Cube Number as Sum of Three Consecutive Odd Squares

Theorem

 * $1331 = 11^3 = 19^2 + 21^2 + 23^2$

No other such sequence of $3$ consecutive odd squares has the same property.

Proof
Any sequence of $3$ consecutive odd integers that have squares that sum to a cube would satisfy:
 * $m^3 = \paren {n - 2}^2 + n^2 + \paren {n + 2}^2$

where $n$ is the middle number of the sequence, with $m, n \in \Z$.

Expanding the :

Substituting $x = 3 m$ and $y = 9 n$:

which is an elliptic curve.

According to LMFDB, this elliptic curve has exactly $5$ lattice points:
 * $\tuple {6, 0}, \tuple {10, \pm 28}, \tuple {33, \pm 189}$

which correspond to these values of $n$:
 * $0, \pm \dfrac {28} 9, \pm 21$

$\pm \dfrac {28} 9$ are not integers.

The other possibilities are:
 * the trivial $0 = \paren {-2}^2 + 0^2 + 2^2$
 * the less trivial $11^3 = \paren {-23}^2 + \paren {-21}^2 + \paren {-19}^2$

The first is not a set of consecutive odd squares.

The second gives the same set of consecutive odd squares.

Hence there are no more solutions.