Competition problem — Number theory

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Number theory Difficulty 5.3 AIME, harder Prove it Philippines

Problem:

A semiprime is a number that is a product of two prime numbers. How many semiprime numbers less than $2018$ can be expressed as $x^{3}-1$ for some natural number $x$ ?

Solution

Solution:

$x^{3}-1 = (x-1)(x^{2}+x+1)$ , so this implies that both $x-1$ and $x^{2}+x+1$ have to be prime. Furthermore, this also means we only have numbers up to $12$ to work on, as $13^{3} > 2018$ . Hence, we only have to check $x = 3, 4, 6, 8, 12$ and determine if $x^{2}+x+1$ is prime.

The values we get are $13, 21, 43, 73, 157$ , and of the $5$ , only $4$ are prime.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.