4
$\begingroup$

I saw this in the NAdigest mailing list, and it was obviously suggested by $70^2 =\sum_{j=1}^{24} j^2 $:


From: Gerhard Opfer gerhard.opfer@uni-hamburg.de

Date: November 06, 2017

Subject: Mathematics, combinatorial

Is it known, whether a square Q of size 70 x 70 can be covered by little squares q_j of size j x j, j=1,2,...,24.

Can one say something about this problem in general.


I don't know.

My first thought was to look at the unit square. However, I realized that it was possible to surround the unit square with larger squares.

The fact that it is possible to square the square (i.e., fill an integer-sided square with distinct integer-sided squares - see https://en.wikipedia.org/wiki/Squaring_the_square) means that some property of 70 and 1 through 24 is needed if it is not possible.

It just might be impossible to square a 70 x 70 square.

Your turn.

$\endgroup$
  • 4
    $\begingroup$ Well, it is one reason that the Leech Lattice works. see SPLAG by Conway and Sloane, or Lattices and Codes by Ebeling. Page 130 in the second edition of Ebeling. $\endgroup$ – Will Jagy Nov 8 '17 at 4:44
  • $\begingroup$ It is impossible, see answers in this question on MO $\endgroup$ – achille hui Nov 8 '17 at 5:47
2
$\begingroup$

According to the wikipedia link you've posted, the smallest (in terms of side length) perfect squared squares are $110\times110$, so it appears no such realization is possible.

This result is due to Ian Gambini in his doctoral thesis, and can be found here (there is no paywall!). It appears to have been determined via computer-aided search.

$\endgroup$
0
$\begingroup$

While the problem as stated is impossible, there is a solution using the third dimension.

In https://m.youtube.com/watch?v=229csazhybg a teacher proves the usual formula for the sum of consecutive squares by representing the sum as a stepped pyramid, then fitting three copies of this figure to make an almost rectangular solid. He uses an averaging argument to get the volume of his solid and divides by three to get his sum.

The teacher misses the fact that he can combine his solid with a center-inverted version of itself to form a true rectangular block, containing six stepped pyramids and having the dimensions $(n)×(n+1)×(2n+1)$. With this modification apply the procedure to $n=24$, thus forming a $24×25×49$ block. Divide this block into six equal layers by cutting orthogonally to the $24$ dimension. Each of these six layers $(4×25×49)$ has the same volume as one of the original stepped pyramids and also can be subdivided into a $2×5×7$ array of $2×5×7$ blocks.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.