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November 19th, 2014, 05:40 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

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November 19th, 2014, 06:03 AM   #2
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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.
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November 19th, 2014, 06:24 AM   #3
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Quote:
Originally Posted by Benit13 View Post
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

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November 19th, 2014, 07:19 AM   #4
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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?
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November 19th, 2014, 09:52 PM   #5
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Originally Posted by CRGreathouse View Post
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
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November 19th, 2014, 10:22 PM   #6
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Quote:
Originally Posted by M_B_S View Post
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.
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November 20th, 2014, 12:26 AM   #7
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Originally Posted by CRGreathouse View Post
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 01:01 AM.
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November 20th, 2014, 06:55 AM   #8
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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?
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November 20th, 2014, 07:20 AM   #9
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Originally Posted by CRGreathouse View Post
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



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November 20th, 2014, 09:26 PM   #10
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Quote:
Originally Posted by M_B_S View Post
$\displaystyle (a+b)^2+((-a^2-a b)/(a-b)+a)^2 = (b+(-a^2-a b)/(a-b))^2$

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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