Triangular Numbers which are also Square
Theorem
Let $A_n$ be the $n$th non-negative integer whose square is also a triangular number.
Then:
- $A_n = \begin{cases} 0 & : n = 0 \\
1 & : n = 1 \\ 6 A_{n - 1} - A_{n - 2} & : n > 1 \end{cases}$
Sequence of Square Triangles
The sequence of triangular numbers which are also square begins:
- $1, 36, 1225, 41 \, 616, 1 \, 413 \, 721, \ldots$
This sequence is A001110 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Their indices are:
- $1, 8, 49, 288, 1681, 9800, \ldots$
This sequence is A001108 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Their roots are:
- $1, 6, 35, 204, 1189, \ldots$
This sequence is A001109 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
Let $n \in \Z_{>0}$ be such that $n^2$ is a triangular number.
Then we have:
\(\ds \exists m \in \Z_{>0}: \, \) | \(\ds n^2\) | \(=\) | \(\ds \dfrac {m \paren {m + 1} } 2\) | Closed Form for Triangular Numbers | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 2 n^2\) | \(=\) | \(\ds m^2 + m\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 m + 1}^2 - 1} 4\) | Completing the Square | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 8 n^2 + 1\) | \(=\) | \(\ds \paren {2 m + 1}^2\) | simplifying | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x^2 - 8 y^2\) | \(=\) | \(\ds 1\) | setting $x = 2 m + 1$ and $y = n$ |
This is a Pellian Equation.
From Pell's Equation: $x^2 - 8 y^2 = 1$, the smallest positive integral solution is:
- $\tuple {x, y} = \tuple {3, 1}$
Thus we have that the sequence of $n$ such that $n^2$ is triangular is the sequence of numerators of the Continued Fraction Expansion of $\sqrt 8$ whose indices are even.
Let $\tuple {x, y} = \tuple {p_k, q_k}$ be a solution to $x^2 - 8 y^2 = 1$.
Then from Continued Fraction Expansion of $\sqrt 8$ we have:
\(\ds p_{k + 1}\) | \(=\) | \(\ds a_{k + 1} p_k + p_{k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 p_k + p_{k - 1}\) | as $a_{k + 1} = 4$ | |||||||||||
\(\ds p_{k + 2}\) | \(=\) | \(\ds a_{k + 2} p_{k + 1} + p_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p_{k + 1} + p_k\) | as $a_{k + 2} = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 p_k + p_{k - 1} + p_k\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds 5 p_k + p_{k - 1}\) | simplfying | |||||||||||
\(\ds p_{k + 3}\) | \(=\) | \(\ds a_{k + 3} p_{k + 2} + p_{k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 p_{k + 2} + p_{k + 1}\) | as $a_{k + 3} = 4$ | |||||||||||
\(\ds p_{k + 4}\) | \(=\) | \(\ds a_{k + 4} p_{k + 3} + p_{k + 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p_{k + 3} + p_{k + 2}\) | as $a_{k + 4} = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 p_{k + 2} + p_{k + 1} + p_{k + 2}\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds 5 p_{k + 2} + p_{k + 1}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 5 p_{k + 2} + \paren {p_{k + 2} - p_k}\) | substituting for $p_{k + 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 6 p_{k + 2} + p_k\) | simplifying |
Hence the result.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$