I'll give this one a go.

The five different shapes for four square tetris (tetrominoes) are

a) Stick. e.g. would cover a1, b1, c1 and d1 on a chessboard

b) L. e.g. would cover a2, a1, b1 and c1.

c) Square. e.g. would cover a2, b2, a1 and b1.

d) T. e.g. would cover b2, a1, b1 and c1.

3) Z. e.g. would cover a2, b2, b1 and c1.

Note that on an ordinary chessboard, all of these except the T would cover two black squares and two white squares. Thus it follows that a solution of 15 Ls and a T covering a chessboard is not possible.

Now consider a chessboard marked with black and white bands, e.g. the a-file is all black, the b-file all white, the c-file black etc.

On this chessboard, the L always covers 3 black squares and a white square, or 3 white squares and a black square. The only other shape for which this is true is the T (this can either cover 2 black and 2 white or a 3/1 split).

So if we try to cover this board with 15 Ls we will end up with a gap consisting of 3 blacks and a white, or 3 whites and a black (I'm assuming one gap, but if the squares are separate there will be still be a 3 and 1 split of colours).

The only shapes that can cover this gap are another L or a T. We know that a T will not fit the gap from our attempt to cover an ordinary chessboard with 15 Ls and a T, thus if there is a solution to the problem of 15 Ls + one other piece to cover an 8x8 chessboard then the 16th piece must be another L.

It is trivial to show that a solution of 16 Ls does indeed exist because 2 Ls can be put together to make a 4x2 rectangle, and thus four Ls can be put together to exactly cover 2 rows of a chessboard.