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Question: for which primes $p$ is $p^2 - p + 1$ a cube number $n^3$? \begin{align*} p^2 - p + 1 &= n^3 \\ p (p - 1) &= (n - 1)(n^2 + n + 1) \end{align*} Per Euclid, $p$ must divide at least one of $n - 1$ and $n^2 + n + 1$. Due to the degrees of the polynomials, $n$ must be smaller than $p$, so $p$ is a factor of $n^2 + n + 1$. Introducing $m \in \mathbb{N}^+$: \begin{align*} p m &= n^2 + n + 1 \\ p (p - 1) &= p m (n - 1) \\ p - 1 &= m (n - 1) \\ p &= m n - m + 1 \\ n^2 + n + 1 &= m^2n - m^2 + m \\ n^2 + (1 - m^2) n + 1 + m^2 - m &= 0 \\ n &= \frac{m^2 - 1 \pm \sqrt{(1 - m^2)^2 - 4(m^2 - m + 1)}}{2} \end{align*} For $n$ to be integer, the discriminant $d$ must be a square number. \begin{align*} d &= 1 - 2 m^2 + m^4 - 4m^2 + 4m - 4 \\ &= m^4 - 6m^2 + 4m - 3 \\ &= (m^2 - 3)^2 + 4m - 12 \end{align*} For $m < 3$, $d$ is negative. For $m \ge 3$, \begin{equation*} (m^2 - 3 + 1)^2 - (m^2 - 3)^2 = 2 m^2 - 5 > 4m - 12 \end{equation*} Therefore $d$ is a square iff $4m - 12 = 0 \leftrightarrow m = 3$. Substituting that back in gives $d = 36$, $n = 4 \pm 3$. $n = 1$ would require the non-prime $p = 1$, and $n = 7$ gives the only prime solution, $p = 19$.
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Sat, 19 Aug 2023 20:55 GMT