Triangular Square Number

In mathematics, a square triangular number (or triangular square number) is a number which is both a triangular number and a perfect square. There are infinitely many square triangular numbers; the first few are:
:0, 1, 36, , , , , , ,

Explicit formulas

Write for the th square triangular number, and write and for the sides of the corresponding square and triangle, so that
:$N_k = s_k^2 = \frac.$
Define the ''triangular root'' of a triangular number to be . From this definition and the quadratic formula,
:$n = \frac.$
Therefore, is triangular ( is an integer) if and only if is square. Consequently, a square number is also triangular if and only if is square, that is, there are numbers and such that . This is an instance of the Pell equation with . All Pell equations have the trivial solution for any ; this is called the zeroth solution, and indexed as . If denotes the th nontrivial solution to any Pell equation for a particular , it can be shown by the method of descent that
:$\begin x_ &= 2x_k x_1 - x_, \\ y_ &= 2y_k x_1 - y_. \end$
Hence there are an infinity of solutions to any Pell equation for which there is one non-trivial one, which holds whenever is not a square. The first non-trivial solution when is easy to find: it is (3,1). A solution to the Pell equation for yields a square triangular number and its square and triangular roots as follows:
:$s_k = y_k , \quad t_k = \frac, \quad N_k = y_k^2.$
Hence, the first square triangular number, derived from (3,1), is 1, and the next, derived from , is 36.

The sequences , and are the OEIS sequences , , and respectively.

In 1778 Leonhard Euler determined the explicit formula

:$N_k = \left( \frac \right)^2.$
Other equivalent formulas (obtained by expanding this formula) that may be convenient include
: $\begin N_k &= \tfrac \left( \left( 1 + \sqrt \right)^ - \left( 1 - \sqrt \right)^ \right)^2 \\ &= \tfrac \left( \left( 1 + \sqrt \right)^-2 + \left( 1 - \sqrt \right)^ \right) \\ &= \tfrac \left( \left( 17 + 12\sqrt \right)^k -2 + \left( 17 - 12\sqrt \right)^k \right). \end$
The corresponding explicit formulas for and are:
:$\begin s_k &= \frac, \\ t_k &= \frac. \end$

Pell's equation

The problem of finding square triangular numbers reduces to Pell's equation in the following way.

Every triangular number is of the form . Therefore we seek integers , such that

:$\frac = s^2.$

Rearranging, this becomes

:$\left(2t+1\right)^2=8s^2+1,$

and then letting and , we get the Diophantine equation

:$x^2 - 2y^2 =1,$

which is an instance of Pell's equation. This particular equation is solved by the Pell numbers as

:$x = P_ + P_, \quad y = P_;$

and therefore all solutions are given by

:$s_k = \frac, \quad t_k = \frac, \quad N_k = \left( \frac \right)^2.$

There are many identities about the Pell numbers, and these translate into identities about the square triangular numbers.

Recurrence relations

There are recurrence relations for the square triangular numbers, as well as for the sides of the square and triangle involved. We have

:$\begin N_k &= 34N_ - N_ + 2,& \textN_0 &= 0\textN_1 = 1; \\ N_k &= \left(6\sqrt - \sqrt\right)^2,& \textN_0 &= 0\textN_1 = 1. \end$

We have

:$\begin s_k &= 6s_ - s_,& \texts_0 &= 0\texts_1 = 1; \\ t_k &= 6t_ - t_ + 2,& \textt_0 &= 0\textt_1 = 1. \end$

Other characterizations

All square triangular numbers have the form , where is a convergent to the continued fraction expansion of .

A. V. Sylwester gave a short proof that there are an infinity of square triangular numbers:

If the th triangular number is square, then so is the larger th triangular number, since:

:$\frac = 4 \, \frac \,\left(2n+1\right)^2.$
As the product of three squares, the right hand side is square. The triangular roots are alternately simultaneously one less than a square and twice a square if is even, and simultaneously a square and one less than twice a square if is odd. Thus,
:49 = 7² = 2 × 5² − 1,
:288 = 17² − 1 = 2 × 12², and
:1681 = 41² = 2 × 29² − 1.
In each case, the two square roots involved multiply to give : , , and .

Additionally:
:$N_k - N_=s_;$
, , and . In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.

The generating function for the square triangular numbers is:
:$\frac = 1 + 36z + 1225 z^2 + \cdots$

Numerical data

As becomes larger, the ratio approaches ≈ , and the ratio of successive square triangular numbers approaches ≈ . The table below shows values of between 0 and 11, which comprehend all square triangular numbers up to .

:

See also

* Cannonball problem, on numbers that are simultaneously square and square pyramidal
* Sixth power, numbers that are simultaneously square and cubical

Notes

External links

Triangular numbers that are also square
at cut-the-knot
*

Michael Dummett's solution

{{Classes of natural numbers, collapsed

Figurate numbers
Integer sequences