Square triangular number: Difference between revisions

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In [[mathematics]], a '''square triangular number''' (or '''triangular square number''') is a number which is both a [[triangular number]] and a [[square number]]. 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}}}}

==Explicit formulas==

Write {{<math|''N''<sub>''k''N_k</submath>}} for the {{mvar|<math>k}}</math>th square triangular number, and write {{math|''s''<submath>''k''s_k</submath>}} and {{math|''t''<submath>''k''t_k</submath>}} 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'' {{=}} {{sfrac|''n''(''n'' + 1)|2}}}} to be {{mvar|n}}. From this definition and the quadratic formula,

:<math>n = \frac{\sqrt{8N + 1} - 1}{2}.</math>

Define the ''triangular root'' of a triangular number {{<math|''>N'' {{=}} {\tfrac{sfrac|''n''(''n'' + 1)|2}}}{t}</math> to be {{mvar|<math>n}}</math>. From this definition and the quadratic formula,

Therefore, {{mvar|N}} is triangular ({{mvar|n}} is an integer) [[if and only if]] {{math|8''N'' + 1}} is square. Consequently, a square number {{math|''M''²}} is also triangular if and only if {{math|8''M''² + 1}} is square, that is, there are numbers {{mvar|x}} and {{mvar|y}} such that {{math|''x''² − 8''y''² {{=}} 1}}. This is an instance of the [[Pell equation]] with {{math|''n'' {{=}} 8}}. All Pell equations have the trivial solution {{math|''x'' {{=}} 1, ''y'' {{=}} 0}} for any {{mvar|n}}; this is called the zeroth solution, and indexed as {{math|(''x''₀, ''y''₀) {{=}} (1,0)}}. If {{math|(''x_k'', ''y_k'')}} denotes the {{mvar|k}}th nontrivial solution to any Pell equation for a particular {{mvar|n}}, it can be shown by the method of descent that

:<math>\begin{align}

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

Therefore, {{mvar|<math>N}}</math> is triangular ({{mvar|<math>n}}</math> is an integer) [[if and only if]] {{<math|8''N'' >8N+ 1}}</math> is square. Consequently, a square number {{<math|''M''<sup>M^2</supmath>}} is also triangular if and only if {{math|8''M''<supmath>8M^2+1</supmath> + 1}} is square, that is, there are numbers {{mvar|<math>x}}</math> and {{mvar|y}} such that {{<math|''x''<sup>2y</supmath> −such that 8''y''<supmath>x^2-8y^2=1</supmath> {{=}} 1}}. 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 {{mvar|<math>n}}</math>; this is called the zeroth solution, and indexed as {{<math|>(''x''₀x_0, ''y''₀y_0) {{=}} (1,0)}}</math>. If {{math|(''x<submath>k</sub>''(x_k, ''y<sub>ky_k)</submath>'')}} denotes the {{mvar|<math>k}}</math>th nontrivial solution to any Pell equation for a particular {{mvar|<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 aninfinitely infinity ofmany solutions to any Pell equation for which there is one non-trivial one, which holdsis true whenever {{mvar|<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<submath>k</sub>''(x_k, ''y<sub>ky_k)</submath>'')}} to the Pell equation for {{<math|''>n'' {{=}} 8}}</math> yields a square triangular number and its square and triangular roots as follows:

:<math>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 (3,1), is 1, and the next, derived from {{nowrap|6 × (3,1) − (1,0) {{=}} (17,6)}}, is 36.

:{{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>}}

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

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

The sequences {{math|''N''<submath>''k''N_k</submath>}}, {{math|''s''<submath>''k''s_k</submath>}} and {{math|''t''<submath>''k''t_k</submath>}} are the [[OEIS]] sequences {{OEIS2C|id=A001110}}, {{OEIS2C|id=A001109}}, and {{OEIS2C|id=A001108}} respectively.

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

Line 28 ⟶ 34:

{{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''<submath>''k''s_k</submath>}} and {{math|''t''<submath>''k''t_k</submath>}} are:<ref name=Euler />{{Rp|13}}

:<math>\begin{align}

:{{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>}}

==Pell's equation==

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

{{cite book | last1 = Barbeau | first1 = Edward | title = Pell's Equation | pages = [https://archive.org/details/pellsequation0000barb/page/16 16]–17 | url=https://archive.org/details/pellsequation0000barb | url-access = registration | access-date = 2009-05-10 |series = Problem Books in Mathematics | publisher = Springer | ___location = New York | year = 2003 | isbn = 978-0-387-95529-2 }}

</ref>

Every triangular number is of the form {{math|{{sfrac|''t''(''t'' + 1)|2}}}}. Therefore we seek integers {{mvar|t}}, {{mvar|s}} such that

:<math>\frac{t(t+1)}{2} = s^2.</math>

Rearranging, this becomes

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

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

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

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

{{cite book |last1=Hardy |first1=G. H. |author-link1=G. H. Hardy |last2=Wright |first2=E. M. |author-link2=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=[https://archive.org/details/introductiontoth00hard/page/210 210] |quote=Theorem 244 |url-access=registration |url=https://archive.org/details/introductiontoth00hard/page/210 }}

</ref>

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

and therefore all solutions are given by

:<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>

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

==Recurrence relations==

Line 75 ⟶ 57:

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)}}

:{{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''<sup>^2c^2</supmath>''c''²}}, where {{<math|>\tfrac{b}{sfrac|''b''|''c''}}}}</math> is a [[Convergent (continued fraction)|convergent]] to the [[continued fraction|continued fraction expansion]] of <math>\sqrt2</math>, the [[square root of 2|{{sqrt|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 {{mvar|<math>n}}</math>th triangular number {{<math|{>\tfrac{sfrac|''n''(''n'' + 1)|}{2}}}}</math> is square, then so is the larger {{<math|4''n''>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>}}