My Math Forum Can you all help with with this Pythagorean Triplets/triples question

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 December 9th, 2015, 06:42 PM #1 Newbie   Joined: Dec 2015 From: Virginia Posts: 5 Thanks: 0 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.
 December 10th, 2015, 03:51 PM #2 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 What are you trying to explain?
 December 10th, 2015, 04:56 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 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.
December 11th, 2015, 09:59 AM   #4
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Quote:
 Originally Posted by v8archie 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

December 13th, 2015, 04:08 AM   #5
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Quote:
 Originally Posted by v8archie 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

 December 13th, 2015, 04:31 AM #6 Newbie   Joined: Dec 2015 From: Virginia Posts: 5 Thanks: 0 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"
 December 13th, 2015, 05:42 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 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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