9 is Only Square which is Sum of 2 Consecutive Positive Cubes

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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 $n$:

\(\displaystyle y^2\) \(=\) \(\displaystyle 2 x^3 + 3 x^2 + 3 x + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 4 y^2\) \(=\) \(\displaystyle 8 x^3 + 12 x^2 + 12 x + 4\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {2 y}^2\) \(=\) \(\displaystyle \paren {2 x}^3 + 3 \paren {4 x^2 + 4 x + 1} + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {2 y}^2\) \(=\) \(\displaystyle \paren {2 x}^3 + 1^3 + 3 \paren {2 x + 1}^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {2 y}^2\) \(=\) \(\displaystyle \paren {2 x + 1} \paren { \paren {2 x}^2 -2 x + 1} + 3 \paren {2 x + 1}^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {2 y}^2\) \(=\) \(\displaystyle \paren {2 x + 1} \paren { \paren {2 x}^2 -2 x + 1 + 3 \paren {2 x + 1} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {2 y}^2\) \(=\) \(\displaystyle \paren {2 x + 1} \paren {4 x^2 + 4x + 1 + 3 }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {2 y}^2\) \(=\) \(\displaystyle \paren {2 x + 1} \paren {\paren {2x + 1}^2 + 3}\)

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:

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

$\blacksquare$


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