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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Martin Sleziak
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1More generally, all integer solutions of $x^2+x+1=y^3$ are known: https://math.stackexchange.com/questions/394240 – Bart Michels May 07 '18 at 06:36
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My answer helped you? – Vladislav Kharlamov May 07 '18 at 06:43
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Really. Thank you very much – Thishanka Alahakoon May 07 '18 at 06:48
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Related How to solve $p^2 - p + 1 = q^3$ over primes?, and quadratic diophantine involving primes. And also this AoS forum thread: https://artofproblemsolving.com/community/c6h35935p224998 – Sil May 07 '18 at 22:21
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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) $
amWhy
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Vladislav Kharlamov
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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 07 '18 at 05:53
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I don't see it yet, either. At best, it follows that $7$ is a cube root, hence $19\equiv 1\pmod 6$ – Hagen von Eitzen May 07 '18 at 05:56
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Theoretically, the last equation should not be difficult to solve and I have some ideas, but without the proper experience, I do not want to assert the correctness of my thoughts – Vladislav Kharlamov May 07 '18 at 06:27