Dissecting the holey octomino into a square

This is a pure dissection problem, with no added twists. Cut the holey octomino (i.e., a square with the middle third removed) into several pieces, and reassemble those pieces into a square with no hole. Aim for as few pieces as possible.

My solution uses 5 pieces. Is it possible to do better?

• I doubt that there is a four piece solution, but that's hard to prove. – greenturtle3141 Mar 18 '17 at 1:24

I seriously doubt that this can be done in 4 pieces or less. It would be a miracle if it was possible, but it obviously isn't a walk in the park to prove. Regardless, to get people started, I have found two solutions that use 5 pieces:

First:

Second:

Well, from my viewpoint this is a four-piece dissection, since parts of each piece don't move relatively to each other. They are even connected, to some extent. However, I would completely agree that there are about 24 pieces in this dissection, from a pragmatic viewpoint.

At least evaluate an hour-long fiddling with MS paint here.

• Not what I had in mind, but clever. (+1) – Franklin Pezzuti Dyer Dec 17 '18 at 14:04
• @Frpzzd - I'm guessing a 'proper' answer is going to involve some tedious case-analysis proof that no 4-piece solution exists, and you'd have better luck on math.SE... but that's just a guess. – deep thought Dec 17 '18 at 14:13
• Very nice! It's even a translation-only dissection, i.e. the pieces are not rotated. – Jaap Scherphuis Dec 17 '18 at 14:29
• I'd say that this is a valid answer, as all the pieces touch each others at corners. – a guy Dec 18 '18 at 8:06
• Gets my vote for "best" answer. Very "puzzling" and a bit out of the box. – BmyGuest Dec 18 '18 at 22:47

greenturtle3141 found two solutions with 5 pieces. Since this ties my own solution, I'm accepting it now. Now that there's an answer, I'll post my own solution:

Here's a dissection into $$5$$ pieces that only uses $$3$$ cuts.