Mate in 4 (HARD)

HINT: Funny little knight move is all it takes! ;-)    

Variation:

ok....

Nice

Nothing is forced.

Actualy it's not forcing. Here is another way for blacks.

How about Rxd8 in ALL the variations? Black has three connected passers until the White Knight goes for a mate.

Mate can be done in 4 if 2...Rg1

I've seen this puzzle on another site (puzzle 12):

http://www.thechesswebsite.com/chess-puzzles/hard-chess-puzzles2.php

You can use the engine to show refutations to other lines...

Hahahaha You guys are very funny  and quite annoying as well  ! 2... Rh1 also ruins the mate in 4 moves! But if you've not seen the solution above yet, I can tell you: 1. Nb4!

TempterMan! More than that if 2...Rh1 then 3.Qg7 will solve this problem in 5 moves. I discuss the solution where the 1st move is Qc3. And if you look with attention i don't contest your solution with 1st move Nb4. So please be carefull with your comments against people.

prativa_prah your solution isnt forcing mate in 4 moves:

Also here is another solution with a variation mate in 2 moves:

If 2...any other move than b6 then mate in 2 moves

Come on guy, you may have many solutions to mate in 5 moves, 6, 7 or whatever in this case, but the challenge of this puzzle is to mate in just 4 moves (most puzzles are like this, you may know that), and the only way to solve this problem - no matter what black replies - is 1. Nb4. For any others first moves you try - even goods - black has ways to delay the checkmate to a minimum of 5 moves. I hope you understand my point.

I didn't mean to be harsh on my last comment, but sorry anyway. I just don't want to discuss this anymore.

But you are right, 1. Qc3 can lead to a mate in 5 moves, no matter what black refutes! Ok?

Why 3.Nd5?? 3.Qd5 was mate already!! :)

GUYS ITS FORCED MATE WHAT ARE YOU TALKING ABOUT

THIS GUY THOUGHT THE KNIGHT PATTERN WAS COOL, AND IT IS

But everyone saying Rd8... just look for the MATE IN 1

 What do we do if we find that it doesn’t work? 

nope

The OP has completely missed the point of the puzzle. The really tricky variation is this:

1. Nb4 c2 2. Qc1 b1=B! (stalemate trap). Now Nd5 doesn't work, black is stalemate. Instead: 3. Nd3! exd3 4. Qh1 mate.