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:
:$m^3 = \ds \sum_{k \mathop = 1}^m \paren {n + k}^2$
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:
:$n^2 + \paren {m + 1} n + \dfrac {\paren {m + 1} \paren {2 m + 1} } 6 - m^2 = 0$
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 Solution to Pell's Equation, 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:
:$(5): \quad x_n + y_n \sqrt {33} = \paren {23 + 4 \sqrt {33} }^n$
for all positive integers $n$.
Using the solution of $(3)$:
:$t = 6, m = 1$
we can obtain another solution of $(3)$ by using:
:$\paren {6 + \sqrt {33} } \paren {x + y \sqrt {33} } = t + m \sqrt {33}$
where:
:$(6): \quad t = 6 x + 33 y, m = x + 6 y$
Thus:
:$t - m \sqrt {33} = \paren {6 - \sqrt {33} } \paren {x - y \sqrt {33} }$
from which:
Thus it is demonstrated that $\tuple {t, m}$ is a solution of $(3)$.
Now let $\tuple {t, m}$ 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:
:$x_n + y_n \sqrt {33} = \paren {23 + 4 \sqrt {33} }^n$
We can set up a recursive process to calculate $\tuple {x_n, y_n}$ of $(4)$ and the corresponding $\tuple {t_n, m_n}$ of $(3)$ as follows:
:$(7): \quad \tuple {x_n, y_n} = \begin{cases}
\tuple {23, 4} & : n = 1 \\
\tuple {23 x_{n - 1} + 132 y_{n - 1}, 23 y_{n - 1}, 4 x_{n - 1} } & : n > 1
\end{cases}$
:$(8): \quad \tuple {t_n, m_n} = \begin{cases}
\tuple {6, 1} & : n = 0 \\
\tuple {23 t_{n - 1} + 132 m_{n - 1}, 23 t_{n - 1}, 4 m_{n - 1} } & : n > 0
\end{cases}$
Using $(8)$, the values of $m$ for $n \ge 1$ are found to be:
:$m_1 = 47, m_2 = 2161, m_3 = 99 \, 359, \ldots$
== Examples ==
== Sources ==
* 
* : $103,823$
Category:Numbers whose Cube equals Sum of Sequence of that many Squares
Category:Cube Numbers
Category:Square Numbers