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October 14th, 2007, 12:34 PM  #1 
Newbie Joined: Oct 2007 Posts: 24 Thanks: 0  A thought... (regarding primes)
Hi, I'm Mark A while ago I was thinking and realised that the difference between 25 and 16 was the same as the sum of 5 and 4. I then wondered if it was a coincidence, so did the same thing with 5 and 6: 36  25 = 11 = 5 + 6 I then tried the same thing with 4 and 6, which didn't work, so I thought I'd try and see if the difference between two square numbers is the same as the sum of their square roots for every two consecutive integers. b = a1 a^2b^2 = a+b a^2(a1)^2 = a+a1 a^2(a1)(a1) = 2a1 a^2a^2+2a1 = 2a1 2a1 = 2a1 a = a What I would like to know is: 1. Is number theory the right forum for this? 2. Is my proof correct? 3. Does this have any use? Thanks 
October 14th, 2007, 12:51 PM  #2 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
1. Yes 2. Yes 3. No But the thing I'm wondering is what's the "(regarding primes)" part about? 
October 14th, 2007, 01:48 PM  #3  
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0  Quote:
Quote:
 
October 15th, 2007, 01:05 AM  #4  
Newbie Joined: Oct 2007 Posts: 24 Thanks: 0  Quote:
Thanks guys  
October 19th, 2007, 09:23 AM  #5 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 
Well, consecutive numbers are always relatively prime. Clearly that's what he meant.

October 19th, 2007, 09:42 AM  #6 
Newbie Joined: Oct 2007 Posts: 24 Thanks: 0 
That does sound like what I meant, although I've never heard it worded like that. *googles* 
October 22nd, 2007, 04:17 PM  #7 
Newbie Joined: Oct 2007 From: Belgrade, Serbia Posts: 2 Thanks: 0 
You can also prove this way: b=a1 => ab=1 a^2b^2=(a+b)(ab)=(a+b)*1=a+b By the other words, the difference between two square numbers is equal to the sum of their square roots (and multiplied by the difference of those squared roots, if they are not consecutive integers). 

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