Generator for Almost Isosceles Pythagorean Triangle

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Theorem

Let $P$ be a Pythagorean triangle whose sides correspond to the Pythagorean triple $T = \tuple {a, b, c}$.

Let the generator for $T$ be $\tuple {m, n}$.


Then:

$P$ is almost isosceles

if and only if

$\tuple {2 m + n, m}$ is the generator for the Pythagorean triple $T'$ of another almost isosceles Pythagorean triangle $P'$.


Proof

By definition of almost isosceles:

$\size {a - b} = 1$


First note that, from Consecutive Integers are Coprime, an almost isosceles Pythagorean triangle is a primitive Pythagorean triangle.

Hence $T$ and $T'$ are primitive Pythagorean triples.


Thus it is established that by Solutions of Pythagorean Equation, both $P$ and $P'$ are of the form:

$\tuple {2 m n, m^2 + n^2, m^2 - n^2}$

for some $m, n \in \Z_{>0}$ where:

$m > n$
$m \perp n$
$m$ and $n$ are of opposite parity.


Necessary Condition

Let $\tuple {2 m + n, m}$ be the generator for the Pythagorean triple $T'$ of the almost isosceles Pythagorean triangle $P'$.


Let $p$ and $q$ be the legs of $P'$.

By Solutions of Pythagorean Equation:

$p = 2 \paren {2 m + n} m$
$q = \paren {2 m + n}^2 - m^2$

Thus:

\(\ds \size {2 \paren {2 m + n} m - \paren {\paren {2 m + n}^2 - m^2} }\) \(=\) \(\ds 1\) Definition of Almost Isosceles Pythagorean Triangle
\(\ds \leadsto \ \ \) \(\ds \size {4 m^2 + 2 m n - \paren {4 m^2 + 4 m n + n^2 - m^2} }\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \size {\paren {m^2 - n^2} - 2 m n}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \size {a - b}\) \(=\) \(\ds 1\) Solutions of Pythagorean Equation

That is, $P$ is almost isosceles.

$\Box$


Sufficient Condition

Let $P$ be almost isosceles.

Without loss of generality, by Solutions of Pythagorean Equation:

$a = 2 m n$
$b = m^2 - n^2$

Thus:

\(\ds \size {a - b}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \size {\paren {m^2 - n^2} - 2 m n}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \size {4 m^2 + 2 m n - \paren {4 m^2 + 4 m n + n^2 - m^2} }\) \(=\) \(\ds 1\) working the above sequence of equations in reverse
\(\ds \leadsto \ \ \) \(\ds \size {2 \paren {2 m + n} m - \paren {\paren {2 m + n}^2 - m^2} }\) \(=\) \(\ds 1\) Solutions of Pythagorean Equation

By Solutions of Pythagorean Equation, $2 \paren {2 m + n} m$ and $\paren {2 m + n}^2 - m^2$ are the legs of a Pythagorean triangle $P'$ whose generator is $\paren {2 m + n, m}$.

But from the above, these legs differ by $1$.

Hence, by definition, $P'$ is an almost isosceles Pythagorean triangle $P'$.

$\blacksquare$


Sequence

The sequence of almost isosceles Pythagorean triangles can be tabulated as follows:

$\begin{array} {r r | r r | r r r}

m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\ \hline 2 & 1 & 4 & 1 & 4 & 3 & 5 \\ 5 & 2 & 25 & 4 & 20 & 21 & 29 \\ 12 & 5 & 144 & 25 & 120 & 119 & 169 \\ 29 & 12 & 841 & 144 & 696 & 697 & 985 \\ 70 & 29 & 4900 & 841 & 4060 & 4059 & 5741 \\ 169 & 70 & 28 \, 561 & 4900 & 23 \, 660 & 23 \, 661 & 33 \, 461 \\ \hline \end{array}$

The sequence of elements of the generators are the Pell numbers:

$1, 2, 5, 12, 29, 70, 169, 408, \ldots$


Sources

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