{i,j\}\!=\! \{2,3\}\!\Rightarrow$ gcd $= 1$
Thus $\,p\ne 2,3,\,$ so $\,p=5.\ $ If $\,5\mid 6k+i,\,6k+j\,$ then $\,5\mid 6k+i-(6k+j) = i-j.\,$ However $\,i,j\in \{\color{#c00}{-1},1,2,3,5\}\equiv \{1,2,3,\color{#c00}4,5\}\pmod 5\,$ are all incongruent mod $5,\,$ so $\ p\ne 5.\ \ $ QED 
