Square triangular number: Difference between revisions

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{{short description|Integer that is both a perfect square and a triangular number}}

{{for|squares of triangular numbers|squared triangular number}}

[[File:TriSquare36square_triangular_number_36.svg|200px234px|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|perfect]], 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:

There are [[Infinity{{bi|left=1.6|infinitely many]] square triangular numbers; the first few are 0, 1, 36, {{val|1225}}, {{val|41616}}, {{val|1413721}}, {{val|48024900}}, {{val|1631432881}}, {{val|55420693056}}, {{val|1882672131025}} {{OEIS|id=A001110}}.}}

==Solution as a Pell equation==

==Explicit formulas==

Write ''N''<submath>''k''N_k</submath> for the ''<math>k''</math>th square triangular number, and write ''s''<submath>''k''s_k</submath> and ''t''<submath>''k''t_k</submath> for the sides of the corresponding square and triangle, so that

:<math>N_k = s_k^2 = \frac{t_k(t_k+1)}{2}.</math>

{{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 = \frac{n(n+1)}{2}</math> to be <math>n</math>. From this definition and the quadratic formula, <math>n = \frac{\sqrt{8N + 1} - 1}{2}.</math> Therefore, <math>N</math> is triangular if and only if <math>8N + 1</math> is square. Consequently, a number <math>M^2</math> is square and triangular if and only if <math>8M^2 + 1</math> is square, i. e., 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, with <math>n=8</math>. All Pell equations have the trivial solution (1,0), for any n; this solution is called the zeroth, and indexed as <math>(x_0,y_0)</math>. If <math> (x_k,y_k)</math> denotes the k'th non-trivial solution to any Pell equation for a particular n, it can be shown by the method of descent that <math>x_{k+1} = 2x_k x_1 - x_{k-1}</math> and <math>y_{k+1} = 2y_k x_1 - y_{k-1}</math>. Hence there are an infinity of solutions to any Pell equation for which there is one non-trivial one, which holds whenever n is not a square. The first non-trivial solution when n=8 is easy to find: it is (3,1). A solution <math>(x_k,y_k)</math> to the Pell equation for n=8 yields a square triangular number and its square and triangular roots as follows: <math>s_k = y_k , t_k = \frac{x_k - 1}{2},</math> and <math>N_k = y_k^2.</math> Hence, the first square triangular number, derived from (3,1), is 1, and the next, derived from (17,6) (=6×(3,1)-(1,0)), is 36.

Define the ''triangular root'' of a triangular number <math>N=\tfrac{n(n+1)}{2}</math> to be <math>n</math>. From this definition and the quadratic formula,

The sequences ''N''_''k'', ''s''_''k'' and ''t''_''k'' are the [[OEIS]] sequences {{OEIS2C|id=A001110}}, {{OEIS2C|id=A001109}}, and {{OEIS2C|id=A001108}} respectively.

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

In 1778 [[Leonhard Euler]] determined the explicit formula<ref name=Dickson>

{{cite book | last1 = Dickson | first1 = Leonard Eugene | authorlink1 = Leonard Eugene Dickson |title = [[History of the Theory of Numbers]] | volume = 2 | publisher = American Mathematical Society | ___location = Providence | year = 1999 |origyear = 1920 | page = 16 | isbn = 978-0-8218-1935-7 }}

</ref><ref name=Euler>

{{cite journal |last=Euler |first=Leonhard |authorlink=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=Latin |accessdate=2009-05-11 |quote=According to the records, it was presented to the St. Petersburg Academy on May 4, 1778.}}

</ref>{{Rp|12–13}}

:<math>N_k = \left( \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{4\sqrt{2}} \right)^2.

</math>

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

: <math>\begin{align}

N_k &= {1 \over 32} \left( ( 1 + \sqrt{2} )^{2k} - ( 1 - \sqrt{2} )^{2k} \right)^2 = {1 \over 32} \left( ( 1 + \sqrt{2} )^{4k}-2 + ( 1 - \sqrt{2} )^{4k} \right) \\

&= {1 \over 32} \left( ( 17 + 12\sqrt{2} )^k -2 + ( 17 - 12\sqrt{2} )^k \right).

\end{align}</math>

The corresponding explicit formulas for ''s''_''k'' and ''t''_''k'' are <ref name=Euler />{{Rp|13}}

:<math> s_k = \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{4\sqrt{2}} </math>

and

:<math> t_k = \frac{(3 + 2\sqrt{2})^k + (3 - 2\sqrt{2})^k - 2}{4}. </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

==Pell's equation==

{{bi|left=1.6|<math>\displaystyle \begin{align}

The problem of finding square triangular numbers reduces to [[Pell's equation]] in the following way.<ref>

x_{k+1} &= 2x_k x_1 - x_{k-1}, \\

{{cite book | last1 = Barbeau | first1 = Edward | title = Pell's Equation | pages = 16–17 | url=https://books.google.com/books?id=FtoFImV5BKMC&pg=PA16 | accessdate = 2009-05-10 |series = Problem Books in Mathematics | publisher = Springer | ___location = New York | year = 2003 | isbn = 978-0-387-95529-2 }}

y_{k+1} &= 2y_k x_1 - y_{k-1}.

</ref>

\end{align}</math>}}

Every triangular number is of the form ''t''(''t'' + 1)/2. Therefore we seek integers ''t'', ''s'' such that

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{t(t+x_k - 1)}{2}, \quad N_k = sy_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>.

With a bit of algebra this becomes

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.

:<math>(2t+1)^2=8s^2+1,</math>

==Explicit formula==

and then letting ''x'' = 2''t'' + 1 and ''y'' = 2''s'', we get the [[Diophantine equation]]

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>x^2 - 2y^2 =1</math>

</math>}}

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

which is an instance of [[Pell's equation]]. This particular equation is solved by the [[Pell number]]s ''P''_''k'' as<ref>

{{cite book |last1=Hardy |first1=G. H. |authorlink1=G. H. Hardy |last2=Wright |first2=E. M. |authorlink2 = E. M. Wright |title=An Introduction to the Theory of Numbers |edition=5th |year=1979 |publisher=Oxford University Press |isbn=0-19-853171-0 |page=210|quote= Theorem 244 }}

</ref>

{{bi|left=1.6|<math>\displaystyle \begin{align}

:<math>x = P_{2k} + P_{2k-1}, \quad y = P_{2k};</math>

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}}

and therefore all solutions are given by

{{bi|left=1.6|<math>\displaystyle \begin{align}

:<math> s_k = \frac{P_{2k}}{2}, \quad t_k = \frac{P_{2k} + P_{2k-1} -1}{2}, \quad N_k = \left( \frac{P_{2k}}{2} \right)^2.</math>

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}.

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

\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}

There are [[recurrence relation]]s for the square triangular numbers, as well as for the sides of the square and triangle involved. We have<ref>{{MathWorld|title=Square Triangular Number|urlname=SquareTriangularNumber}}</ref>{{Rp|(12)}}

N_k &= 34N_{k-1} - N_{k-2} + 2,& \text{with }N_0 &= 0\text{ and }N_1 = 1; \\

:<math>N_k &= 34N_\left(6\sqrt{N_{k-1}} - \sqrt{N_{k-2} + }\right)^2,& \text{ with }N_0 &= 0\text{ and }N_1 = 1.</math>

\end{align}</math>}}

:<math>N_k = \left(6\sqrt{N_{k-1}} - \sqrt{N_{k-2}}\right)^2,\text{ with }N_0 = 0\text{ and }N_1 = 1.</math>

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

{{bi|left=1.6|<math>\displaystyle \begin{align}

:<math>s_k = 6s_{k-1} - s_{k-2},\text{ with }s_0 = 0\text{ and }s_1 = 1;</math>

s_k &= 6s_{k-1} - s_{k-2},& \text{with }s_0 &= 0\text{ and }s_1 = 1; \\

:<math>t_k &= 6t_{k-1} - t_{k-2} + 2,& \text{ with }t_0 &= 0\text{ and }t_1 = 1.</math>

\end{align}</math>}}

==Other characterizations==

All square triangular numbers have the form ''b''<supmath>b^2c^2</sup>''c''<sup>2</supmath>, where ''<math>\tfrac{b'' }{c}</ ''c''math> is a [[Convergent (continued fraction)|convergent]] to the [[simple continued fraction|continued fraction expansion]] forof <math>\sqrt2</math>, the [[square root of 2]].<ref name=Ball>

{{cite book | last1 = Ball | first1 = W. W. Rouse |authorlink1author-link1 = W. W. Rouse Ball | last2 = Coxeter | first2 = H. S. M. |authorlink2author-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 aninfinitely infinity ofmany square triangular numbers,: toIf the wit:<refmath>n</math>th name=Sylwestertriangular number <math>\tfrac{n(n+1)}{2}</math> is square, then so is the larger <math>4n(n+1)</math>th triangular number, since:

{{cite journal |last=Pietenpol |first=J. L. |author2=A. V. Sylwester |author3=Erwin Just |author4=R. M 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>

{{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>}}

If the triangular number ''n''(''n''+1)/2 is square, then so is the larger triangular number

:<math>\frac{\bigl( 4n(n+1) \bigr) \bigl( 4n(n+1)+1 \bigr)}{2} = 2^2 \, \frac{n(n+1)}{2} \,(2n+1)^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>

We know this result has to be a square, because it is a product of three squares: 2^2 (by the exponent), (n(n+1))/2 (the n'th triangular number, by proof assumption), and the (2n+1)^2 (by the exponent). The product of any numbers that are squares is naturally going to result in another square. This can be seen from the fact that a necessary and sufficient condition for a number to be square is that there should be only even powers of primes in its prime factorisation, and multiplying two square numbers preserves this property in the product.

{{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}}

The triangular roots <math>t_k</math> are alternately simultaneously one less than a square and twice a square, if k is even, and simultaneously a square and one less than twice a square, if k is odd. Thus, <math>49 = 7^2 = 2*5^2 - 1, 288 = 17^2 - 1 = 2 * 12^2</math>, and <math>1681 = 41^2 = 2 * 29^2 - 1.</math> In each case, the two square roots involved multiply to give <math>s_k: 5 * 7 = 35, 12 * 17 = 204, </math> and <math>29 * 41 = 1189.</math>{{citation needed|date=December 2014}}

<math>N_k - N_{k-1}=s_{2k-1}: 36 - 1 = 35, 1225 - 36 = 1189,</math> and <math>41616 - 1225 = 40391.</math> In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.{{citation needed|date=December 2014}}

The generating function for the square triangular numbers is:<ref>

{{cite web |first=Simon |last=Plouffe |authorlink=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 |format=PDF |date=August 1992 |accessdate=2009-05-11 }}

</ref>

:<math>\frac{1+z}{(1-z)(z^2 - 34z + 1)} = 1 + 36z + 1225 z^2 + \cdots.</math>

==Numerical data==

As <math>k</math> becomes larger, the ratio <math>t_k/s_k</math> approaches <math>\sqrt{2} \approx 1.41421356</math> and the ratio of successive square triangular numbers approaches <math> (1+\sqrt{2})^4 = 17+12\sqrt{2} \approx 33.970562748</math>. The table below shows values of <math>k</math> between 0 and 11, which comprehend all square triangular numbers up to <math>100\,000\,000</math>.

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>

{| class="wikitable" border="1"

:<math>\frac{1+z}{(1-z)\left(z^2 - 34z + 1\right)} = 1 + 36z + 1225 z^2 + \cdots</math>

|-

! <math>k</math>

! <math>N_k</math>

! <math>s_k</math>

! <math>t_k</math>

! <math>t_k/s_k</math>

! <math> N_k/N_{k-1}</math>

|-

|<math>0</math>

|<math>0</math>

|<math>0</math>

|<math>0</math>

|

|

|-

|<math>1</math>

|<math>1</math>

|<math>1</math>

|<math>1</math>

|<math>1.00000000</math>

|

|-

|<math>2</math>

|<math>36</math>

|<math>6</math>

|<math>8</math>

|<math>1.33333333</math>

|<math>36.000000000</math>

|-

|<math>3</math>

|<math>1\,225</math>

|<math>35</math>

|<math>49</math>

|<math>1.40000000</math>

|<math>34.027777778</math>

|-

|<math>4</math>

|<math>41\,616</math>

|<math>204</math>

|<math>288</math>

|<math>1.41176471</math>

|<math>33.972244898</math>

|-

|<math>5</math>

|<math>1\,413\,721</math>

|<math>1\,189</math>

|<math>1\,681</math>

|<math>1.41379310</math>

|<math>33.970612265</math>

|-

|<math>6</math>

|<math>48\,024\,900</math>

|<math>6\,930</math>

|<math>9\,800</math>

|<math>1.41414141</math>

|<math>33.970564206</math>

|-

|<math>7</math>

|<math>1\,631\,432\,881</math>

|<math>40\,391</math>

|<math>57\,121</math>

|<math>1.41420118</math>

|<math>33.970562791</math>

|-

|<math>8</math>

|<math>55\,420\,693\,056</math>

|<math>235\,416</math>

|<math>332\,928</math>

|<math>1.41421144</math>

|<math>33.970562750</math>

|-

|<math>9</math>

|<math>1\,882\,672\,131\,025</math>

|<math>1\,372\,105</math>

|<math>1\,940\,449</math>

|<math>1.41421320</math>

|<math>33.970562749</math>

|-

|<math>10</math>

|<math>63\,955\,431\,761\,796</math>

|<math>7\,997\,214</math>

|<math>11\,309\,768</math>

|<math>1.41421350</math>

|<math>33.970562748</math>

|-

|<math>11</math>

|<math>2\,172\,602\,007\,770\,041</math>

|<math>46\,611\,179</math>

|<math>65\,918\,161</math>

|<math>1.41421355</math>

|<math>33.970562748</math>

|}<!-- The table was generated in 23-jul-2016 using a Python script available at http://pastebin.com/sWyesrR8 -->

==See also==

*[[Cannonball problem]], on numbers that are simultaneously square and square pyramidal

*[[Sixth power]], numbers that are simultaneously square and cubical