Sum of Cubes of 3 Consecutive Integers which is Square

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Theorem

The following sequences of $3$ consecutive (strictly) positive integers have cubes that sum to a square:

$1, 2, 3$
$23, 24, 25$

No other such sequence of $3$ consecutive positive integers has the same property.


However, if we allow sequences containing zero and negative integers, we also have:

$-1, 0, 1$
$0, 1, 2$


Proof

\(\ds 1^3 + 2^3 + 3^3\) \(=\) \(\ds 1 + 8 + 27\)
\(\ds \) \(=\) \(\ds 36\)
\(\ds \) \(=\) \(\ds 6^2\)
\(\ds 23^3 + 24^3 + 25^3\) \(=\) \(\ds 12 \, 167 + 13 \, 824 + 15 \, 625\)
\(\ds \) \(=\) \(\ds 41 \, 616\)
\(\ds \) \(=\) \(\ds 204^2\)
\(\ds \paren {-1}^3 + 0^3 + 1^3\) \(=\) \(\ds -1 + 0 + 1\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds 0^2\)
\(\ds 0^3 + 1^3 + 2^3\) \(=\) \(\ds 0 + 1 + 8\)
\(\ds \) \(=\) \(\ds 9\)
\(\ds \) \(=\) \(\ds 3^2\)


Any sequence of $3$ consecutive integers that have cubes that sum to a square would satisfy:

$m^2 = \paren {n - 1}^3 + n^3 + \paren {n + 1}^3$

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


Expanding the right hand side:

\(\ds m^2\) \(=\) \(\ds \paren {n - 1}^3 + n^3 + \paren {n + 1}^3\)
\(\ds \) \(=\) \(\ds n^3 - 3 n^2 + 3 n - 27 + n^3 + n^3 + 3 n^2 + 3 n + 27\) Cube of Sum, Cube of Difference
\(\ds \) \(=\) \(\ds 3 n^3 + 6 n\)

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

\(\ds \paren {\frac y 3}^2\) \(=\) \(\ds 3 \paren {\frac x 3}^3 + 6 \paren {\frac x 3}\)
\(\ds \leadsto \ \ \) \(\ds \frac {y^2} 9\) \(=\) \(\ds \frac {x^3} 9 + 2 x\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(=\) \(\ds x^3 + 18 x\)

which is an elliptic curve.

According to LMFDB, this elliptic curve has exactly $7$ lattice points:

$\tuple {0, 0}, \tuple {3, \pm 9}, \tuple {6, \pm 18}, \tuple {72, \pm 612}$

which correspond to these values of $n$:

$0, 1, 2, 24$

Hence there are no more solutions.

$\blacksquare$


Also see

Sources