View source for Square triangular number

{{short description|Integer that is both a perfect square and a triangular number}}
{{for|squares of triangular numbers|squared triangular number}}
[[File:square_triangular_number_36.svg|234px|thumb|Square triangular number 36 depicted as a triangular number and as a square number.]]

In [[mathematics]], a '''square triangular number''' (or '''triangular square number''') is a number which is both a [[triangular number]] and a [[square number]], in other words, the sum of all integers from <math>1</math> to <math>n</math> has a square root that is an integer. There are [[Infinity|infinitely many]] square triangular numbers; the first few are:
{{bi|left=1.6|0, 1, 36, {{val|1225}}, {{val|41616}}, {{val|1413721}}, {{val|48024900}}, {{val|1631432881}}, {{val|55420693056}}, {{val|1882672131025}} {{OEIS|id=A001110}}}}

{| class="wikitable" style="margin:auto auto auto auto; front-size:95%; float:right;"
|-
! N <br> A001110 !! s<sup>2</sup> = N <br> A001109 !! t(t+1)/2 = N <br> A001108
|-
| 0 || 0 || 0
|-
| 1 || 1 || 1
|-
| 36 || 6 || 8
|-
| 1225 || 35 || 49
|-
| 41616 || 204 || 288
|-
| 1413721 || 1189 || 1681
|-
| 48024900 || 6930 || 9800
|}

==Solution as a Pell equation==

Write <math>N_k</math> for the <math>k</math>th square triangular number, and write <math>s_k</math> and <math>t_k</math> for the sides of the corresponding square and triangle, so that 

{{bi|left=1.6|<math>\displaystyle N_k = s_k^2 = \frac{t_k(t_k+1)}{2}.</math>}}

Define the ''triangular root'' of a triangular number <math>N=\tfrac{n(n+1)}{2}</math> to be <math>n</math>. In the form of the quadratic equation, <math>n^2 + n - 2N = 0</math>. From the quadratic formula,

{{bi|left=1.6|<math>\displaystyle n = \frac{\sqrt{8N + 1} - 1}{2}.</math>}}

Therefore, <math>N</math> is triangular (<math>n</math> is an integer) [[if and only if]] <math>8N+1</math> is square. Consequently, a square number <math>M^2</math> is also triangular if and only if <math>8M^2+1</math> is square, that is, there are numbers <math>x</math> and <math>y</math> such that <math>x^2-8y^2=1</math>. This is an instance of the [[Pell equation]] <math>x^2-ny^2=1</math> with <math>n=8</math>. All Pell equations have the trivial solution <math>x=1,y=0</math> for any <math>n</math>; this is called the zeroth solution, and indexed as <math>(x_0,y_0)=(1,0)</math>. If <math>(x_k,y_k)</math> denotes the <math>k</math>th nontrivial solution to any Pell equation for a particular <math>n</math>, it can be shown by the method of descent that the next solution is
{{bi|left=1.6|<math>\displaystyle \begin{align}
x_{k+1} &= 2x_k x_1 - x_{k-1}, \\
y_{k+1} &= 2y_k x_1 - y_{k-1}.
\end{align}</math>}}
Hence there are infinitely many solutions to any Pell equation for which there is one non-trivial one, which is true whenever <math>n</math> is not a square. The first non-trivial solution when <math>n=8</math> is easy to find: it is <math>(3,1)</math>. A solution <math>(x_k,y_k)</math> to the Pell equation for <math>n=8</math> yields a square triangular number and its square and triangular roots as follows:

{{bi|left=1.6|<math>\displaystyle s_k = y_k , \quad t_k = \frac{x_k - 1}{2}, \quad N_k = y_k^2.</math>}}

Hence, the first square triangular number, derived from <math>(3,1)</math>, is <math>1</math>, and the next, derived from <math>6\cdot (3,1)-(1,0)=(17,6)</math>, is <math>36</math>.

The sequences <math>N_k</math>, <math>s_k</math> and <math>t_k</math> are the [[OEIS]] sequences {{OEIS2C|id=A001110}}, {{OEIS2C|id=A001109}}, and {{OEIS2C|id=A001108}} respectively.

==Explicit formula==
In 1778 [[Leonhard Euler]] determined the explicit formula<ref name=Dickson>
{{cite book | last1 = Dickson | first1 = Leonard Eugene | author-link1 = Leonard Eugene Dickson |title = [[History of the Theory of Numbers]] | volume = 2 | publisher = American Mathematical Society | location = Providence | year = 1999 |orig-year = 1920 | page = 16 | isbn = 978-0-8218-1935-7 }}
</ref><ref name=Euler>
{{cite journal |last=Euler |first=Leonhard |author-link=Leonhard Euler |year=1813 |title=Regula facilis problemata Diophantea per numeros integros expedite resolvendi (An easy rule for Diophantine problems which are to be resolved quickly by integral numbers)  |journal=Mémoires de l'Académie des Sciences de St.-Pétersbourg |volume= 4 |pages=3–17 |url=http://math.dartmouth.edu/~euler/pages/E739.html |language=la |access-date=2009-05-11 |quote=According to the records, it was presented to the St. Petersburg Academy on May 4, 1778.}}
</ref>{{Rp|12–13}}

{{bi|left=1.6|<math>\displaystyle N_k = \left( \frac{\left(3 + 2\sqrt{2}\right)^k - \left(3 - 2\sqrt{2}\right)^k}{4\sqrt{2}} \right)^2.
</math>}}

Other equivalent formulas (obtained by expanding this formula) that may be convenient include

{{bi|left=1.6|<math>\displaystyle \begin{align} 
N_k &= \tfrac{1}{32} \left( \left( 1 + \sqrt{2} \right)^{2k} - \left( 1 - \sqrt{2} \right)^{2k} \right)^2 \\
&= \tfrac{1}{32} \left( \left( 1 + \sqrt{2} \right)^{4k}-2 + \left( 1 - \sqrt{2} \right)^{4k} \right) \\
&= \tfrac{1}{32} \left( \left( 17 + 12\sqrt{2} \right)^k -2 + \left( 17 - 12\sqrt{2} \right)^k \right).
\end{align}</math>}}

The corresponding explicit formulas for <math>s_k</math> and <math>t_k</math> are:<ref name=Euler />{{Rp|13}}

{{bi|left=1.6|<math>\displaystyle \begin{align}
s_k &= \frac{\left(3 + 2\sqrt{2}\right)^k - \left(3 - 2\sqrt{2}\right)^k}{4\sqrt{2}}, \\
t_k &= \frac{\left(3 + 2\sqrt{2}\right)^k + \left(3 - 2\sqrt{2}\right)^k - 2}{4}.
\end{align}</math>}}

==Recurrence relations==
The solution to the Pell equation can be expressed as a [[recurrence relation]] for the equation's solutions. This can be translated into recurrence equations that directly express the square triangular numbers, as well as the sides of the square and triangle involved.  We have<ref>{{MathWorld|title=Square Triangular Number|urlname=SquareTriangularNumber}}</ref>{{Rp|(12)}}

{{bi|left=1.6|<math>\displaystyle \begin{align}
N_k &= 34N_{k-1} - N_{k-2} + 2,& \text{with }N_0 &= 0\text{ and }N_1 = 1; \\
N_k &= \left(6\sqrt{N_{k-1}} - \sqrt{N_{k-2}}\right)^2,& \text{with }N_0 &= 0\text{ and }N_1 = 1.
\end{align}</math>}}

We have<ref name=Dickson /><ref name=Euler />{{Rp|13}}

{{bi|left=1.6|<math>\displaystyle \begin{align}
s_k &= 6s_{k-1} - s_{k-2},& \text{with }s_0 &= 0\text{ and }s_1 = 1; \\
t_k &= 6t_{k-1} - t_{k-2} + 2,& \text{with }t_0 &= 0\text{ and }t_1 = 1.
\end{align}</math>}}

==Other characterizations==

All square triangular numbers have the form <math>b^2c^2</math>, where <math>\tfrac{b}{c}</math> is a [[Convergent (continued fraction)|convergent]] to the [[simple continued fraction|continued fraction expansion]] of <math>\sqrt2</math>, the [[square root of 2]].<ref name=Ball>
{{cite book | last1 = Ball | first1 = W. W. Rouse |author-link1 = W. W. Rouse Ball | last2 = Coxeter | first2 = H. S. M. |author-link2 = Harold Scott MacDonald Coxeter | title = Mathematical Recreations and Essays | url = https://archive.org/details/mathematicalrecr00coxe | url-access = limited | publisher = Dover Publications | location = New York | year = 1987 | page = [https://archive.org/details/mathematicalrecr00coxe/page/n72 59]| isbn = 978-0-486-25357-2 }}
</ref>

A. V. Sylwester gave a short proof that there are infinitely many square triangular numbers: If the <math>n</math>th triangular number <math>\tfrac{n(n+1)}{2}</math> is square, then so is the larger <math>4n(n+1)</math>th triangular number, since:

{{bi|left=1.6|<math>\displaystyle\frac{\bigl( 4n(n+1) \bigr) \bigl( 4n(n+1)+1 \bigr)}{2} = 4 \, \frac{n(n+1)}{2} \,\left(2n+1\right)^2.</math>}}

The left hand side of this equation is in the form of a triangular number, and as the product of three squares, the right hand side is square.<ref name=Sylwester>
{{cite journal |last1=Pietenpol |first1=J. L. |first2=A. V. |last2=Sylwester |first3=Erwin |last3=Just |first4=R. M. |last4=Warten  |date=February 1962 |title=Elementary Problems and Solutions: E 1473, Square Triangular Numbers |journal=American Mathematical Monthly |volume=69 |issue=2 |pages=168–169 |issn=0002-9890  |jstor=2312558|publisher=Mathematical Association of America | doi = 10.2307/2312558}}
</ref>

The [[generating function]] for the square triangular numbers is:<ref>{{cite web |first=Simon |last=Plouffe |author-link=Simon Plouffe |title=1031 Generating Functions |url=http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf |publisher=University of Quebec, Laboratoire de combinatoire et d'informatique mathématique |page=A.129 |date=August 1992 |access-date=2009-05-11 |archive-date=2012-08-20 |archive-url=https://web.archive.org/web/20120820012535/http://www.plouffe.fr/simon/articles/FonctionsGeneratrices.pdf |url-status=dead }}</ref>
:<math>\frac{1+z}{(1-z)\left(z^2 - 34z + 1\right)} = 1 + 36z + 1225 z^2 + \cdots</math>

==See also==
*[[Cannonball problem]], on numbers that are simultaneously square and square pyramidal
*[[Sixth power]], numbers that are simultaneously square and cubical

==Notes==
{{reflist}}

==External links==
* [http://www.cut-the-knot.org/do_you_know/triSquare.shtml Triangular numbers that are also square] at [[cut-the-knot]]
* {{MathWorld|title=Square Triangular Number|urlname=SquareTriangularNumber}}
* [http://opinionator.blogs.nytimes.com/2012/01/04/remembering-michael-dummett/ Michael Dummett's solution]

{{Classes of natural numbers|collapsed}}

[[Category:Figurate numbers]]
[[Category:Integer sequences]]