Pythagorean Triangle from Sum of Reciprocals of Consecutive Same Parity Integers

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Theorem

Let $a, b \in \Z_{>0}$ be (strictly) positive integers such that they are consecutively of the same parity.

Let $\dfrac p q = \dfrac 1 a + \dfrac 1 b$.


Then $p$ and $q$ are the legs of a Pythagorean triangle.


Proof

Let $a$ and $b$ both be even.

Then:

\(\ds a\) \(=\) \(\ds 2 n\)
\(\ds b\) \(=\) \(\ds 2 \paren {n + 1}\) for some $n \in \Z_{>0}$
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 a + \dfrac 1 b\) \(=\) \(\ds \dfrac 1 {2 n} + \dfrac 1 {2 \paren {n + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {n + \paren {n + 1} } {2 n \paren {n + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {2 n + 1} {2 n \paren {n + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {n^2 + 2 n + 1 - n^2} {2 n \paren {n + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {n + 1}^2 - n^2} {2 n \paren {n + 1} }\)

From Solutions of Pythagorean Equation, $\tuple {n, n + 1}$ form the generator for the primitive Pythagorean triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m = n + 1$.

The legs of the resulting primitive Pythagorean triangle are $2 n \paren {n + 1}$ and $\paren {n + 1}^2 - n^2$.


Let $a$ and $b$ both be odd.

Then:

\(\ds a\) \(=\) \(\ds 2 n + 1\)
\(\ds b\) \(=\) \(\ds 2 \paren {n + 1} + 1\) for some $n \in \Z_{>0}$
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 a + \dfrac 1 b\) \(=\) \(\ds \dfrac 1 {2 n + 1} + \dfrac 1 {2 \paren {n + 1} + 1}\)
\(\ds \) \(=\) \(\ds \dfrac {4 n + 4} {4 n^2 + 8 n + 3}\)
\(\ds \) \(=\) \(\ds \dfrac {2 \paren {2 n + 2} \times 1} {\paren {2 n + 2}^2 - 1}\)

From Solutions of Pythagorean Equation, $\tuple {2 n + 2, 1}$ form the generator for the primitive Pythagorean triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m = 2 n + 2$ and $n = 1$.

The legs of the resulting primitive Pythagorean triangle are $2 \paren {2 n + 2}$ and $\paren {2 n + 2}^2 - 1$.

$\blacksquare$


Sources

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