Sum of Cubes of 5 Consecutive Integers which is Square

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:


 * $0, 1, 2, 3, 4$


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

Proof
Then we also have:

and finally the degenerate case:

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 :

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

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.

Also see

 * Sum of Cubes of 3 Consecutive Integers which is Square