Cube Number as Sum of Three Consecutive Odd Squares

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Theorem

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

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


Proof

\(\ds 19^2 + 21^2 + 23^2\) \(=\) \(\ds 361 + 441 + 529\)
\(\ds \) \(=\) \(\ds 1331\)

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 right hand side:

\(\ds m^3\) \(=\) \(\ds \paren {n - 2}^2 + n^2 + \paren {n + 2}^2\)
\(\ds \) \(=\) \(\ds n^2 - 4 n + 4 + n^2 + n^2 + 4 n + 4\) Square of Sum, Square of Difference
\(\ds \) \(=\) \(\ds 3 n^2 + 8\)

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

\(\ds \paren {\frac x 3}^3\) \(=\) \(\ds 3 \paren {\frac y 9}^2 + 8\)
\(\ds \leadsto \ \ \) \(\ds \frac {x^3} {27}\) \(=\) \(\ds \frac {y^2} {27} + 8\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(=\) \(\ds x^3 - 216\)

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.

$\blacksquare$


Historical Note

In his Curious and Interesting Numbers, 2nd ed. of $1997$, David Wells attributes this result to Michal Stajsczak, but gives no context.


Sources

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