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October 14th, 2007, 12:34 PM   #1
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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 = a-1

a^2-b^2 = a+b
a^2-(a-1)^2 = a+a-1
a^2-(a-1)(a-1) = 2a-1
a^2-a^2+2a-1 = 2a-1
2a-1 = 2a-1
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
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October 14th, 2007, 12:51 PM   #2
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1. Yes
2. Yes
3. No



But the thing I'm wondering is what's the "(regarding primes)" part about?
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October 14th, 2007, 01:48 PM   #3
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Quote:
3. Does this have any use?
Quote:
3. No
But then again, a lot of things in number theory don't have much use in real life, at least not now. However, I still love it anyway.
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October 15th, 2007, 01:05 AM   #4
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Quote:
Originally Posted by milin
1. Yes
2. Yes
3. No



But the thing I'm wondering is what's the "(regarding primes)" part about?
I know from previous experience in other forums that not-very-descriptive thread titles are annoying and tend to get less views than ones that at least hint at what's inside

Thanks guys
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October 19th, 2007, 09:23 AM   #5
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Well, consecutive numbers are always relatively prime. Clearly that's what he meant.
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October 19th, 2007, 09:42 AM   #6
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That does sound like what I meant, although I've never heard it worded like that.

*googles*
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October 22nd, 2007, 04:17 PM   #7
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You can also prove this way:
b=a-1 => a-b=1
a^2-b^2=(a+b)(a-b)=(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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