Pythagorean quilts

Question

The King requests Pythagoras to his palace to discuss an important matter.

After the usual formal greetings the King asks:
- I have been told that you have a marvelous formula about adding squares together.

Pythagoras:
- Highness, what I discovered is that in a right triangle with sides A, B and C, if you add the square of A to the square of B, you get the square of C.

- So you can add two squares and make a square?

- Well... Excellence, it's not exactly that, but there are in fact geometrical constructs to decompose two squares and form a single one.

- Good. Here is the thing. I have a large quilt and a smaller quilt, both square. I need you to combine them into a large square quilt.

- Certainly, Greatest. I can cut this square diagonally, cut a triangle there, turn it around and move it up, ...

- No, no, no! You can't do triangles. You can't cut diagonally. Can't you see the quilts are made of small square pictures of our goddess? These pictures must remain intact and property oriented.

- I see. Your Magnificence is most lucky, because this quilt is 12x12 squares, that smaller quilt is 5x5 squares. This is 144 squares plus 25 that makes 169 squares, and that is exactly what you need for a quilt of 13x13 squares.
I can split the small quilt into 25 squares and stich them on two sides of the large quilt, making a larger one.

- That is too many pieces. The sewing will be done as per your instructions but by expert tailors. And you see, their price is based on the number of pieces they put together, regardless of the length of the seam. With all these pieces they will get a fortune for little work. That will spoil them. Tell me, what is the smallest number of pieces that you can cut these quilts into, so that they can be stiched back into a larger square quilt, with all squares intact and property oriented?

- ... I think I will need to think.

TL;DR
You have a square of 12x12 unit squares and another of 5x5 unit squares. Cutting along the grey lines, you want to split these into N pieces and, without rotating or flipping any one, just moving them, form a 13x13 square.

What is the smallest possible N, i.e. the smallest number of pieces?

Show how it can be done.

It is less than 6.
I mention 6 because I found a number of different solutions with that count.

Answer 1

The optimal solution is

4 pieces,

which is achievable (for example) like this:

For another (or perhaps, the other) way to achieve the minimal number of pieces, you can check out OP's self-answer below.

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Here's how I got there:

Studying the situation, we can instantly see that there won't be a solution with fewer than 4 pieces: there can be no piece big enough to include more than one of the 4 corners of the 13x13 square.

So we plonk down the 5x5 in a corner, and then we have (surprise, surprise) even more corners than before. If we want to keep up with the optimum pace, we know how they must be connected:

This, very nicely, leaves us with a white piece that fits on the 12x12 square:

The problem is, of course, that the 16 squares needed for the long parts are in exactly the wrong shape on the 12x12 side. Luckily, that's easy to fix:

Even though it doesn't exactly fit, we can just cram in one of the pieces (I chose the red one) in the corner of the 12x12 square. Where it overlaps the white piece (1 in the next picture), we shrink the white portion in the 13x13 square (2), and add the missing squares to the red piece:

We then bring back the added red squares to the 12x12 side, and since they all fit in the 4x4 corner area, we are done with the red piece.

Then, we repeat the procedure (taking a couple more iterations, since the red piece is taking up the corner) for the other piece, and that's just about it!

Answer 2

Just for reference here is my intended answer.

I thought it is so weird it would take you days to crack. As it happens you didn't have to, you came up with a simpler and even more logical solution.

N=4

An there is another one

Bass's solution and mine are actually three of a same series. Here is the third one, midway between his and mine.

Answer 3

N =

5

The cuts

The rearrangement

Answer 4

Here is the almost (?) optimal solution, where $N$ is:

$6$

The cuts:

Answer 5

Cut two strips 1 x 5 from the 5 x 5 square and one strip 1 x 3 and place them on the right side of the 12 x 12 square. Then cut two strips 1 x 3 from the remaining square and three strips 1 x 2 and put them on top of the 12 x 12 square. Cuts to be done as shown on the drawing. Total cuts 7. So N=7.