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For any given three positive integers $a < b < c$ and $n=2$ then it can be derived that

$$\frac{c^2 - ( a^2 + b^2 ) + {a+b-c}^2 }{ (c-a)(c-b)} = 2 $$

Then the condition $$\frac{(a+b-c)^2}{(c-a)(c-b)} = 2$$ to be satisfied to produce the Pythagorean-triples.

alkabary
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user245958
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  • I assume you intende $a^2+b^2=c^2$. There is a typo in the final condition; it should read $a+b-c^2$, with no parentheses. As stated, it always evaluates to a negative number (for pythagorean triples), so I wonder if a similar error crept into the first derivatoin? –  Jun 06 '15 at 04:47
  • It reminds curves triangular number. The formula there. http://math.stackexchange.com/questions/794510/curves-triangular-numbers – individ Jun 06 '15 at 05:06
  • @CuddlyCuttlefish The typo is not in the final condition: it's the first line which should be $(a+b-c)^2$ rather than $a+b-c^2$. – Erick Wong Jun 06 '15 at 06:42
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    @user245958 What is your actual question? – Erick Wong Jun 06 '15 at 06:42

1 Answers1

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Yes, your result does hold for any Pythagorean triple. Here is the proof.

For relatively prime Pythagorean triplets, $(a, b, c) =(2mn, m^2-n^2, m^2+n^2) $ with $m > n$.

For this,

$\begin{array}\\ \frac{(a+b-c)^2}{(c-a)(c-b)} &=\frac{(2mn+(m^2-n^2)-(m^2+n^2))^2}{(m^2+n^2-2mn)(m^2+n^2-(m^2-n^2))}\\ &=\frac{(2mn-2n^2)^2}{(m-n)^2(2n^2)}\\ &=\frac{4n^2(m-n)^2}{(m-n)^2(2n^2)}\\ &= 2 \end{array} $

marty cohen
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