Pythagorean Triangle with Sides in Arithmetic Sequence

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Theorem

The $3-4-5$ triangle is the only Pythagorean triangle such that:

the lengths of whose sides are in arithmetic sequence

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 sequence:

$b - a = c - b$

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:

\(\ds 2 m n - \paren {m^2 - n^2}\) \(=\) \(\ds m^2 + n^2 - 2 m n\) Definition of Arithmetic Sequence
\(\ds \leadsto \ \ \) \(\ds 4 m n\) \(=\) \(\ds 2 m^2\) adding $2 m n + m^2 - n^2$ to both sides
\(\ds \leadsto \ \ \) \(\ds 2 n\) \(=\) \(\ds m\) dividing both sides by $2 m$

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:

\(\ds m^2 - n^2 - 2 m n\) \(=\) \(\ds m^2 + n^2 - \paren {m^2 - n^2}\) Definition of Arithmetic Sequence
\(\ds \leadsto \ \ \) \(\ds m^2 - n^2 - 2 m n\) \(=\) \(\ds 2 n^2\) simplifying right hand side
\(\ds \leadsto \ \ \) \(\ds m^2 - 2 m n - 3 n^2\) \(=\) \(\ds 0\) subtracting $2 n^2$ from both sides

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 sequence and be a primitive Pythagorean triple.


Hence follows the result.

$\blacksquare$


Sources