# $p^2-p+1$ is a perfect cube of a prime [closed]

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?

## closed as off-topic by Saad, John B, Namaste, Xander Henderson, Trevor GunnMay 9 '18 at 3:53

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• HINT It can be rewritten as $$p_{1}(p_{1}-1) = (p_{2}-1)(p_{2}^2+p_{2}+1)$$
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 $(k = 3; y = 1), (k = 3; y = 7)$
• $gcd(p_1, p_2-1) = ...?$ – Vladislav Kharlamov May 7 '18 at 5:50
• I think you will understand what happens if you write your solution like this: $19*(19-1) = (7-1)(7^2+7+1)$ – Vladislav Kharlamov May 7 '18 at 5:53