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November 21st, 2014, 10:16 AM   #21
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So I have a new one :

$\displaystyle (n^2-1)^2+(2n)^2=(n^2+1)^2$


Now its perfect!


M_B_S
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November 21st, 2014, 10:33 AM   #22
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Apart from the list of Pythagorean triples (such as 5, 12, 13) that it can't generate.
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November 21st, 2014, 01:50 PM   #23
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Quote:
Originally Posted by v8archie View Post
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}
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November 21st, 2014, 02:36 PM   #24
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That has at most two solutions.

Last edited by v8archie; November 21st, 2014 at 02:38 PM.
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November 21st, 2014, 03:54 PM   #25
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the new triple?

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
You beat the Pythagorean. Congratulations!
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