20170624, 19:30  #144 
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
111100_{2} Posts 
tiling an 11x11 square with squares from the 4sq rep (10,4,2,1)
I made the tiling of an 11x11 square. I started with the 4sq rep (10,4,2,1). Now using combinations of nonsquares, we can find a factor by adding 2 terms 10+1=11. In this picture, it will be shown that one can tile the whole 11x11 square with squares derived from the 4sq rep (10,4,2,1).
One has to remember that 11=5+5+1, 3+3+1+1+1+1+1=11, 5+4+2=11 and 5+4+2=11 and 2+2+2+2+3=11. So 10+1=11 becomes in the final picture 5+5+1=11. 
20170624, 19:52  #145  
Aug 2006
3·1,993 Posts 
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20170624, 19:54  #146  
Aug 2006
3×1,993 Posts 
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20170624, 19:55  #147  
Aug 2006
1011101011011_{2} Posts 
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20170624, 20:07  #148  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×4,787 Posts 
Quote:
And here is my picture for 13^2=10^2+7^2+4^2+2^2. In fact it works for any set of squares. We will just cut them all into 1x1 pieces. One simply has to remember that 7=1+1+1+1+1+1+1, 4=1+1+1+1, and 2=1+1. Elementary, Watson! 

20170624, 20:12  #149 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
I somewhat understand, after all I linked to the numberphile video. But, I also get the point, that factoring a number with known factors, allows you to plan ahead. This allows you to make sure you use squares that fit such that their side lengths add to a known factor. The problem is, in a general factorization, baring any knowledge of the type of side length we will be dealing with ( aka factor types like 2kp+1 for mersennes), we can't say which sums will give us a factor so we must try basically all of them. edit: Also the problem of n^2 = n^2*1^2 shows up like Batalov talked about.
Last fiddled with by science_man_88 on 20170624 at 20:30 
20170624, 20:34  #150  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
111100_{2} Posts 
Quote:
In the original 4sq rep used, (10,4,2,1), by just adding 10+1=11, we get a factor. But of course we cannot tile with this rep as is. But if we expand 10^2=(5,5,5,5) and 4^2+2^2=3^3+3^2+1+1, then we would be able to tile the whole square. Each side will give the same factor. 

20170624, 20:46  #151 
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 

20170624, 21:16  #152  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·269 Posts 
Quote:
Here is my 2 cents, FWIW. It has absolutely nothing to do with 4 squares and there is no justification to start from 5. I just applied the geometric tiling algorithm and started with 2x6^2 I ended up tiling a 14x13 rectangle. I think if one starts from N/2 rounded down and works its way down to N/sqrt(N), then he will find a deterministic factor if any. I also think it will be very very slow. But if I am missing something, it won't be the first time. Last fiddled with by a1call on 20170624 at 21:20 

20170624, 21:43  #153  
Aug 2006
3·1,993 Posts 
Quote:


20170624, 21:49  #154  
Feb 2017
Nowhere
3·1,669 Posts 
Quote:
a^2 + b^2 + c^2 + d^2 = N exactly. As long as a^2 + b^2 + c^2 + d^2 is close to N, go ahead and form the sums of some or all of a, b, c, d, a^2, b^2, c^2, and d^2, and take the gcd with N. (There are 127 nonempty sums.) I hypothesize that, there being so many possibilities, the chances are good that any factor of N under 100 will show up fairly quickly. I tried this idea with N = 1457 = 31*47, and the results were better than I expected. Even though a^2 + b^2 + c^2 + d^2 never hit N right on the nose, I still got at least one of the 15 sums of a, b, c, and d alone having a common factor with N in 6 of 8 tries. If anyone would care to have a go at my examples N = 77, 1457, 3901, 4661, 6557, 7597, 8549, and 9869 using this approach, have at them! I cheerfully proclaim that this approach really is nothing more than blazing away with a scattergun, in hopes that some of the buckshot will hit. The advantage over the OP's idea is, reloading is a whole lot quicker ;) 

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