Prove that $ n^3 + 5n$ is divisible by 6 for all $n\in \textbf{N}$ Prove that $ n^3 + 5n  $ is divisible by 6 for all $ n \in \textbf{N} $.  I provide my proof below.  

We prove that $ n^3 + 5n  $ is divisible by 6 for all $ n \in \textbf{N} $.  
Base case:  Observe that if $ n = 1 $, then $ n^3 + 5n = 1 + 5 = 6 $.  
So the base case holds.  
Inductive step:  Assume that $ k^3 + 5k $ is divisible by 6 for some $ k\in\textbf{N} $. 
We show that $ (k+1)^3 + 5(k+1) $ is divisible by 6.  
Since 6 divides $ k^3 + 5k $, it follows that $ k^3 + 5k = 6q $ for some integer $ q $.  
Obverse that  
$ (k+1)^3 + 5(k+1) = k^3 + 3k^2 + 3k + 1 + 5k + 5 = k^3 + 3k^2 + 8k+ 6 $ 
$ = 6(q+1) + 3k^2 + 3k  $. 
We break this up into 2 cases to consider odd and even values of $ k $: 
Case 1.  Suppose $ k $ is even.  Then $ k=2w $ for some integer $ w $.  Thus 
$ 6(q+1) + 3k^2 + 3k =6(q+1) + 3(4w^2)+3(2w) = 6(q+1) + 6(2w^2) + 6w $ 
$ = 6(q+2w^2+w+1) $. 
Case 2.  Consider $ k$ is odd.  Then $ k =2v + 1 $ for some integer $ v $.  Then 
$ 6(q+1) + 3k^2 + 3k = 6(q+1) + 3(2v+1)^2 + 3(2v+1) $ 
$ = 6(q+1) + 3(4v^2+4v+1) + 6v+3 = 6(q+1) + 6(2v^2+2v)+6(v+1)$
$ = 6 (2v^2+3v+2+q) $. 
So the claim holds for all $ n \in \textbf{N} $.
