Sum of Cubes of 5 Consecutive Integers which is Square

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Theorem

The following sequences of $5$ consecutive (strictly) positive integers have cubes that sum to squares:

$1, 2, 3, 4, 5$
$25, 26, 27, 28, 29$
$96, 97, 98, 99, 100$
$118, 119, 120, 121, 122$

No other such sequence of $5$ consecutive positive integers has the same property.


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

$0, 1, 2, 3, 4$
$-2, -1, 0, 1, 2$


This sequence is A126203 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

\(\ds 1^3 + 2^3 + 3^3 + 4^3 + 5^3\) \(=\) \(\ds 1 + 8 + 27 + 64 + 125\)
\(\ds \) \(=\) \(\ds 225\)
\(\ds \) \(=\) \(\ds 15^2\) also see Sum of Sequence of Cubes


\(\ds 25^3 + 26^3 + 27^3 + 28^3 + 29^3\) \(=\) \(\ds 15 \, 625 + 17 \, 576 + 19 \, 683 + 21 \, 952 + 24 \, 389\)
\(\ds \) \(=\) \(\ds 99 \, 225\)
\(\ds \) \(=\) \(\ds 315^2\)


\(\ds 96^3 + 97^3 + 98^3 + 99^3 + 100^3\) \(=\) \(\ds 884 \, 736 + 912 \, 673 + 941 \, 192 + 970 \, 299 + 1 \, 000 \, 000\)
\(\ds \) \(=\) \(\ds 4 \, 708 \, 900\)
\(\ds \) \(=\) \(\ds 2170^2\)


\(\ds 118^3 + 119^3 + 120^3 + 121^3 + 122^3\) \(=\) \(\ds 1 \, 643 \, 032 + 1 \, 685 \, 159 + 1 \, 728 \, 000 + 1 \, 771 \, 561 + 1 \, 815 \, 848\)
\(\ds \) \(=\) \(\ds 8 \, 643 \, 600\)
\(\ds \) \(=\) \(\ds 2940^2\)


Then we also have:


\(\ds 0^3 + 1^3 + 2^3 + 3^3 + 4^3\) \(=\) \(\ds 0 + 1 + 8 + 27 + 64\)
\(\ds \) \(=\) \(\ds 100\)
\(\ds \) \(=\) \(\ds 10^2\) also see Sum of Sequence of Cubes


and finally the degenerate case:

\(\ds \paren {-2}^3 + \paren {-1}^3 + 0^3 + 1^3 + 2^3\) \(=\) \(\ds \paren {-8} + \paren {-1} + 0 + 1 + 8\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds 0^2\)


Any sequence of $5$ consecutive integers that have cubes that sum to a square would satisfy:

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

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


Expanding the right hand side:

\(\ds m^2\) \(=\) \(\ds \paren {n - 2}^3 + \paren {n - 1}^3 + n^3 + \paren {n + 1}^3 + \paren {n + 2}^3\)
\(\ds \) \(=\) \(\ds n^3 - 6 n^2 + 12 n - 8 + n^3 - 3 n^2 + 3 n - 27 + n^3 + n^3 + 3 n^2 + 3 n + 27 + n^3 + 6 n^2 + 12 n + 8\) Cube of Sum, Cube of Difference
\(\ds \) \(=\) \(\ds 5 n^3 + 30 n\)

Substituting $y = 5 m$ and $x = 5 n$:

\(\ds \paren {\frac y 5}^2\) \(=\) \(\ds 5 \paren {\frac x 5}^3 + 30 \paren {\frac x 5}\)
\(\ds \leadsto \ \ \) \(\ds \frac {y^2} {25}\) \(=\) \(\ds \frac {x^3} {25} + 6 x\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(=\) \(\ds x^3 + 150 x\)

which is an elliptic curve.

According to LMFDB, this elliptic curve has exactly $13$ lattice points:

$\tuple {0, 0}, \tuple {10, \pm 50}, \tuple {15, \pm 75}, \tuple {24, \pm 132}, \tuple {135, \pm 1575}, \tuple {490, \pm 10 \, 850}, \tuple {600, \pm 14 \, 700}$

which correspond to these values of $n$:

$0, 2, 3, \dfrac {24} 5, 27, 98, 120$

Note that $\dfrac {24} 5$ is not an integer.

Hence there are no more solutions.

$\blacksquare$


Also see


Historical Note

This result was originally published by Édouard Lucas in $1873$, where he wrote:

The sum of the cubes of $5$ consecutive numbers is never equal to a square, except for the solutions of which the middle numbers are $2$, $3$, $27$, $98$ or $120$.

Subsequently it was reported by Leonard Eugene Dickson, who made a mistake by omitting the $27$.

The erroneous sequence is still mistakenly published on occasion.


Sources

  • The LMFDB Collaboration, The L-functions and Modular Forms Database, Elliptic Curve 57600/bt/2, $2013$ [Online; accessed 31-Mar-2022]