492 is Sum of 3 Cubes in 3 Ways

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Theorem

$492$ can be expressed as the sum of $3$ cubes, either positive or negative in $3$ known ways.

\(\ds 492\) \(=\) \(\ds 50^3 + \paren {-19}^3 + \paren {-49}^3\)
\(\ds \) \(=\) \(\ds 123 \, 134^3 + 9179^3 + \paren {-123 \, 151}^3\)
\(\ds \) \(=\) \(\ds 1 \, 793 \, 337 \, 644^3 + \paren {-81 \, 3701 \, 167}^3 + \paren {-1 \, 735 \, 662 \, 109}^3\)


Proof

Brute force.


Also see


Historical Note

Andreas-Stephan Elsenhans and Jörg Jahnel reported in $2009$ on a systematic investigation they performed on all the solutions to the equation:

$x^3 + y^3 + z^3 = n$

for all $0 < n < 1000$ and such that $\size x, \size y, \size z \le 10^{14}$.


Within that range, they discovered that there are exactly $3$ solutions for $n = 492$.

Of all the numbers from $0$ to $1000$, most have many more than $3$ such ways, although a few still have no known solutions.


The full results can be found at https://www.uni-math.gwdg.de/jahnel/Arbeiten/Liste/threecubes_20070419.txt.


Sources