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 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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $5$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $13$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $5$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $13$