Determine with a proof all prime numbers p such that p$^2$-p+1 is a cube of a prime number.
By trial and error method 19$^2$-19+1=7$^3$
Is it the only p?
How should I prove it?
2026-03-25 06:05:28.1774418728
$p^2-p+1$ is a perfect cube of a prime
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If you rewrite as: $ (p_{1}-1) = k(p_{2}-1),$ then everything reduces to a solution to the given equation:
$$(k y - k + 1) k = y^2 + y + 1$$
I myself, unfortunately, could not solve this equation, but if to believe WolframAlf, its integer solution $ (k = 3; y = 1), (k = 3; y = 7) $