November 19th, 2014, 04:40 AM  #1 
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  Formular for all Pythagorean Triple
$\displaystyle (a+b)^2+((a^2a b)/(ab)+a)^2 = (b+(a^2a b)/(ab))^2$ b>a ; a,b in N M_B_S 
November 19th, 2014, 05:03 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
The LHS is equal to the RHS for all $\displaystyle a$ and $\displaystyle b$, but you need to explain why this formula is relevant for Pythagorean triples: $\displaystyle a^2 + b^2 = c^2$ and how to use it. 
November 19th, 2014, 05:24 AM  #3  
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  Quote:
a=1 and b = 3 so b>a <=> 3>1 put it in $\displaystyle (a+b)^2+((a^2a b)/(ab)+a)^2 = (b+(a^2a b)/(ab))^2$ you get 4² +3² = 5² with n=(a+b) , b>a you generate all Triples M_B_S  
November 19th, 2014, 06:19 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
The identity holds for all $a\ne b$ but it doesn't produce integers unless b  a divides 2ab. Can you prove that this covers all Pythagorean triples? 
November 19th, 2014, 08:52 PM  #5  
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  Quote:
A) n = odd choose (a) even and (b) odd so that (ab) = 1 B) n = even choose (a) and (b) so that (ab) = 2 a, b generate (c) := (a²ab)/(ab) , b>a (a+b)²+(a+c)² = (b+c)²  
November 19th, 2014, 09:22 PM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  This method always makes the second and third members of the triple within 1 or 2, so triples like (20, 21, 29) never occur.

November 19th, 2014, 11:26 PM  #7  
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  Quote:
So (a+b)^2+((a^2a b)/(ab)+a)^2 = (29)^2 L:= here (a+b)^2+((a^2a b)/(ab)+a)^2 = (29)^2  WolframAlpha My formular fits (20,21,29) Here is a better one: (a+b)^2+((a^2+a b)/(ab)a)^2 = ((a^2+a b)/(ab)b)^2 Last edited by M_B_S; November 20th, 2014 at 12:01 AM.  
November 20th, 2014, 05:55 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
That still doesn't show that all Pythagorean triples are of that form. I can find infinitely many Pythagorean triples that your 20141120 05:52 method does not generate. Showing that just one of those can be expressed in your form doesn't show that they all can be. Do you have an alternate proof?

November 20th, 2014, 06:20 AM  #9  
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  Quote:
https://en.wikipedia.org/wiki/Talk:P...gorean_triples M.BeckerSievert (M_B_S)  
November 20th, 2014, 08:26 PM  #10  
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20  Quote:
$\displaystyle (a+b)^2+[(a^2ab+a^2ab)/(ab)]^2=[(abb^2a^2ab)/(ab)]^2$ To get rid of denominator $\displaystyle (b^2a^2)^2+(2ab)^2=[(b^2+a^2)]^2$ We obtained Euclid's formula for triples. But what is the reason to present it for fractions ($\displaystyle 2ab$ and $\displaystyle b^2+a^2$ are coprime with $\displaystyle ba$) and why minuses are needed?  

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