20110613, 14:44  #1 
May 2011
France
7·23 Posts 
Magic squaress
Sometimes you can write a value like the sum of two square
i/e 16+25= 51.... a= b²+c² Are b and c unics? John 
20110613, 14:53  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:
It is a fairly simple elementary exercize to determine when a number is the sum of two squares. It is an exercize for a 1st year number theory class to determine when the representation is unique. May I suggest that you perform some numerical experimentation to see if you can hypothesize when the representation is unique. 

20110613, 14:58  #3 
Jun 2003
2221_{8} Posts 

20110613, 15:03  #4 
Nov 2003
2^{2}×5×373 Posts 

20110613, 15:24  #5 
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
"Magic Squaress" sound's like the ideal title to bestow on
Alice Through the Looking Glass Last fiddled with by davieddy on 20110613 at 15:32 
20110613, 16:08  #6 
Oct 2010
191 Posts 
51?
Last fiddled with by Ralf Recker on 20110613 at 16:09 
20110613, 16:36  #7 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 

20110613, 18:01  #8 
May 2011
France
10100001_{2} Posts 
Sorry
http://www.mersenneforum.org/showthread.php?t=15623
In fact i need to find 5 polynoms a²+b²+c²+d².....i²+j² =Z Is there a math mehod? thanks Last fiddled with by JohnFullspeed on 20110613 at 18:16 Reason: typo errors 
20110613, 18:32  #9  
Nov 2003
2^{2}·5·373 Posts 
Quote:
is asked. What are you looking for? 5 polynomials for what? You have only defined an integer Z as the sum of 10 squares. Last fiddled with by R.D. Silverman on 20110613 at 18:36 

20110613, 19:05  #10 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
magic squares wouldn't ever have 10 squares either as they are n by n with a magic constant so I can't so far make any sense of it either.

20110613, 19:06  #11 
Aug 2006
13511_{8} Posts 

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