Knights covering a 10x10 chess board

Here is a solution using

knights.

I believe this is minimal, but I do not yet have a proof.

The obvious lower bound is 12, as 11 knights can cover at most 88 squares, so there is at least $100-88-11=1$ square uncovered and empty.

This lower bound can be improved slightly if you take into account that a knight covering a corner can cover at most 6 squares. Suppose you use 12 knights. Four of the knights cover 6 squares (including one corner each), and the other 8 knights cover at most 8 squares, and that gives a total of at most $4*6+8*8+12=100$ squares that are covered or non-empty. However, there is no slack at all. For a solution with 12 knights to be possible, every square must be covered exactly once, no knight attacks another, and every knight covers 6 or 8 squares. When you try to place the knights to cover the squares near a corner of the board, it is soon obvious that you will have squares that are covered multiple times.

So at least 13 knights are necessary. There is unfortunately still a gap between this lower bound and what I believe to be the actual minimum.

I also tried to prove it by looking at this pattern:

x x . . . . . . . x
. x . . . . . . x x
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
x x . . . . . . x .
x . . . . . . . x x

Each of the twelve marked squares must be covered or contain a knight. However, no two of them can be covered/filled by a single knight (i.e. no two marked squares are one or two knight moves apart), so at least 12 knights are necessary. I have not yet been able to find a similar pattern with more marked squares.