Square triangular number

For squares of triangular numbers, see squared triangular number.

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 infinitely many square triangular numbers; the first few are:

Square triangular number 36 depicted as a triangular number and as a square number.

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025 (sequence A001110 in the OEIS)

Explicit formulas

Write $N_{k}$ for the $k$th square triangular number, and write $s_{k}$ and $t_{k}$ for the sides of the corresponding square and triangle, so that

$\displaystyle N_{k}=s_{k}^{2}={\frac {t_{k}(t_{k}+1)}{2}}.$

Define the triangular root of a triangular number $N={\tfrac {n(n+1)}{t}}$ to be $n$. From this definition and the quadratic formula,

$\displaystyle n={\frac {{\sqrt {8N+1}}-1}{2}}.$

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

$\displaystyle {\begin{aligned}x_{k+1}&=2x_{k}x_{1}-x_{k-1},\\y_{k+1}&=2y_{k}x_{1}-y_{k-1}.\end{aligned}}$

Hence there are infinitely many solutions to any Pell equation for which there is one non-trivial one, which is true whenever $n$ is not a square. The first non-trivial solution when $n=8$ is easy to find: it is $(3,1)$. A solution $(x_{k},y_{k})$ to the Pell equation for $n=8$ yields a square triangular number and its square and triangular roots as follows:

$\displaystyle s_{k}=y_{k},\quad t_{k}={\frac {x_{k}-1}{2}},\quad N_{k}=y_{k}^{2}.$

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

The sequences $N_{k}$, $s_{k}$ and $t_{k}$ are the OEIS sequences OEIS: A001110, OEIS: A001109, and OEIS: A001108 respectively.

In 1778 Leonhard Euler determined the explicit formula

$\displaystyle N_{k}=\left({\frac {\left(3+2{\sqrt {2}}\right)^{k}-\left(3-2{\sqrt {2}}\right)^{k}}{4{\sqrt {2}}}}\right)^{2}.$

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

$\displaystyle {\begin{aligned}N_{k}&={\tfrac {1}{32}}\left(\left(1+{\sqrt {2}}\right)^{2k}-\left(1-{\sqrt {2}}\right)^{2k}\right)^{2}\\&={\tfrac {1}{32}}\left(\left(1+{\sqrt {2}}\right)^{4k}-2+\left(1-{\sqrt {2}}\right)^{4k}\right)\\&={\tfrac {1}{32}}\left(\left(17+12{\sqrt {2}}\right)^{k}-2+\left(17-12{\sqrt {2}}\right)^{k}\right).\end{aligned}}$

The corresponding explicit formulas for $s_{k}$ and $t_{k}$ are:

$\displaystyle {\begin{aligned}s_{k}&={\frac {\left(3+2{\sqrt {2}}\right)^{k}-\left(3-2{\sqrt {2}}\right)^{k}}{4{\sqrt {2}}}},\\t_{k}&={\frac {\left(3+2{\sqrt {2}}\right)^{k}+\left(3-2{\sqrt {2}}\right)^{k}-2}{4}}.\end{aligned}}$

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

$\displaystyle {\begin{aligned}N_{k}&=34N_{k-1}-N_{k-2}+2,&{\text{with }}N_{0}&=0{\text{ and }}N_{1}=1;\\N_{k}&=\left(6{\sqrt {N_{k-1}}}-{\sqrt {N_{k-2}}}\right)^{2},&{\text{with }}N_{0}&=0{\text{ and }}N_{1}=1.\end{aligned}}$

We have

$\displaystyle {\begin{aligned}s_{k}&=6s_{k-1}-s_{k-2},&{\text{with }}s_{0}&=0{\text{ and }}s_{1}=1;\\t_{k}&=6t_{k-1}-t_{k-2}+2,&{\text{with }}t_{0}&=0{\text{ and }}t_{1}=1.\end{aligned}}$

Other characterizations

All square triangular numbers have the form $b^{2}c^{2}$, where ${\tfrac {b}{c}}$ is a convergent to the continued fraction expansion of ${\sqrt {2}}$, the square root of 2.

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

$\displaystyle {\frac {{\bigl (}4n(n+1){\bigr )}{\bigl (}4n(n+1)+1{\bigr )}}{2}}=4\,{\frac {n(n+1)}{2}}\,\left(2n+1\right)^{2}.$

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.

The generating function for the square triangular numbers is:

${\frac {1+z}{(1-z)\left(z^{2}-34z+1\right)}}=1+36z+1225z^{2}+\cdots$

See also

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

External links

• Triangular numbers that are also square at cut-the-knot
• Weisstein, Eric W. "Square Triangular Number". MathWorld.
• Michael Dummett's solution