Magic Squares
 

Christian Boyer posted these puzzles [URL="http://www.multimagie.com/English/MagicSquaresEnigmasE.pdf"]Magic Squares[/URL] on his site.

Thank you for your time

Mathew

Toshihiro Shirakawa has solved 2 of the enigmas:

Main Enigma #5 solved!
Small Enigma #3a solved!

His solutions can be viewed here
[URL="http://www.multimagie.com/"]Solutions[/URL]

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 :smile: So halfway between semimagic and magic, is there a name for that?

[COLOR="Red"]103[sup]2[/sup][/COLOR] 302[sup]2[/sup] 394[sup]2[/sup]
446[sup]2[/sup] [COLOR="Red"]233[sup]2[/sup][/COLOR] 62[sup]2[/sup]
218[sup]2[/sup] 334[sup]2[/sup] [COLOR="Red"]313[sup]2[/sup][/COLOR]

Sum=257,049 (red diagonal sum = 162,867)

[QUOTE=ATH;213861]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 :smile: So halfway between semimagic and magic, is there a name for that?

[COLOR="Red"]103[sup]2[/sup][/COLOR] 302[sup]2[/sup] 394[sup]2[/sup]
446[sup]2[/sup] [COLOR="Red"]233[sup]2[/sup][/COLOR] 62[sup]2[/sup]
218[sup]2[/sup] 334[sup]2[/sup] [COLOR="Red"]313[sup]2[/sup][/COLOR]

Sum=257,049 (red diagonal sum = 162,867)[/QUOTE]

well could it fit demisemimagic square ?

[QUOTE=ATH;213861]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 :smile: So halfway between semimagic and magic, is there a name for that?

[COLOR="Red"]103[sup]2[/sup][/COLOR] 302[sup]2[/sup] 394[sup]2[/sup]
446[sup]2[/sup] [COLOR="Red"]233[sup]2[/sup][/COLOR] 62[sup]2[/sup]
218[sup]2[/sup] 334[sup]2[/sup] [COLOR="Red"]313[sup]2[/sup][/COLOR]

Sum=257,049 (red diagonal sum = 162,867)[/QUOTE]

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 507[SUP]2[/SUP]

Keep up the good work!

Thank you for your efforts

Mathew

for the 3x3 magic square of squares solve a[SUP]2[/SUP] + b[SUP]2[/SUP]=c[SUP]2[/SUP] 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 ?

[URL="http://www.research.att.com/~njas/sequences/A009003"]this[/URL] might help

See this research by Lee Morgenstern: [URL="http://home.earthlink.net/~morgenstern/magic/sq3.htm"]http://home.earthlink.net/~morgenstern/magic/sq3.htm[/URL]

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
[/CODE]

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.

a^2+b^2+d^2=e^2

a^2+b^2=c^2
c^2+d^2=e^2

[URL="http://en.wikipedia.org/wiki/Pythagorean_quadruple"]this[/URL] 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.

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

sorry b equation should do - and c should do + messed up