Problem of the Month (June 2005)

This month we generalize the reptile problem by asking which instances of

a12 + a22 + . . . + an2 = b12 + b22 + . . . + bm2

can be demonstrated by tilings of similar shapes. That is, does there exist a shape with the property that copies of size a1 through an can tile the same region as copies of size b1 through bm?

Can you find one of the missing tilings? Can you prove that they do not exist? How about for larger sums? We continue the tradition of offering a $10 prize to the most numerous new contributions.


ANSWERS

Gavin Theobald found tilings for every formula of the form:

a2 + b2 = c2 + d2 (using right triangles)

Claudio Baiocchi generalized some existing tilings to give tilings for these formulas:

(2N+1)2 + 3 * 12 = (N+2)2 + 3 * N2 (using equilateral triangles)
(5N)2 = (5N-1)2 + 2N * 22 + (2N-1) * 12
(5N+1)2 = (5N)2 + 2N * 22 + (2N+1) * 12
(5N+4)2 = (5N)2 + (2N+1) * 42 + (2N) * 22
(6N)2 = (6N-1)2 + 2N * 22 + (4N-1) * 12
(6N+1)2 = (6N)2 + 2N * 22 + (4N+1) * 12
(7N)2 = (7N-1)2 + 2N * 22 + (6N-1) * 12
(7N+1)2 = (7N)2 + 2N * 22 + (6N+1) * 12 (using trapezoids)

George Sicherman added these tilings:

a2 + b2 + c2 = s2 + (a-s)2 + (s-b)2 + (s-c)2 (where 2s = a+b+c) (using equilateral triangles)

Here are the known tilings:

SumEquationPacking
42=1111
93=221?
93=211111
1031=2211
123111=222
1332=2221
164=32111
1741=322
18411=33?
1833=3221?
1833=222211
194111=331

George Sicherman

2042=3311
224211=332
255=43
255=4221
2543=4221
255=332111?
2651=3322
27511=333?
27422111=333
274311=333?
285111=3331

George Sicherman

285111=4222
284222=3331
2952=33311

George Sicherman

315211=3332?
315111111=3332
3252111=44?
3251111111=44

Claudio Baiocchi

3244=33321
33522=333211

Claudio Baiocchi

33522=441
3453=4411
3453=3332111

George Sicherman

3544111=33322

George Sicherman

3552211=4331

George Sicherman

366=5311?
366=522111
366=441111
366=43311?
366=4322111
366=333221?
366=33222211
365311=442

George Sicherman

3761=53111?
3761=5222?
3761=433111
3761=43222
3761=3332211
38611=532
38611=52221
38611=4332
3852221=4332

George Sicherman

396111=333222

Tino Jonker

4062=53211

George Sicherman

4062=433211
41621=54?
41621=443
426211=541

George Sicherman

4362111=4333

George Sicherman

4362111=533?
44622=5331
44622=43331
44622=443111
445331=44222

George Sicherman

4563=542?
4563=53311?
4553311=4432

Claudio Baiocchi

46631=5421

Claudio Baiocchi

4863111=444

George Sicherman

497=632?
497=62221?
497=62211111
497=5422?
497=5421111?
497=5411111111?
497=533211?
497=53222111
497=5222222?
497=5222221111
497=4441?
497=4333211
497=43222222
497=4222222221?
497=4222222211111
497=333332?
49632=5422

George Sicherman

49632=4441

George Sicherman

5071=6321

George Sicherman

5071=55

Gavin Theobald

5071=54221

George Sicherman

5071=4433
506321=55

George Sicherman

5055=44411?
51711=5431

George Sicherman

51551=444111?
527111=64?
527111=62222
527111=5333

Claudio Baiocchi

527111=4442

George Sicherman

527111=43333?
5264=5511

George Sicherman

525333=4442

George Sicherman

5372=54222?
5372=54221111
5372=44421

George Sicherman

5372=433331

Claudio Baiocchi

54721=6411

George Sicherman

54721=5432

Claudio Baiocchi

54552=444211

George Sicherman

5672111=63311

George Sicherman

57552111=4443

George Sicherman

5873=64211

George Sicherman

5873=5522

George Sicherman

5873=5441

George Sicherman

587221=6332

George Sicherman

586332=5441

George Sicherman

Corey Plover noticed that if fractals are allowed, then these regions are easier to find. He gives a fractal to illustrate 32=22+22+12:

Claudio Baiocchi noticed that 32=22+22+12 has a solution on the projective plane:

The winner of the $10 prize this month is Claudio Baiocchi.


If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 6/14/08.