Here is my attempt at the solution. I don't have any mathematical proof that this is the minimum no. of knights, but the steps I followed suggest that.
In the figures below, the yellow cells denote the knight's location, and the corresponding numbers denote their covering cells.
I first divided the square into 4 identical squares of 4X4 each (grey cells are the axes of symmetry). Then I tried to fill up the outer corner of each of those boxes (placing 1-4th knight): Next, I want to fill-up the grey cells. For this, I first put the 5th knight to the 4th one's left. This covers the right half of the grey cells completely, and half of the top half grey cells:Keeping in mind the symmetry, I continue to do the same till the 8th knight Now, I have to put at least 1 knight inside every small 4X4 square. Note that the centre spot is empty; and from the symmetry it is clear that it will be populated by all those 4 knights in each 4X4 grid. That leads the following (placing of 9-12th knight): The rest is easily filled up by visual inspection, giving me the final result (placing of 13 & 14th knight): Thus, the minimal solution requires 14 knights