Generator for Almost Isosceles Pythagorean Triangle

Theorem
Let $P$ be a Pythagorean triangle whose sides correspond to the Pythagorean triple $T = \left({a, b, c}\right)$.

Let the generator for $T$ be $\left({m, n}\right)$.

Then:
 * $P$ is almost isosceles


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

Proof
By definition of almost isosceles:
 * $\left\vert{a - b}\right\vert = 1$

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:
 * $\left({2 m n, m^2 + n^2, m^2 - n^2}\right)$

for some $m, n \in \Z_{>0}$ where:
 * $m > n$
 * $m \perp n$
 * $m$ and $n$ are of opposite parity.

Necessary Condition
Let $\left({2 m + n, m}\right)$ 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:
 * $p = 2 \left({2 m + n}\right) m$
 * $q = \left({2 m + n}\right)^2 - m^2$

Thus:

That is, $P$ is almost isosceles.

Sufficient Condition
Let $P$ be almost isosceles.

, by Solutions of Pythagorean Equation:
 * $a = 2 m n$
 * $b = m^2 - n^2$

Thus:

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

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

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