Pythagorean Triangle whose Hypotenuse and Leg differ by 1

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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

if and only if:

the generator for $T$ is of the form $G = \tuple {n, n + 1}$ 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:

$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$

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:

$G = \set {n, n + 1}$

Recall that from Solutions of Pythagorean Equation, $T$ is of the form:

$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$

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:

\(\displaystyle 2 m n\) \(=\) \(\displaystyle 2 n \paren {n + 1}\)
\(\displaystyle m^2 - n^2\) \(=\) \(\displaystyle \paren {n + 1}^2 - n^2\)
\(\displaystyle m^2 - n^2\) \(=\) \(\displaystyle n^2 + 2 n + 1 - n^2\)
\(\displaystyle \) \(=\) \(\displaystyle 2 n + 1\)
\(\displaystyle m^2 + n^2\) \(=\) \(\displaystyle n^2 + 2 n + 1 + n^2\)
\(\displaystyle \) \(=\) \(\displaystyle 2 n^2 + 2 n + 1\)
\(\displaystyle \) \(=\) \(\displaystyle 2 n \paren {n + 1} + 1\)

Thus it is seen that $P$ is a Pythagorean triangle where:

one legs is of length $2 n \paren {n + 1}$
the hypotenuse is of length $2 n \paren {n + 1 + 1$

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

$\Box$


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:

\(\displaystyle h\) \(=\) \(\displaystyle b + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle m^2 + n^2\) \(=\) \(\displaystyle 2 m n + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle m^2 - 2 m n + n^2\) \(=\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {m - n}^2\) \(=\) \(\displaystyle 1\) Square of Difference
\(\displaystyle \leadsto \ \ \) \(\displaystyle m - n\) \(=\) \(\displaystyle 1\) Definition of Square Root, and we have that $m > n$
\(\displaystyle \leadsto \ \ \) \(\displaystyle m\) \(=\) \(\displaystyle n + 1\)

Hence the result.

$\blacksquare$


Sequence

The sequence of Pythagorean triangles whose hypotenuse and leg differ by 1 can be tabulated as follows:

$\begin{array} {r r | r r | r r r} m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\ \hline 2 & 1 & 4 & 1 & 4 & 3 & 5 \\ 3 & 2 & 9 & 4 & 12 & 5 & 13 \\ 4 & 3 & 16 & 9 & 24 & 7 & 25 \\ 5 & 4 & 25 & 16 & 40 & 9 & 41 \\ 6 & 5 & 36 & 25 & 60 & 11 & 61 \\ 7 & 6 & 49 & 36 & 84 & 13 & 85 \\ \hline \end{array}$


Sources