Pythagorean Triangle whose Hypotenuse and Leg differ by 1

Theorem
Let $P$ be a Pythagorean triangle whose sides correspond to the Pythagorean triple $T$.

Then:
 * the hypotenuse of $P$ is $1$ greater than one of its legs


 * the generator for $T$ is of the form $G = \left({n, n + 1}\right)$ where $n \in \Z_{> 0}$ is a (strictly) positive integer.
 * the generator for $T$ is of the form $G = \left({n, n + 1}\right)$ where $n \in \Z_{> 0}$ is a (strictly) positive integer.

Proof
We have from Solutions of Pythagorean Equation that the set of all primitive Pythagorean triples is generated by:
 * $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$

where $m, n \in \Z$ such that:
 * $m, n \in \Z_{>0}$ are (strictly) positive integers
 * $m \perp n$, that is, $m$ and $n$ are coprime
 * $m$ and $n$ are of opposite parity
 * $m > n$.

Necessary Condition
Let the generator $G$ for $T$ be of the form:
 * $G = \left\{ {n, n + 1}\right\}$

Recall that from Solutions of Pythagorean Equation, $T$ is of the form:
 * $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$

where $m, n \in \Z$ such that:
 * $m, n \in \Z_{>0}$ are (strictly) positive integers
 * $m \perp n$, that is, $m$ and $n$ are coprime
 * $m$ and $n$ are of opposite parity
 * $m > n$.

We are given that $n \in \Z_{>0}$ such that $m = n + 1$.

We have that Consecutive Integers are Coprime.

They are also of opposite parity.

Therefore the Pythagorean triple generated by $n$ and $n + 1$ is primitive.

So:

Thus it is seen that $P$ is a Pythagorean triangle where:
 * one legs is of length $2 n \left({n + 1}\right)$
 * the hypotenuse is of length $2 n \left({n + 1}\right) + 1$

Thus $P$ is a Pythagorean triangle whose hypotenuse of $P$ is $1$ greater than one of its legs.

Sufficient Condition
Let the hypotenuse of $P$ be $1$ greater than one of its legs.

Let the legs of $P$ be $a$ and $b$.

Let the hypotenuse of $P$ be $h$ such that $h = b + 1$.

By Consecutive Integers are Coprime, $h$ is coprime to $b$.

Then by Elements of Primitive Pythagorean Triple are Pairwise Coprime, $P$ is primitive.

Because $h$ and $b$ are of opposite parity, it follows that $b$ is even.

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

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

Thus:

Hence the result.