1.Rxd8+ Nxd8 2.Rc7#
I'm not sure if this is the answer...
@sonicbash206: 1... Nxd8 gives check to White.
@fpr_1997: And why would White be able to do en passant? Can you prove that Black's last move must have been c7-c5?
Blacks last move was c7-c5, proof follows (let me know if I missed anything):
Clearly the bB, bR, bK or the bNs haven't moved last (no room or illegal checks). Black is missing four units. Two of them died on the b- and c-files. White is also missing four units. The original white LSB died on f1 leaving three white units for useful captures. The new white LSB must have promoted on g8. This must have happened with the h-pawn walking to h7 and capturing a piece on g8 (otherwise promoting requires 3 captured black units but only 2 are available). The promotion and getting the LSB out takes two captured white pieces on g6 and h5. So the bP at h5 hasn't moved last. The other captured piece must have been the promoted f-pawn (the white DSB obviously won't suffice). Promoting the f-pawn takes one capture, so all the missing black units are accounted for.
We only have the white DSB left for capture. It has been captured by the black b-pawn to get to the c-file on a dark square, meaning the Pc2 has gotten there with the move c3-c2. So the Pc2 hasn't moved last. (The black a-pawn has promoted at a1.)
The promoted white f-pawn and the missing black DSB mean that the Pe5 hasn't moved last (e6-e5 is impossible and there are no white units left for capture). The pawn on c5 has moved last, but not by capture and c6-c5 is impossible. Therefore the last move was c7-c5.
Mate in 2:
1. bxc6 (threat cxd7#)
1... dxc6+ 2. Qxc6#
1... d6 2. Bf5#
1... Nc5 2. Rc7#
(12+12) White to move, mate in 2.
Alexander Kislyak
Shakhmaty v SSSR, 1974