How to solve $p^2 - p + 1 = q^3$ over primes?
I checked that $p$ and $q$ must be odd and I don't know what to do next. I don't have any experience with such equations.
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Xam
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user4201961
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2Rewrite as $p(p-1)=q^3-1=(q-1)(q^2+q+1)$. – vadim123 Feb 23 '17 at 19:51
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2write your equation in the form $$p(p-1)=(q-1)(q^2+q+1)$$ – Dr. Sonnhard Graubner Feb 23 '17 at 19:51
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1$p$ is a primitive sixth root of $1$ modulo $q$, so $q-1$ must be divisible by $6$ . – Thomas Andrews Feb 23 '17 at 20:07
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@vadim123 $p$ must divide $(q - 1)$ or $(q^2 + q + 1)$ (only one of these two, because $(p - 1)$ is not a prime and is smaller than $p$). I really can't see what's next. – user4201961 Feb 23 '17 at 20:14
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1Better to study the given problem in the ring of Eisenstein integers $\mathbb{Z}[\omega]$, that is a UFD. – Jack D'Aurizio Feb 23 '17 at 20:19
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It is also interesting to point out that the problem can be put in the following form $$ (4q)^3 - (8p-4)^2 = 48$$ and there probably are just a finite number of solutions of $a^3-b^2=48$. – Jack D'Aurizio Feb 23 '17 at 20:27
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3More general https://artofproblemsolving.com/community/c6h35935p224998 – arberavdullahu Feb 23 '17 at 20:39
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The hidden answer below is not true because $(p,q)=(19,7)$ is a solution. – Ataulfo Jun 07 '22 at 10:58
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COMMENT.-Primes $p$ of the form $6x-1$ cannot be solution: in fact $$(6x-1)^2-(6x-1)+1=36x^2-18x+3=3(12x^2-6x+1)$$ so $q=3$ which is not compatible with $12x^2-6x+1=3^2$.
It follows $p$ must be of the form $6x+1$ so we have $$(6x+1)^2-(6x+1)+1=36x^2+6x+1=q^3$$ Solving the equation, where $X=6x$, $$X^2+X+1-q^3=0$$ we have $$2X=-1+\sqrt{4q^3-3}$$ and the equation $$4q^3-3=z^2$$ seems to have $(z,q)=(37,7)$ as only solution which gives $\boxed{(p,q)=(19,7)}$.
Is it $(p,q)=(19,7)$ the only solution?
Ataulfo
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FWIW: $q \not \equiv 3 \bmod 3 \Rightarrow q^3 \equiv \pm 1 \bmod 9$. But $p^2-p+1 \not \equiv \pm 1 \bmod 9$ unless $p\equiv 1 \bmod 9$. Also $q \not \equiv 0 \bmod 7 \Rightarrow q^3 \equiv \pm 1 \bmod 7$. But $p^2-p+1 \not \equiv \pm 1 \bmod 7$ unless $p\equiv (0,1,4) \bmod 7$. $p\equiv 0 \bmod 7 \Rightarrow p=7$, which is not a solution. Thus, any solutions other than $p=19, q=7$, if they exist, will have the form $109+126k$ or $127+126k$. A quick check yields no further easy solutions. – Keith Backman Jun 07 '22 at 18:02