Outside-the-box mate-in-one challenge

Question

I've received an (I would say impossible) mate-in-one challenge and I have already checked all movements, but couldn't solve it yet. It came from a trustable source and they say you need to think outside the box, so certainly there's a solution. Can someone help me to find it out?

 [FEN "2q1rb2/pR3pkr/3p2pN/2nPp1QP/1nB1P1b1/6N1/PBP3P1/5RK1 w - - 0 1"]

1. Qg6 Kh8 (1. Qf6 Kh6) (1. Qe5 Re5) (1. Be5 Re5) (1. Ngf5 Kh8) (1. Rff7 Kh8) 2. Qg8

White moves, and it is pretty close to win. Qxg6, Qg8 could be a two-movement winning sequence. However, as said in the challenge statement, it is possible to be victorious in the very next movement, so I have been wondering how!

The original statement is:

Today's puzzle is one of the most trickiest chess problems you'll ever see. All you need is to think outside the box. White to move and mate in one! Yes, in one move. No, no extra pieces are appearing on the board likewise no removal. It's achieved with one unique valid movement, are you able to solve it?

There's just 6 possible checks white can do (I mean, i just saw 6). All them are included in above board followed by how black escape from it.

Can anyone find the solution?

Also also also also, for honesty here, I googled the FEN to find the answer.

Answer 1

I checked with Stockfish and indeed there is no normal mate in one for White, and there isn't a mate in one for Black either. So this obviously a trick/joke problem.

The trick is this: What last move of Black would allow a mate in one?

Solution:

The mate in one is “1. dxe6#” en passant! This move takes away the Black king's only flight square on h8, which is what stops any other attempts by White.

Explanation:

The joke is that it can't be proved that Black's last move was “0... e5”. The accepted convention in chess problems is that an en passant capture is only allowed on the first move of the solution if it can be proved that the captured Black pawn made the double-step as Black's last move. It is obvious that Black has plenty of possible last moves other then e5, hence why this a trick problem.

Answer 2

It's an "en passant" problem.
Black's last move had to have been 0..., e5. Then when white captures the pawn en passant with 1. dxe6 e.p., black is mated by the white bishop on b2 since the black f7 pawn is pinned by the white b7 rook and can't interpose to block the bishop.

Answer 3

Black touched the white queen rook but moved their own rook from d8. Therefore, they have to undo the rook move, putting that rook back on d8. And capture the white rook with the Queen. Then B x P is mate. Because the board is upside down. Well, okay, if the board is upside down, it wasn't White's Queen rook, but the King Rook. NO matter.

Answer 4

Black's last move was 1.....e7-e5, & so the mate in one is achieved with 2.de6 en passant (ep). Stockfish will have to do better. I have featured this study on my YouTube channel, plus a lot more. Happy solving! NB. I note in the 'spoiler' that it is of the opinion that Black had many alternatives to 1....e7-e5. Not so, because White's last move was evidently 1.Bb2+ (B at c1 to b2, as per my video on YouTube from 11 months back), requiring either the given reply or 1...f6 (2.Bf6# to finish). This proves my answer is not repeating others, but, on the contrary, identifying the correct sequence & that no 'trick' or 'spoil' exists. To sum up & leave no room for doubt (thus vindicating the composer),if the Black e-pawn were on e6 then 1...f6 would not be possible (f-pawn pinned by the Rook at b7, hence no alternative move at all), & that e6-e5 would not enable White to deliver mate in one & so e7-e5 is the only possible & logical reply to 1.Bb2+ to bring into being a # in 1.