Numbers whose Cube equals Sum of Sequence of that many Squares/Examples/99,359

Example of Numbers whose Cube equals Sum of Sequence of that many Squares

$\ds 99 \, 359^3 = \sum_{k \mathop = 1}^{99 \, 359} \paren {45 \, 449 + k}^2$

$\ds m^3 = \sum_{k \mathop = 1}^m \paren {n + k}^2$

for $m = 99 \, 359$ and for $n$ given by:

$n = \dfrac {m + 1} 2 \pm \dfrac 1 6 \sqrt {33 m^2 + 3}$

So in this instance:

$\ds n$

$=$

$\ds -\dfrac {99 \, 359 + 1} 2 \pm \frac 1 6 \sqrt {33 \times 99 \, 359^2 + 3}$

$\ds $

$=$

$\ds -49 \, 680 \pm \frac 1 6 \sqrt {33 \times 9 \, 872 \, 210 \, 881 + 3}$

$\ds $

$=$

$\ds -49 \, 680 \pm \frac 1 6 \sqrt {325 \, 782 \, 959 \, 076}$

$\ds $

$=$

$\ds -49 \, 680 \pm \frac {570 \, 774} 6$

$\ds $

$=$

$\ds -49 \, 680 \pm 95 \, 129$

This gives:

$n = 45 \, 449, n = -144 \, 809$

which leads to the two solutions:

$\ds 99 \, 359^3$

$=$

$\ds 45 \, 450^2 + 45 \, 451^2 + \cdots + 144 \, 808^2$

$\ds $

$=$

$\ds \paren {-144 \, 808}^2 + \paren {-144 \, 807}^2 + \cdots + \paren {-45 \, 450}^2$

$\blacksquare$