Theorem: Sum of Cubes of 5 Consecutive Integers which is Square

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:
:$0, 1, 2, 3, 4$
:$-2, -1, 0, 1, 2$
{{OEIS|A126203}}

Proof:
{{begin-eqn}}
{{eqn | l = 1^3 + 2^3 + 3^3 + 4^3 + 5^3
      | r = 1 + 8 + 27 + 64 + 125
      | c = 
}}
{{eqn | r = 225
      | c = 
}}
{{eqn | r = 15^2
      | c = also see Sum of Sequence of Cubes
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 25^3 + 26^3 + 27^3 + 28^3 + 29^3
      | r = 15 \, 625 + 17 \, 576 + 19 \, 683 + 21 \, 952 + 24 \, 389
      | c = 
}}
{{eqn | r = 99 \, 225
      | c = 
}}
{{eqn | r = 315^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 96^3 + 97^3 + 98^3 + 99^3 + 100^3
      | r = 884 \, 736 + 912 \, 673 + 941 \, 192 + 970 \, 299 + 1 \, 000 \, 000
      | c = 
}}
{{eqn | r = 4 \, 708 \, 900
      | c = 
}}
{{eqn | r = 2170^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 118^3 + 119^3 + 120^3 + 121^3 + 122^3
      | r = 1 \, 643 \, 032 + 1 \, 685 \, 159 + 1 \, 728 \, 000 + 1 \, 771 \, 561 + 1 \, 815 \, 848
      | c = 
}}
{{eqn | r = 8 \, 643 \, 600
      | c = 
}}
{{eqn | r = 2940^2
      | c = 
}}
{{end-eqn}}
Then we also have:
{{begin-eqn}}
{{eqn | l = 0^3 + 1^3 + 2^3 + 3^3 + 4^3
      | r = 0 + 1 + 8 + 27 + 64
      | c = 
}}
{{eqn | r = 100
      | c = 
}}
{{eqn | r = 10^2
      | c = also see Sum of Sequence of Cubes
}}
{{end-eqn}}
and finally the degenerate case:
{{begin-eqn}}
{{eqn | l = \paren {-2}^3 + \paren {-1}^3 + 0^3 + 1^3 + 2^3
      | r = \paren {-8} + \paren {-1} + 0 + 1 + 8
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{eqn | r = 0^2
      | c = 
}}
{{end-eqn}}
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 {{RHS}}:
{{begin-eqn}}
{{eqn | l = m^2
      | r = \paren {n - 2}^3 + \paren {n - 1}^3 + n^3 + \paren {n + 1}^3 + \paren {n + 2}^3
}}
{{eqn | r = 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
      | c = Cube of Sum, Cube of Difference
}}
{{eqn | r = 5 n^3 + 30 n
}}
{{end-eqn}}
Substituting $y = 5 m$ and $x = 5 n$:
{{begin-eqn}}
{{eqn | l = \paren {\frac y 5}^2
      | r = 5 \paren {\frac x 5}^3 + 30 \paren {\frac x 5}
}}
{{eqn | ll= \leadsto
      | l = \frac {y^2} {25}
      | r = \frac {x^3} {25} + 6 x
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = x^3 + 150 x
}}
{{end-eqn}}
which is an elliptic curve.
According to [https://www.lmfdb.org/EllipticCurve/Q/57600/bt/2 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.
{{qed}}