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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?

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closed as off-topic by Saad, John B, Namaste, Xander Henderson, Trevor Gunn May 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) $
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  • $\begingroup$ $ gcd(p_1, p_2-1) = ...? $ $\endgroup$ – Vladislav Kharlamov May 7 '18 at 5:50
  • $\begingroup$ I think you will understand what happens if you write your solution like this: $19*(19-1) = (7-1)(7^2+7+1)$ $\endgroup$ – Vladislav Kharlamov May 7 '18 at 5:53
  • $\begingroup$ I proved that it should be 1 but still no proof $\endgroup$ – Thishanka Alahakoon May 7 '18 at 5:53
  • $\begingroup$ If their common divisor is 1, what follows from this? $\endgroup$ – Vladislav Kharlamov May 7 '18 at 5:54
  • $\begingroup$ That the other factor of R.H.S. should be divisible by P1 $\endgroup$ – Thishanka Alahakoon May 7 '18 at 5:55

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