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Let $p,q$ be two distinct prime number such that $p^n=q^2+q+1$, where $n \in \Bbb{N}$. I want to proof that, this forces $n=1$.

Rima
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    The RHS polynomial $x^2+x+1$ is irreducible over the integers. Now conclude. – Parcly Taxel Sep 15 '16 at 13:42
  • can you explain me any more? I dont know anything about RHS polynomial, please – Rima Sep 15 '16 at 13:43
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    @ParclyTaxel:conclude how? $x^2+x+1=\Phi_3(x)$, hence every prime factor of $x^2+x+1$ is $3$ or a prime $\equiv 1\pmod{3}$, but we do not know much about the multiplicity of such prime factors. – Jack D'Aurizio Sep 15 '16 at 13:53
  • @ParclyTaxel Yes, $x^2 + x + 1$ is irreducible over the integers, so there isn't a "common" factorisation to all the different numbers $q^2 + q + 1$, the way $q^2 - 1$ can always be factored as $(q-1)(q+1)$. That doesn't stop $q^2 + q + 1$ from having different non-trivial factorisations for different $q$. Most of all, that in itself doesn't stop $q^2 + q + 1$ from being a power of a prime for some $q$. – Arthur Sep 15 '16 at 13:57
  • ...well I should be scurrying off from here. I'm actually taking a module in number theory this semester, but I have the more important topic of NMR spectroscopy. – Parcly Taxel Sep 15 '16 at 13:58
  • please hint me!! I dont understand your answers ! – Rima Sep 15 '16 at 13:58
  • Fact: $3^n=q^2+q+1$ has no solutions, hence if $p^n=q^2+q+1$ with $n\geq 2$, $q$ has to be a prime $\equiv 2\pmod{3}$. – Jack D'Aurizio Sep 15 '16 at 14:00
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    Now it is probably useful to work in the ring of Eistenstein integers, that is a UFD. – Jack D'Aurizio Sep 15 '16 at 14:00
  • this fact $3^n=q^2+q+1$ has no solutions, is in a book? can you address me please? – Rima Sep 15 '16 at 14:01
  • @Rima: that is simple to prove, since it is equivalent to $$ 4\cdot 3^n-3 = (2q+1)^2$$ – Jack D'Aurizio Sep 15 '16 at 14:11
  • Why do you want to proof this? – miracle173 Sep 15 '16 at 14:29
  • because it is a part of my algebraic problem! – Rima Sep 15 '16 at 14:33

2 Answers2

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Erdös and Selfridge proved that a product of consecutive integers can never be a perfect power. That is, the equation $$ x(x + 1)\cdots (x + (m - 1)) = y^n $$ has no solutions in positive integers $x,y,m,n$ with $m,n > 1$. Also for $$ x(x + 1)\cdots (x + (m - 1))+1 = y^n $$ all solutions are explicitly known, see here, and the reference to the paper of N. Abe. For $m=2$ we obtain the above equation $x^2+x+1=y^n$. There is no integer solution for $n>1$.

A related problem can be found in Ribenboim's book, where he gives all solutions of the Diophantine equation $$ y^2=1+x+x^2+\cdots +x^k. $$

Dietrich Burde
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I borrowed two books on contest problems; this one is An Introduction to Diophantine Equations, by Titu Andreescu, Dorin Andrica, and Ion Cucurezeanu. Most of what is needed for your problem, probably all, is in section 4.2, The Ring of Integers of $\mathbb Q[\sqrt d].$ This is mentioned above by Jack, suggesting working in the Eisenstein integers $\mathbb Z[\sqrt {-3}],$ which is a UFD. This is a pretty good start...

We have $x^2 + x + 1 = p^n;$ I am not sure yet whether the restriction that $x$ be prime matters. We have, page 168, the unit $$ \omega = \frac{-1 + \sqrt {-3}}{2}, $$ which is a cube root of $1.$ The main detail is that $\sqrt {-3}$ is irreducible in $\mathbb Z[\sqrt {-3}].$ On page 173 we have some examples of finding gcd of two quantities. Also on page 177, problem 2 is similar to yours, solution takes pages 315-319. Note that $$ \omega^2 = -1 - \omega,$$ $$ \omega - \omega^2 = \sqrt {-3}. $$ Alright, we have factorization $$ (x - \omega)(x - \omega^2) = p^n. $$ We need to know $$ \gcd(x - \omega, x - \omega^2). $$ If some $\delta$ in the ring divides both, then it divides $ \omega - \omega^2 = \sqrt {-3}. $ However, this is irreducible, so $$ \gcd(x - \omega, x - \omega^2)= 1. $$ It follows that each factor is such a power, that is $$ x - \omega = (a + b \omega)^n, $$ for ordinary $a,b \in \mathbb Z.$ NOTE: remind me to put in alternatives $$ x - \omega = \omega(a + b \omega)^n, \; \; \; x - \omega = \omega^2(a + b \omega)^n. $$ What I have so far is the implication $ b = \pm 1; $ let me do exponent $n=5.$

$$ x - \omega = (a + b \omega)^5, $$ $$ x - \omega = a^5 + 5 a^4 b \omega + 10 a^3 b^2 \omega^2 + 10 a^2 b^3 + 5 a b^4 \omega + b^5 \omega^2. $$ Remember $\omega^2 = -1 - \omega,$ so that $$ x - \omega = (a^5 - 10 a^3 b^2 + 10 a^2 b^3 - b^5) + (5 a^4 b - 10 a^3 b^2 + 5 a b^4 - b^5 ) \omega. $$ The coefficient of $\omega$ needs to be $-1,$ that is $$ (5 a^4 b - 10 a^3 b^2 + 5 a b^4 - b^5 ) = -1 $$ and $$ b(5 a^4 - 10 a^3 b + 5 a b^3 - b^4 ) = -1. $$ Well, $b$ is an ordinary integer, and $b$ divides $-1.$ that is, $$ b = \pm 1. $$ Next, we need either $$ 5 a^4 - 10 a^3 + 5 a - 1 = -1, $$ or $$ 5 a^4 + 10 a^3 - 5 a - 1 = -1. $$ Either way, divide out by $5a,$ $$ a^3 + 2 a^2 - 1 = 0 $$ OR $$ a^3 - 2 a^2 + 1 = 0. $$ The rational root theorem tells us that $a = \pm 1$ as well.

Will Jagy
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