My Math Forum Formular for all Pythagorean Triple

 Number Theory Number Theory Math Forum

 November 21st, 2014, 09:16 AM #21 Senior Member   Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory So I have a new one : $\displaystyle (n^2-1)^2+(2n)^2=(n^2+1)^2$ Now its perfect! M_B_S
 November 21st, 2014, 09:33 AM #22 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra Apart from the list of Pythagorean triples (such as 5, 12, 13) that it can't generate.
November 21st, 2014, 12:50 PM   #23
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Joined: Nov 2013
From: Germany

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Math Focus: Number Theory
Quote:
 Originally Posted by v8archie Apart from the list of Pythagorean triples (such as 5, 12, 13) that it can't generate.

$\displaystyle (n^n-1)^n+(2n)^n=(n^n+1)^n$
L:= {1,2}

 November 21st, 2014, 01:36 PM #24 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra That has at most two solutions. Last edited by v8archie; November 21st, 2014 at 01:38 PM.
November 21st, 2014, 02:54 PM   #25
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Joined: Oct 2014
From: Lublin, POLAND

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the new triple?

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
You beat the Pythagorean. Congratulations!

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