How to conclude 4 + 4k is divisible by 8 in proof by induction? Note: This problem is from Discrete Mathematics and Its Applications [7th ed, prob 35, pg 330].
Problem:
a) Use mathematical induction to prove that $n^2$ - 1 is divisible by 8 whenever n is an odd positive integer. 
My work:
I defined predicate P(n) as $n^2$- 1 is divisible by 8
My basis step was showing P(1), as 1 is the first odd positive integer
P(1)  -  8 | $1^2$ - 1 or 8 | 0 which is true
My inductive hypothesis is to assume P(k) for some odd positive integer k and use that to show P(k+2)
So inductive hypothesis - P(k) - $k^2$ - 1 = 8c where s is some integer
Now the next odd positive integer is at k+2 so we have 
8| (k + 2)$^2$ - 1 
8 | k$^2$ - 1 + 4k + 4 
I was able to recognize the  k$^2$ - 1 segment from my inductive step so I made an immediate substitution for that, now we have 
8 | 8c + 4k + 4. 
I recognized that the 8c portion was divisible by 8 but what mathematical steps can you go to show that 4k + 4 is also divisible by 8?
 A: If it is insisted that induction is used, it's enough after checking the base case to note that
$$
(2k+3)^2-1=4k^2+12k+8=(8k+8)+4k^2+4k=\color{blue}{8(k+1)}+\color{red}{[(2k+1)^2-1]}.
$$
The $\color{blue}{\text{blue}}$ part is clearly divisible by $8$ whereas the induction hypothesis says $8$ divides the $\color{red}{\text{red}}$ part.
A: Hint:  Remember that you defined $k$ to be an odd integer $k$.  If $k$ is odd then you can write it as...  So $4k + 4$ can be written as...
Very good work otherwise.  Just keep track of what pieces of information you defined earlier on.
A: Since $k$ is odd, let $k = 2p + 1$ where $p$ is an integer, so you will be able to complete the proof.
A: Another proof (without induction)
Set $n=2k+1$ for some $k\in \mathbb N$. Then we want to see that $(2k+1)^2-1$ is divisible by 8
We have $(2k+1)^2-1=4k^2+4k=4k(k+1)$ depending on the value of $k$ we have that $k$ is even or $(k+1)$ is even. So a factor of the RHS is a product of $4$ and an even number, which is clearly divisible by $8$.
A: Let $n = 2k+1$ and $n^2-1$ is divisible by 8 $\tag1$.
All you have to prove is for $n+2 = 2k+3$,  $(n+2)^2-1$ is divisible by 8$\tag2$
From 1, $(2k+1)^2-1 = 4k^2 +4 = 8q$
From 2, $4k^2 + 12k + 9 -1 = 4k^2+4k+8k+8 = 8q+8k+8$ which is divisible by 8.
