Hearted Youtube comments on Mathologer (@Mathologer) channel.
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First challenge: The chessboard cannot always be tiled after removing 2 black & 2 green. Suppose we remove the two black squares adjacent to the upper-left corner and the two green squares adjacent to the lower-left corner. Then the left corner squares are no longer adjacent to any other squares, so the board cannot be tiled.
EDIT:
Second challenge: if m and n are odd, then ⌈m/2⌉ = (m+1)/2 and ⌈n/2⌉ = (n+1)/2. Now the j=(m+1)/2, k=(n+1)/2 term of the product is 4cos²(π/2) + 4cos²(π/2) = 0, so the product is 0. Moreover, if m is not odd, then 0 ≤ j < (m+1)/2 in all terms of the product, hence 0 ≤ jπ/(m+1) < π/2 in all terms, so cos²(jπ/(m+1)) > 0 in all terms, so each term of the product is nonzero. This means the formula gives a nonzero answer whenever m is even -- symmetrically, the answer is nonzero whenever n is even. Thus, the formula returns 0 if and only if m and n are both odd.
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