Numbers whose Cube equals Sum of Sequence of that many Squares/Examples/47

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

 * $\displaystyle 47^3 = \sum_{k \mathop = 1}^{47} \left({21 + k}\right)^2$

Proof
From Numbers whose Cube equals Sum of Sequence of that many Squares:
 * $\displaystyle m^3 = \sum_{k \mathop = 1}^m \left({n + k}\right)^2$

for $m = 47$ and for $n$ given by:


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

So in this instance:

This gives:
 * $n = 21, n = -69$

which leads to the two solutions:

Hence we have: