Is there a simple perfect squaring of a 1366 by 768 rectangle?

So, a simple perfect squaring of a rectangle is a tiling of that rectangle by squares whose side lengths are all distinct integers. Additionally, not subset of the squares must form a smaller rectangle. My question is if there is a simple perfect squaring of a 1366 by 768 rectangle?

We could try reducing it to a simpler problem by splitting the rectangle into a square and a smaller rectangle, but then we need to ensure that the side lengths are different, and that the their combination is simple. So we basically back to where we started.

P.S. If you are wondering why 1366 by 768, that's the dimension of my monitor, which I am trying to artistically square (hence the tag).

• Just out of curiosity: are you making a wallpaper image? :) – lisyarus Mar 8 at 7:18
• @lisyarus I will. I am not much of an artist, but given the sequence of squares involved I could use a program to produce an image. I plan on making the edges black and the rest transparent, and then putting it on top of another (non-math related) wallpaper, but I'll definitely post the basic square tiling image if anyone is interested. – PyRulez Mar 8 at 7:24
• Well you can have at most $146$ squares as $\sum_{k=1}^{147}k^2>1366\times768$. A brute force search shouldn't be too hard? – Servaes Mar 8 at 13:07
• @Servaes well, $146!$ is greater than the number of atoms in the universe, so it might take a little while. :P – PyRulez Mar 8 at 13:23
• Sure, so is $146\uparrow\uparrow\uparrow\uparrow146$, but how is that relevant? Also, this shows that the largest squares is bigger than $146\times146$, and I'm sure better upper bounds than $768$ aren't hard to come by. – Servaes Mar 8 at 13:50

I looked through some of the data on Squaring.Net (actually, copies at the Internet Archive, since the site is currently down), and there is no $$1366\times768$$ simple perfect squared rectangle of order $$17$$ or less; there may be one of higher order, but that's harder to check. But there is a $$1354\times764$$ simple perfect squared rectangle of order $$17$$, so if you can accept a six-pixel border on the left and right and a two-pixel border at the top and bottom, this will do the job. It looks like this: (Click on the image for the full-size version.)

The Bouwkamp code for this rectangle dissection is $$17\ 1354\ 764\ (389, 403, 311, 251)\ (60, 191)\ (83, 157, 131)\ (375, 14)\ (9, 74)\ (361, 65)\ (322)\ (296)$$.

EDIT: As Servaes points out, I overlooked an even better $$1366\times766$$ alternative, also of order $$17$$: Its Bouwkamp code is $$17\ 1366\ 766\ (419, 288, 258, 401) (115, 143) (203, 85) (200) (179, 365) (347, 72) (275) (193, 7) (186)$$.

• How did you search the site for that? I'm taking a look, and the navigation is confusing me. – PyRulez Mar 8 at 13:25
• @PyRulez Home -> Simple Perfect Squared Rectangles -> Simple Perfects by aspect (which gets you here); then download the .csv file. – FredH Mar 8 at 13:30
• There seems to be a $1366\times766$ rectangle in the linked file as well. – Servaes Mar 8 at 15:22
• @Servaes I missed that one. It's definitely better than the one I did notice. I'll add it. – FredH Mar 8 at 15:45
• I find the $1366\times766$ arrangement especially aesthetically pleasing. I should hope a similar configuration exists for the $1366\times768$ rectangle... – Servaes Mar 9 at 1:13