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November 19th, 2014, 04:40 AM   #1
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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
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 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
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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
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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
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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
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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
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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
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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?

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