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December 9th, 2015, 07:42 PM   #1
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Can you all help with with this Pythagorean Triplets/triples question

I have an oral exam next week. I need to be able to explain this to my professor
b^2 = [(s^2 - t^2)/2]^2 and b^2 = [(s^2 + t^2)/2]^2

Can you all please show me how to distribute these the long way?

If this is in the wrong area, can someone move it to the correct area, thanks.
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December 10th, 2015, 04:51 PM   #2
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What are you trying to explain?
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December 10th, 2015, 05:56 PM   #3
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I looks like a method of generating Pythagorean triples as per Joseph H. Silverman's A Friendly Introduction to Number Theory. The book's homepage can be found here.
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December 11th, 2015, 10:59 AM   #4
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Quote:
Originally Posted by v8archie View Post
I looks like a method of generating Pythagorean triples as per Joseph H. Silverman's A Friendly Introduction to Number Theory. The book's homepage can be found here.
That answers my question as well. Thanks!

-Dan
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December 13th, 2015, 05:08 AM   #5
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Quote:
Originally Posted by v8archie View Post
I looks like a method of generating Pythagorean triples as per Joseph H. Silverman's A Friendly Introduction to Number Theory. The book's homepage can be found here.
THANK YOU. This helps me a lot. I appreciate it
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December 13th, 2015, 05:31 AM   #6
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One more question

Where are the "S" and "T" derived from in this formula for finding a Pythagorean Triplet/triple?

"So, now that we know what a Pythagorean Triplet is, here is a formula which creates them:

Do the following:

(1) Choose two integers, s and t, which satisfy:
(a) s and t are both odd
(b) s>t>0

(2) Then:
Let a = st
Let b = (s2 - t2)/2
Let c = (s2 + t2)/2"
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December 13th, 2015, 06:42 AM   #7
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That is clearly shown in the document. Since $a^2=c^2-b^2=(c-b)(c+b)$ and both $(c-b)$ and $(c+b)$ are square, we set $s^2=c+b$ and $t^2=c-b$.
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