9 is Only Square which is Sum of 2 Consecutive Positive Cubes
Theorem
Discounting the trivial solution:
- $1^2 = 1 = 0^3 + 1^3$
$9$ is the only square number which is the sum of $2$ consecutive positive cube numbers:
- $3^2 = 9 = 1^3 + 2^3$
Proof
The expression for the $n$th square number is:
- $n^2$
for $n \in \Z$.
The expression for the $n$th cube number is:
- $n^3$
again, for $n \in \Z$.
Therefore the closed-form expression for the $n$th sum of two consecutive cubes is:
- $n^3 + \paren {n + 1}^3$
This simplifies to:
- $2 n^3 + 3 n^2 + 3 n + 1$
Equate the two expressions with a variable replacing $n$:
\(\ds y^2\) | \(=\) | \(\ds 2 x^3 + 3 x^2 + 3 x + 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 4 y^2\) | \(=\) | \(\ds 8 x^3 + 12 x^2 + 12 x + 4\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {2 y}^2\) | \(=\) | \(\ds \paren {2 x}^3 + 3 \paren {4 x^2 + 4 x + 1} + 1\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 x}^3 + 1^3 + 3 \paren {2 x + 1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 x + 1} \paren { \paren {2 x}^2 -2 x + 1} + 3 \paren {2 x + 1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 x + 1} \paren { \paren {2 x}^2 -2 x + 1 + 3 \paren {2 x + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 x + 1} \paren {4 x^2 + 4x + 1 + 3 }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 x + 1} \paren {\paren {2x + 1}^2 + 3}\) |
Substituting:
- $u = 2 x + 1$
- $v = 2 y$
the equation becomes:
- $v^2 = u^3 + 3 u$
We then apply Integral Points of Elliptic Curve $y^2 = x^3 + 3 x$:
The elliptic curve:
- $y^2 = x^3 + 3x$
has exactly $7$ lattice points:
- $\tuple {0, 0}, \tuple {1, \pm 2}, \tuple {3, \pm 6}, \tuple {12, \pm 42}$
We then write them in terms of $x$ and $y$:
- $\tuple {-\frac 1 2, 0}, \tuple {0, \pm 1}, \tuple {1, \pm 3}, \tuple {\frac {11} 2, \pm 21}$
Due to the restrictions on the variables, solutions with non-integer inputs are invalid.
This leaves $4$ solutions:
- $\tuple {0, \pm 1}, \tuple {1, \pm 3}$
as follows:
- $\paren {\pm1}^2 = 1 = 0^3 + 1^3$
- $\paren {\pm3}^2 = 9 = 1^3 + 2^3$
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $9$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $9$