A very hard mate in 1!!

If the Black king is on e4 and his e2 pawn can promote on e1,  than Qe5#  is checkmate.  so diagonal one square from the white king. Flip the board idea.  But a pitty of the coordinates. The white Queen is on h8 and not on a1 then.

eddysallin's reply in post #37, after 4 years, is still the best one though it isn't completely correct.

Either of the black pawns d3, d6 or d7 may be removed to make the position legal. Though it is just a fun problem, the logical justification is that removing one of these pawns is the minimum change required to make the position legal. If you do something else like turning the board 90 degrees or removing a white unit, it is not enough to make the position legal. You'd still have to take off a black pawn on top of it!

Nevertheless, the problem wasn't created to show just three "mates in one" but to show eight "mates in one" as can be verified by removing each one of the black pawns. Which means the composer overlooked the restrictions analyzed by eddysallin. Conclusion: it is a failed problem.

By chance I made a similar problem some time ago (by chance it is the third time in a few days I start a sentence this way) which I will link to when I find it in the forum. To my knowledge, that one was valid.

Guys, I appreciate your enthusiasm. But I think it's perfectly legit and as soon as I saw this I realized it was obviously a retro problem.  I didn't bother to solve it because I don't like retro problems, but it's a perfectly legitimate kind of problem.

:/

You shouldn't have bothered to comment either since you clearly have no clue what you are talking about. It is a retro-problem, it is a non-standard retro-problem and it is totally flawed. But you are not going to understand that because you don't do retro-problems, right?

Here is the link to a correct version of the same idea as in the current topic:

https://www.chess.com/forum/view/more-puzzles/mate-in-1-puzzles-1

(first diagram)

Bxe6