We have: {{begin-eqn}} {{eqn | n = 1       | l = m^3       | r = \sum_{k \mathop = 1}^m \left({n + k}\right)^2       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^m \left({n^2 + 2 n k + k^2}\right)       | c =  }} {{eqn | r = n^2 \sum_{k \mathop = 1}^m 1 + 2 n \sum_{k \mathop = 1}^m k +  \sum_{k \mathop = 1}^m k^2       | c =  }} {{eqn | r = m n^2 + 2 n \frac {m \left({m + 1}\right)} 2 + \frac {m \left({m + 1}\right) \left({2 m + 1}\right)} 6       | c = Closed Form for Triangular Numbers, Sum of Sequence of Squares }} {{eqn | ll= \leadsto       | l = m^2       | r = n^2 + n \left({m + 1}\right) + \frac {\left({m + 1}\right) \left({2 m + 1}\right)} 6       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = n       | r = \dfrac {-\left({m + 1}\right) \pm \sqrt{\left({m + 1}\right)^2 - 4 \left({\dfrac {\left({m + 1}\right) \left({2 m + 1}\right)} 6 - m^2}\right)} } 2       | c =  }} {{eqn | r = - \dfrac {m + 1} 2 \pm \frac 1 2 \sqrt {m^2 + 2 m + 1 - 2 \left({\dfrac {2 m^2 + 3 m + 1} 3}\right) - 4 m^2}       | c =  }} {{eqn | r = - \dfrac {m + 1} 2 \pm \frac 1 2 \sqrt {5 m^2 + 2 m + 1 - \dfrac {4 m^2 + 6 m + 2} 3}       | c =  }} {{eqn | r = - \dfrac {m + 1} 2 \pm \frac 1 2 \sqrt {\frac {15 m^2 + 6 m + 3 - 4 m^2 - 6 m - 2 + 6 m^2} 3}       | c =  }} {{eqn | r = - \dfrac {m + 1} 2 \pm \frac 1 2 \sqrt {\frac {11 m^2 + 1} 3}       | c =  }} {{eqn | r = - \dfrac {m + 1} 2 \pm \frac 1 2 \sqrt {\frac {33 m^2 + 3} 9}       | c =  }} {{eqn | r = - \dfrac {m + 1} 2 \pm \frac 1 6 \sqrt {33 m^2 + 3}       | c =  }} {{end-eqn}} 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 {{iff}} $t$ 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: {{begin-eqn}} {{eqn | l = t^2 - 33 m^2       | r = \left({t - m \sqrt {33} }\right) \left({t + m \sqrt {33} }\right)       | c =  }} {{eqn | r = \left({6 - \sqrt {33} }\right) \left({6 + \sqrt {33} }\right) \left({x - y \sqrt {33} }\right) \left({x + y \sqrt {33} }\right)       | c =  }} {{eqn | r = \left({6^2 - 1 \times 33}\right) \left({x^2 - 33 y^2}\right)       | c =  }} {{eqn | r = 3 \times 1       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} 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: :$m_1 = 47, m_2 = 2161, m_3 = 99 \, 359, \ldots$ {{qed}}