Demonstrating the Pythagorean Quadruple

$6\times6 + 6\times6 + 7\times7 = 11\times11$

Using the pieces shown in the $11\times11$ square:

The objective:

  1. Arrange the pink pieces (four enneominoes) into a $6\times6$

  2. Arrange the blue pieces (six hexominoes) into a $6\times6$

  3. Arrange the orange pieces (seven heptominoes) into a $7\times7$

  4. Arrange all 17 pieces into a $11\times11$ but with no like colours touching, even at a corner.

All four tilings are unique. Hand tiling puzzle please, a computer will just spoil it for you. Flipping pieces is allowed.

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  • $\begingroup$ can the tiles be flipped? thanks $\endgroup$ Oct 15 '19 at 6:55
  • $\begingroup$ Are we certain the third task is possible? I have been trying off and on for a while and am getting terribly close, but never exact. $\endgroup$ Oct 15 '19 at 12:36
  • $\begingroup$ Yes flipping of pieces is allowed. Added to puzzle statement. $\endgroup$ Oct 15 '19 at 13:36

Partial Answer - finished Task 1

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Sooo much NSFW! But thanks anyway :)

Task 1

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Task 2

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Task 3

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Task 4 - not solved yet.

  • 1
    $\begingroup$ Beautifully illustrated! Go for the last one! +1 $\endgroup$
    – Stiv
    Oct 15 '19 at 16:14

Like @Omega Krypton I have solved one part, Task 2:



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