My Math Forum Formular for all Pythagorean Triple

 Number Theory Number Theory Math Forum

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

$\displaystyle (a+b)^2+((-a^2-a b)/(a-b)+a)^2 = (b+(-a^2-a b)/(a-b))^2$

b>a ; a,b in N

M_B_S
Attached Images
 Pythagorean Triple.gif (1.6 KB, 7 views)

 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:
 Originally Posted by Benit13 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.
choose a and b free with b > a

a=1 and b = 3 so b>a <=> 3>1

put it in

$\displaystyle (a+b)^2+((-a^2-a b)/(a-b)+a)^2 = (b+(-a^2-a b)/(a-b))^2$

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:
 Originally Posted by CRGreathouse 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?
Let (a+b) = n every natural number

A) n = odd choose (a) even and (b) odd so that (a-b) = -1

B) n = even choose (a) and (b) so that (a-b) = -2

a, b generate (c) := (-a²-ab)/(a-b) , 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
Quote:
 Originally Posted by M_B_S Let (a+b) = n every natural number A) n = odd choose (a) even and (b) odd so that (a-b) = -1 B) n = even choose (a) and (b) so that (a-b) = -2 a, b generate (c) := (-a²-ab)/(a-b) , b>a (a+b)²+(a+c)² = (b+c)²
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:
 Originally Posted by CRGreathouse This method always makes the second and third members of the triple within 1 or 2, so triples like (20, 21, 29) never occur.
Yes you are correct

So

(a+b)^2+((-a^2-a b)/(a-b)+a)^2 = (29)^2

L:= here

(a+b)^2+((-a^2-a b)/(a-b)+a)^2 = (29)^2 - Wolfram|Alpha

My formular fits (20,21,29)

Here is a better one:

(a+b)^2+((a^2+a b)/(a-b)-a)^2 = ((a^2+a b)/(a-b)-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 2014-11-20 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:
 Originally Posted by CRGreathouse That still doesn't show that all Pythagorean triples are of that form. I can find infinitely many Pythagorean triples that your 2014-11-20 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?
My Formula was proofed correct here:

https://en.wikipedia.org/wiki/Talk:P...gorean_triples

M.Becker-Sievert (M_B_S)

November 20th, 2014, 08:26 PM   #10
Senior Member

Joined: Sep 2010

Posts: 221
Thanks: 20

Quote:
 Originally Posted by M_B_S $\displaystyle (a+b)^2+((-a^2-a b)/(a-b)+a)^2 = (b+(-a^2-a b)/(a-b))^2$ b>a ; a,b in N M_B_S
Some transformations
$\displaystyle (a+b)^2+[(-a^2-ab+a^2-ab)/(a-b)]^2=[(ab-b^2-a^2-ab)/(a-b)]^2$
To get rid of denominator
$\displaystyle (b^2-a^2)^2+(-2ab)^2=[-(b^2+a^2)]^2$
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 $\displaystyle b-a$) and why minuses are needed?

 Tags formular, pythagorean, triple

,

,

,

# math formular

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Denis New Users 2 April 21st, 2013 09:02 PM heomoi Calculus 0 April 6th, 2009 05:55 PM bigbang Elementary Math 1 April 29th, 2008 11:12 AM flyinghat Advanced Statistics 2 January 17th, 2008 10:04 PM Kiranpreet Geometry 1 August 2nd, 2007 06:33 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top