Pythagorean Triangle with Sides in Arithmetic Sequence

Theorem
The $3-4-5$ triangle is the only Pythagorean triangle such that:
 * the lengths of whose sides are in arithmetic progression

and:
 * the lengths of whose sides form a primitive Pythagorean triple.

Proof
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle such that $a < b < c$.

Let $a, b, c$ be in arithmetic progression:
 * $b - a = c - b$

Let $a, b, c$ form a primitive Pythagorean triple:
 * $a \perp b$

By definition of primitive Pythagorean triple, $a, b, c$ are in the form:
 * $2 m n, m^2 - n^2, m^2 + n^2$

We have that $m^2 + n^2$ is always the hypotenuse.

There are two cases:
 * $(1): \quad 2 m n > m^2 - n^2$, as in, for example, $3, 4, 5$, where $m = 2, n = 1$.
 * $(2): \quad 2 m n < m^2 - n^2$, as in, for example, $8-15-17$, where $m = 4, n = 1$.

First, let $2 m n > m^2 - n^2$:
 * $a = m^2 - n^2$
 * $b = 2 m n$
 * $c = m^2 + n^2$

Then:

From Solutions of Pythagorean Equation: Primitive, $m$ and $n$ must be coprime.

Hence $n = 1$ and $m = 2$ are the only $m$ and $n$ which fulfil the requirements.

This leads to the $3-4-5$ triangle.

Now let $2 m n < m^2 - n^2$:
 * $a = 2 m n$
 * $b = m^2 - n^2$
 * $c = m^2 + n^2$

Then:

In order for $a, b, c$ to form a primitive Pythagorean triple, then $m$ and $n$ must be of opposite parity.

If $m$ is even, then $m^2 - 2 m n$ is even.

But then $3 n^2$ is even, which makes $n$ even.

Otherwise, if $m$ is odd, then $m^2 - 2 m n$ is odd.

But then $3 n^2$ is odd, which makes $n$ odd.

So when $2 m n < m^2 - n^2$, $a, b, c$ cannot be both in arithmetic progression and be a primitive Pythagorean triple.

Hence follows the result.