Triangular Numbers which are Sum of Two Cubes

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Theorem

The sequence of triangular numbers which are the sum of $2$ cubes begins:

$28, 91, 351, 2926, 8001, 46971, 58653, 93528, 97461, \dots$

This sequence is A113958 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

Can be demonstrated by brute force.

For example:

\(\ds 28\) \(=\) \(\ds 1 + 27\)
\(\ds \) \(=\) \(\ds 1^3 + 3^3\)
\(\ds \) \(=\) \(\ds \dfrac {7 \paren {7 + 1} } 2\)


\(\ds 91\) \(=\) \(\ds 27 + 64\)
\(\ds \) \(=\) \(\ds 3^3 + 4^3\)
\(\ds \) \(=\) \(\ds \dfrac {13 \paren {13 + 1} } 2\)


\(\ds 351\) \(=\) \(\ds 125 + 216\)
\(\ds \) \(=\) \(\ds 5^3 + 6^3\)
\(\ds \) \(=\) \(\ds \dfrac {26 \paren {26 + 1} } 2\)


\(\ds 2976\) \(=\) \(\ds 729 + 2197\)
\(\ds \) \(=\) \(\ds 5^3 + 6^3\)
\(\ds \) \(=\) \(\ds \dfrac {76 \paren {76 + 1} } 2\)


\(\ds 8001\) \(=\) \(\ds 1 + 8000\)
\(\ds \) \(=\) \(\ds 1^3 + 20^3\)
\(\ds \) \(=\) \(\ds \dfrac {126 \paren {126 + 1} } 2\)


\(\ds 46 \, 971\) \(=\) \(\ds 4096 + 42 \, 875\)
\(\ds \) \(=\) \(\ds 16^3 + 35^3\)
\(\ds \) \(=\) \(\ds \dfrac {306 \paren {306 + 1} } 2\)


\(\ds 58 \, 653\) \(=\) \(\ds 8000 + 50 \, 653\)
\(\ds \) \(=\) \(\ds 20^3 + 37^3\)
\(\ds \) \(=\) \(\ds \dfrac {342 \paren {343 + 1} } 2\)


\(\ds 93 \, 528\) \(=\) \(\ds 42 \, 875 + 50 \, 653\)
\(\ds \) \(=\) \(\ds 35^3 + 37^3\)
\(\ds \) \(=\) \(\ds \dfrac {432 \paren {433 + 1} } 2\)


\(\ds 97 \, 461\) \(=\) \(\ds 125 + 97 \, 336\)
\(\ds \) \(=\) \(\ds 5^3 + 46^3\)
\(\ds \) \(=\) \(\ds \dfrac {441 \paren {442 + 1} } 2\)

$\blacksquare$


Sources

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