Numbers whose Cube equals Sum of Sequence of that many Squares/Examples/2161

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

$\ds 2161^3 = \sum_{k \mathop = 1}^{2161} \paren {988 + k}^2$

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

for $m = 2161$ and for $n$ given by:

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

So in this instance:

$\ds n$

$=$

$\ds -\dfrac {2161 + 1} 2 \pm \frac 1 6 \sqrt {33 \times 2161^2 + 3}$

$\ds $

$=$

$\ds -1081 \pm \frac 1 6 \sqrt {33 \times 4 \, 669 \, 921 + 3}$

$\ds $

$=$

$\ds -1081 \pm \frac 1 6 \sqrt {154 \, 107 \, 396}$

$\ds $

$=$

$\ds -1081 \pm \frac {12 \, 414} 6$

$\ds $

$=$

$\ds -1081 \pm 2069$

This gives:

$n = 988, n = -3150$

which leads to the two solutions:

$\ds 2161^3$

$=$

$\ds 989^2 + 990^2 + \cdots + 3149^2$

$\ds $

$=$

$\ds \paren {-3149}^2 + \paren {-3148}^2 + \cdots + \paren {-989}^2$

$\blacksquare$