Primes, divisors and congruences

Note that

\(\begin{align*} (a + b)^3 & = a(a^2 + 3ab + 3b^2) + b^3 \\ & = a(a^2 + 3ab + 3b^2 - 1) + a + b^3 \end{align*} \\)

holds. Since \(a + b^3\) is divisible by \(a^2 + 3ab + 3b^2 - 1\), it follows that \((a + b)^3\) is also divisible by \(a^2 + 3ab + 3b^2 - 1\). Assume for the sake of contradiction that \(a^2 + 3ab + 3b^2 - 1\) is not divisible by the cube of an integer greater than \(1\). It follows that \(a^2 + 3ab + 3b^2 - 1\) divides \((a + b)^2\). This implies that \[ a^2 + 2ab + b^2 \geq a^2 + 3ab + 3b^2 - 1, \] which simplifies to \[ ab + 2b^2 \leq 1, \] which is impossible since \(a\) and \(b\) are positive integers. This proves that \(a^2 + 3ab + 3b^2 - 1\) is divisible by the cube of an integer greater than \(1\).