The Aperiodical 

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33 can be written as the sum of three cubes

By Peter Rowlett and Christian Lawson-Perfect. Posted March 11, 2019

It was an open question whether 33 could be written as the sum of three cubes. Thanks to Andrew R. Booker, it now isn’t.

$$\begin{array}{c}(8866128975287528{)}^3\\ +\\ (-8778405442862239{)}^3\\ +\\ (-2736111468807040{)}^3\\ =\\ 33\end{array}$$