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 2010-04-13, 22:25 #1 Mathew     Nov 2009 2×52×7 Posts Magic Squares Christian Boyer posted these puzzles Magic Squares on his site. Thank you for your time Mathew
 2010-04-26, 02:43 #2 Mathew     Nov 2009 2×52×7 Posts Toshihiro Shirakawa has solved 2 of the enigmas: Main Enigma #5 solved! Small Enigma #3a solved! His solutions can be viewed here Solutions
 2010-05-03, 14:34 #3 ATH Einyen     Dec 2003 Denmark C6A16 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? 1032 3022 3942 4462 2332 622 2182 3342 3132 Sum=257,049 (red diagonal sum = 162,867) Last fiddled with by ATH on 2010-05-03 at 14:36
2010-05-03, 23:55   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by ATH 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? 1032 3022 3942 4462 2332 622 2182 3342 3132 Sum=257,049 (red diagonal sum = 162,867)
well could it fit demisemimagic square ?

2010-05-06, 01:23   #5
Mathew

Nov 2009

5368 Posts

Quote:
 Originally Posted by ATH 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? 1032 3022 3942 4462 2332 622 2182 3342 3132 Sum=257,049 (red diagonal sum = 162,867)
ATH,

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 5072

Keep up the good work!

Mathew

 2010-05-11, 23:34 #6 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 Posts for the 3x3 magic square of squares solve a2 + b2=c2 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 2010-05-12 at 00:29
 2010-05-12, 13:29 #7 ATH 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-(a-b) c c+(a-b) c-b c+(a+b) c-a So we need all these numbers to be square: c, c±a, c±b, c±(a+b), c±(a-b) 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±(a-b) also need to be square, which they are not, so far. Last fiddled with by ATH on 2010-05-12 at 13:32
 2010-05-12, 14:36 #8 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts a^2+b^2+d^2=e^2 a^2+b^2=c^2 c^2+d^2=e^2
 2010-05-12, 14:43 #9 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 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 2010-05-12 at 15:01
 2010-06-09, 23:42 #10 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 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 2010-06-09 at 23:48
 2010-06-11, 00:15 #11 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 Posts sorry b equation should do - and c should do + messed up Last fiddled with by science_man_88 on 2010-06-11 at 00:26

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