Prime Equal to Sum of Digits of Cube

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Theorem

The only prime number which is equal to the sum of the digits of its cube is $17$:


Proof

We have that:

\(\displaystyle 17^3\) \(=\) \(\displaystyle 4913\)
\(\displaystyle 17\) \(=\) \(\displaystyle 4 + 9 + 1 + 3\)


From Positive Integers Equal to Sum of Digits of Cube, the complete set of positive integers with this property are:

$0, 1, 8, 17, 18, 26, 27$

Of these, only $17$ is prime.

$\blacksquare$


Sources