Pythagorean Triangle whose Hypotenuse and Leg differ by 1

Theorem
The set of Pythagorean triangles such that the hypotenuse is $1$ greater than one of the legs is unlimited.

All such Pythagorean triangles are primitive.

Proof
From Solutions of Pythagorean Equation, 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
 * $m > n$.

Consider $n \in \Z$ 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 the primitive Pythagorean triples generated by $n, n + 1$ forms a primitive Pythagorean triangle where:
 * one legs is of length $2 n \left({n + 1}\right)$
 * the hypotenuse is of length $2 n \left({n + 1}\right) + 1$

This holds for all $n$.

Thus there are as many such Pythagorean triangles as there are natural numbers.

From Solutions of Pythagorean Equation, any non-primitive Pythagorean triple is of the form:
 * $\left({2 k m n, k \left({m^2 - n^2}\right), k \left({m^2 + n^2}\right)}\right)$

for some $k > 1$.

Thus a non-primitive Pythagorean triangle has its hypotenuse at least $k$ more than its legs.

The result follows.