Mate in ... ?

In this position a mate in 1 is infact possible.
There is no unique solution to this puzzle, BUT the correct answer is either cxd6# or fxg6# (en passant)!
If you take a closer look at the position, Black can only have moved d7-d5 or g7-g5 on his last move, since everything else (as you can check) was illegal for Black.

That's a retro-analytical problem designed to trap even experts.

For non-experts, first note the rule regarding potential en passant captures in a problem diagram. The rule is that such an e.p. capture is illegal unless it can be proved – by a type of backward reasoning called retro-analysis – that the previous move must have been a double-step by the pawn to be captured. See my blog for an example.

The OP problem has two potential solutions: 1.cxd6 e.p. mate which is allowed only IF the last move was necessarily ...d7-d5, and 1.fxg6 e.p. mate which is allowed only IF the last move was necessarily ...g7-g5. We can rule out alternative last moves by Black through a process of elimination. Black's last move couldn't have been ...Nc4-a5, ...d6-d5, or ...f6xg5 because these would imply Black was checking White while it was Black's turn – an impossibility.

What about ...e6xd5, ...g6-g5 ...h5xg4, ...Kd7-e7, ...Kd8-e7, or ...Ke8-e7, where the BK could also have just captured something on e7? If you can demonstrate why each of these is also an impossible last move, well done!

Finally, you are left with ...d7-d5 and ...g7-g5. Are both possible last moves or just one of them, meaning is the solution 1.cxd6 e.p. or 1.fxg6 e.p. mate? That is the tricky question that even experts can get wrong!

Discovered check.

@Rocky64  Thanks for the profound explanation! 

Because it's prettier. The full game is left as an exercise, but the position occurs quite frequently in the Bongcloud. 

d7-d5 is ruled out because the c8-bishop needed to get out to be captured by a pawn.  The a and b pawns had three captures to get to the c file.  The two extra f pawns required two more captures.  The h pawn had to promote to be one of the five captures Black's pawns had to make.  That is six White captures and the c8 bishop was one of the only six pieces Black is missing.  d7-d5 could not have been the last move.  That leaves only  Rf6-h6+ g7-g5.

Next we have to see if the position with the rook on f6 can be reached.  a few moves earlier the g4 pawn could have been on g5, the b3 bishop could have been on a2 and the a5 knight was free to roam between b3 and a5 while everything was being set up.  The previous moves could have been:

Bb1-a2 Nb3-a5

Ba2-b3 g5-g4

Rf6-g6+ g7-g5

Well done, @jetoba!

So...

That is a failed en passant puzzle

Indeed jetoba and Rocky! I missed it because I thought the white captures wouldn't matter. Never trust your instincts in a retro-problem, always play the proof game!

@Arisktotle. Heh, I missed it too and thought it was a PRA problem. Got a surprise when I looked up the solution!

Two trophies to the author!

Wow, I'm very impressed. This goes deeper than I initially thought it would.. and is another lesson to me not to draw premature conclusions.

@jetoba  Even though I have not yet fully checked your arguments in order to rule out d7-d5, they sound convincing at first glance and I will give them a more intense look over the weekend.

Thank you for sharing your ideas on this puzzle!

Rook on h6 to f6, then rook on a8 to e8 mate! 2 moves.

There were various mates in two.  The real question was whether or not there was a mate in one.  The arguments of the thread finally boiled down to being unable to prove that en passant was definitely available.

f6+, Kxf7, Bxd5#

2 moves

and no I didn't copy anyone.