Numbers whose Cube equals Sum of Sequence of that many Squares

Theorem
The integers $m$ in the following sequence all have the property that $m^3$ is equal to the sum of $m$ consecutive squares:


 * $m^3 = \displaystyle \sum_{k \mathop = 1}^m \left({n + k}\right)^2$

for some $n \in \Z_{\ge 0}$:


 * $0, 1, 47, 2161, 99 \, 359, 4 \, 568 \, 353, \ldots$

Proof
We have:

Thus we have the quadratic equation:


 * $n^2 + \left({m + 1}\right) n + \dfrac {\left({m + 1}\right) \left({2 m + 1}\right)} 6 - m^2 = 0$

From Solution to Quadratic Equation:

Let $t := +\sqrt {33 m^2 + 3}$.

We are given that $m$ is an integer.

Let $n$ be an integer.

Then $t$ is a rational number which is the square root of an integer.

Therefore $t$ is an integer.

Now let $t$ be an integer.

Then $3$ is a divisor of $t^2$.

Thus $3$ is a divisor of $t$.

It follows that $\dfrac t 3$ and $m + 1$ have the same parity.

Thus either $\dfrac {m + 1} 2$ and $\dfrac t 6$ are both integers or both half-integers.

Hence $n$ is an integer

Thus it has been demonstrated that $n$ is an integer $t$ is an integer.

Thus, finding the solutions of $(1)$ is equivalent to finding the solutions to the Diophantine equation:
 * $(3): \quad t^2 - 33m^2 = 3$

We first note the degenerate solution:
 * $t = 6, m = 1$

Consider Pell's Equation:
 * $(4): \quad x^2 - 33 y^2 = 1$

By working it out (or looking it up), the first positive solution to $(4)$ is:
 * $x = 23, y = 4$

Thus all the solutions to $(4)$ are:
 * $x = 1, y = 0$

and:
 * $x = \pm x_n, y = \pm y_n$

where:
 * $(5): \quad x_n + y_n \sqrt {33} = \left({23 + 4 \sqrt {33} }\right)^n$

for all positive integers $n$.

Using the solution of $(3)$:
 * $t = 6, m = 1$

we can obtain another solution of $(3)$ by using:
 * $\left({6 + \sqrt {33} }\right) \left({x + y \sqrt {33} }\right) = t + m \sqrt {33}$

where:
 * $(6): \quad t = 6 x + 33 y, m = x + 6 y$

Thus:
 * $t - m \sqrt {33} = \left({6 - \sqrt {33} }\right) \left({x - y \sqrt {33} }\right)$

from which:

Thus it is demonstrated that $\left({t, m}\right)$ is a solution of $(3)$.

Now let $\left({t, m}\right)$ be any solution of $(3)$.

Let:
 * $x = 2 t - 11 m$
 * $y = \dfrac {6 m - t} 3$

We have that:
 * $t^2 - 33 m^2 = 3$

and so:
 * $3$ is a divisor of $t^2$

and so:
 * $3$ is a divisor of $t$

and so $x$ and $y$ are both integers.

$x$ and $y$ are seen to be solutions to $(4)$, and:
 * $t = 6 x + 33 y$
 * $m = x + 6 y$

Thus from $(5)$ and $(6)$ it follows that the solutions of $(3)$ with $m > 1$ are obtained from $x = \pm x_n, y = \pm y_n$ in $(5)$.

It follows further that all values of $m$ in such solutions are odd.

The trivial solution $x = 1, y - 0$ of $(4)$ corresponds to $m = 1, t = 6$ of $(3)$.

Thus we have that all the values of $m$ are given by:
 * $m_n = x_n + 6 y_n$

where:
 * $x_n + y_n \sqrt {33} = \left({23 + 4 \sqrt {33} }\right)^n$

We can set up a recursive process to calculate $\left({x_n, y_n}\right)$ of $(4)$ and the corresponding $\left({t_n, m_n}\right)$ of $(3)$ as follows:


 * $(7): \quad \left({x_n, y_n}\right) = \begin{cases}

\left({23, 4}\right) & : n = 1 \\ \left({23 x_{n - 1} + 132 y_{n - 1}, 23 y_{n - 1}, 4 x_{n - 1} }\right) & : n > 1 \end{cases}$


 * $(8): \quad \left({t_n, m_n}\right) = \begin{cases}

\left({6, 1}\right) & : n = 0 \\ \left({23 t_{n - 1} + 132 m_{n - 1}, 23 t_{n - 1}, 4 m_{n - 1} }\right) : & n > 0 \end{cases}$

Using $(8)$, the values of $m$ for $n \ge 1$ are found to be:
 * $m_1 = 47, m_2 = 2161, m_3 = 99 \, 359, \ldots$