How many different knight's moves are there on an $n \times n$ chessboard? I have found there to be $48$ total moves on a $4 \times 4$ board... and $96$ on a $5 \times5$... but I can not see the relevance to each other in terms of a "$n  \times n$" board. 
By "moves" I am referring to from each space the amount of moves available from it.
Example
 A: A knight move can go in one of $8$ directions: if we think of squares on the chessboard as being given coordinates $(x,y)$ with $1 \le x,y \le n$, then our options are to add one of $$\{(2,1), (2,-1), (1,2), (1,-2), (-1,2), (-1,-2), (-2,1), (-2,-1)\}$$ to $(x,y)$.
For each direction, there are $(n-1)(n-2)$ ways to choose a valid starting square without going off an $n \times n$ board:


*

*If we're adding $1$ to a coordinate, it can be one of $\{1,2,\dots,n-1\}$, and if we're subtracting $1$, it can be one of $\{2,3,\dots,n\}$, for $n-1$ choices either way.

*If we're adding $2$ to a coordinate, it can be one of $\{1,2,\dots,n-2\}$, and if we're subtracting $2$, it can be one of $\{3,4,\dots,n\}$, for $n-2$ choices either way.


So the total number of knight moves is $8(n-1)(n-2)$.
A: Note:
I am not counting
different moves
that end on the same square.
On an $n x n$ board:
On the 4 corners there are 2,
for 8 total.
On the squares on the border
adjacent to the corners
there are 3
for $3\cdot 8 = 24$.
For the squares 
adjacent diagonally
there are 4
for $4\cdot 4 = 16$.
For the $n-4$
on each border 
not yet counted
there are 4
for $4\cdot 4\cdot (n-4)
=16n-64$.
For the 2
at offset $(1, 1)$
from each corner
there are
4 for
$4 \cdot 4 \cdot 2
= 32
$.
For the
$n-4$ that are one from the border
not yet counted
there are 
6 for
$6\cdot 4 \cdot (n-4)
= 24n-96$.
For the remaining
$(n-4)^2$
not in the two border rows,
each has 8
for $8\cdot (n-4)^2$.
Add these up to get the total
number of knight moves.
This might even be correct,
but it should be easy
to correct any errors.
A: Extending your example ... 
For the top row you get 2,3,4,4,....,4,4,3,2
For the second row you get 3,4,6,6,....,6,6,4,3
For the third 4,6,8,8,...,8,8,6,4
For the fourth also 4,6,8,8,...,8,8,6,4
...
[and then the mirror of that for the bottom 4 rows]
So:
With $n\ge 4$ you get:
$2+3+(n-4)*4+3+2=10 +4*(n-4)$ for rows 1 and $n$
$3+4+(n-4)*6+4+3=14+6*(n-4)$ for rows 2 and $n$
$(n-4)*(4+6+(n-4)*8+6+4=(n-4)*(20+8*(n-4))$ for all other rows
For a total of 
$$2*(10+14)+ (2*(4+6)+20)*(n-4)+ 8*(n-4)^2 = $$
$$48+40*(n-4)+8*(n-4)^2$$
Check: for $n=4$ we indeed get 48. For $n=5$, we get $48+40+8=96$ (So your 98 was a bit off)
