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 $n$:

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 $y^2 = x^3 + 3 x$:

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:


 * $\paren {\pm1}^2 = 1 = 0^3 + 1^3$
 * $\paren {\pm3}^2 = 9 = 1^3 + 2^3$