20100413, 22:25  #1 
Nov 2009
2×5^{2}×7 Posts 
Magic Squares

20100503, 14:34  #3 
Einyen
Dec 2003
Denmark
C6A_{16} Posts 
I didn't solve main enigma #1 but I found a semi magic square where 1 of the 2 diagonals also has the correct sum So halfway between semimagic and magic, is there a name for that?
103^{2} 302^{2} 394^{2} 446^{2} 233^{2} 62^{2} 218^{2} 334^{2} 313^{2} Sum=257,049 (red diagonal sum = 162,867) Last fiddled with by ATH on 20100503 at 14:36 
20100503, 23:55  #4  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:


20100506, 01:23  #5  
Nov 2009
536_{8} Posts 
Quote:
Good work, unfortunately this has already been discovered. This is in the Lucas 3x3 Semi Magic Squares of Squares family: with the following values p=4 q=9 r=11 s=17 plugging those values in Lucas's equation and you will get your magic sum of 507^{2} Keep up the good work! Thank you for your efforts Mathew 

20100511, 23:34  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
for the 3x3 magic square of squares solve a^{2} + b^{2}=c^{2} such that a and b take 4 different values each
then fill in the center: a,a,a, b,x,a, b,b,b, so : 1) is there a a^2 + b^2 = c^2 such that c^2 can be represented 4 ways with a and b each being unique. 2) can you plug this value into a^2+b^2=c^2 to get a value you can plug in the center and still give a square ? this might help Last fiddled with by science_man_88 on 20100512 at 00:29 
20100512, 13:29  #7 
Einyen
Dec 2003
Denmark
2·7·227 Posts 
See this research by Lee Morgenstern: http://home.earthlink.net/~morgenstern/magic/sq3.htm
We can represent the square like: Code:
c+a c(a+b) c+b c(ab) c c+(ab) cb c+(a+b) ca I made a program to search for a solution to this but no luck. There are many square c which have many candidates a and b where c±a and c±b is also square, but the "hard" part is that c±(a+b) and c±(ab) also need to be square, which they are not, so far. Last fiddled with by ATH on 20100512 at 13:32 
20100512, 14:36  #8 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
a^2+b^2+d^2=e^2
a^2+b^2=c^2 c^2+d^2=e^2 
20100512, 14:43  #9 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
this is basically what you have to solve in the 3*3 case and try for d such that one of the other three gets repeated in 4 but no others repeat except the d value. though I'm probably wrong.
Last fiddled with by science_man_88 on 20100512 at 15:01 
20100609, 23:42  #10 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
what I think is that most pythagorean triples (if not all) follow the formulae:
b=1/2n*a^2 + n^2/2n c=1/2n*a^2  n^2/2n so basically to solve this (or have a chance with pythagorean triples) we need to find a c that works for 4 n values that hopefully will give 4 unique pairs of a and b that c can then be plugged in as b to find another a or as a and solve for a new b to go in the middle 25 is the lowest repeating c so far but I only found it twice in my lists. (and so the search continues). is there software to check when multiple equations get the same y value ? if not should I try and make some ? so far I can figure: 3,4,5 6,8,10 9,12,15 12,16,20 15,20,25 5,12,13 10,24,26 15,36,39 20,48,52 25,60,65 7,24,25 14,48,50 21,72,75 anyway I'm bored of typing it out and more still Last fiddled with by science_man_88 on 20100609 at 23:48 
20100611, 00:15  #11 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
sorry b equation should do  and c should do + messed up
Last fiddled with by science_man_88 on 20100611 at 00:26 
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