Square triangular number

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

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

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

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

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

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

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

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

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

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

In 1778 Leonhard Euler determined the explicit formula^{[1][2]: 12–13}

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

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

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

The corresponding explicit formulas for ${\displaystyle s_{k}}$ and ${\displaystyle t_{k}}$ are:^[2]: 13

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

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^[3]: (12)

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

We have^[1][2]: 13

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

All square triangular numbers have the form ${\displaystyle b^{2}c^{2}}$, where ${\displaystyle {\tfrac {b}{c}}}$ is a convergent to the continued fraction expansion of ${\displaystyle {\sqrt {2}}}$, the square root of 2.^[4]

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

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

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.^[5]

The generating function for the square triangular numbers is:^[6]

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

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