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This following Problem is from Pell equation chapters exercise

Let $y>3$ positive integer numbers, show that following diophantine equation $$x^2+x+1=7^y\tag{1}$$ has no integer solutions.

I tried write the equation $$(2x+1)^2+3=4\cdot 7^y$$ if $y=2k$ then we have $$(2\cdot 7^k+2x+1)(2\cdot 7^k-2x-1)=3$$ this case has no integer.

But $y$ is odd number, How to prove equation (1) has no integer solutions for $x,y (y>3)$? Any help would be appreciated.

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    $\begingroup$ I don't know any elementary solution right now. The only way I can see utilizes the unique factorization property of Eisensteins integers. $\endgroup$ Aug 20, 2015 at 8:31
  • $\begingroup$ can you post your solution? PS:This problem is from pell equation excise. $\endgroup$
    – user246384
    Aug 20, 2015 at 8:32
  • $\begingroup$ what book is this from ? $\endgroup$
    – mercio
    Aug 20, 2015 at 9:44
  • $\begingroup$ Note,, the equation here is a special case of the equation $\frac{x^n-1}{x-1}=y^q$ for integers $x,y > 1$, $n > 2$, $q\ge2$ (for $n=3$ and $y=7$). It is conjectured that this only has finitely many solutions - possibly only 3 solutions. See Bugeaud, Mignotte & Roy (www-irma.u-strasbg.fr/~bugeaud/travaux/edgardef.ps). There, it is stated Ljunggren and Nagell have proved this for $3\vert n$ and $4\vert n$, which covers the case $n=3,y=7$ which you are asking about. $\endgroup$ Aug 30, 2015 at 1:36

6 Answers 6

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Continuing from @Batominovski's answer,

the problem asks us to find all the values taking $\pm 1$ in the three following recurrent linear sequences satisfying $a_n = 5a_{n-1} - 7a_{n-2}$,

  • $1,2,3,1,-16,-87,-323,\ldots$
  • $0,1,5,18,55,149,360,\ldots$
  • $1,3,8,19,39,62,37,\ldots$

Looking at the recurrence relation mod $18$, we have $a_n \equiv 5a_{n-1}+11a_{n-2} \equiv 5(5a_{n-2}+11a_{n-3}) + 11a_{n-2} \equiv a_{n-3}$.

Looking at the first three terms of each sequence modulo $18$, we can already discard $2/3$ of all the terms and also remove the possibility of a $-1$ appearing, which reduces the problem to finding the $1$ values in the linear recurrent sequences satisfying $a_n = 20a_{n-1}-343_{n-2}$

  • $1,1,-323,-6803,\ldots$
  • $1,55,757,-3725,\ldots$
  • $1,19,37,-5777,\ldots$

Now each sequence is a linear combination of the coefficient sequences of $(2-\omega)^{3n} = (1-18\omega)^n = (10 - 9 \sqrt{-3})^n$, that is, the two sequences

  • $1,10,-143,-6290,\ldots$
  • $0,-9,-180,-513, \ldots$

(with integer coefficents in all three cases)

With the binomial theorem, we can expand $(1+9(1-\sqrt{-3}))^n = 1 + 9(1-\sqrt{-3})n + 81(1-\sqrt{-3})^2n(n-1)/2 + \ldots$.

If we place ourselves in $\Bbb Z_3[\sqrt{-3}]$ and we notice that $v_3(9^k/k!) >= 3k/2 \to \infty$, we can reorder the summation and get $1 + [9(1-\sqrt{-3}) + O(81)]n + O(81)n^2 + \ldots$ i.e. each coefficient sequence can be extended as a function $\Bbb Z_3 \mapsto \Bbb Z_3$ which is a power series in $n$ with coefficients in $\Bbb Z_3$

Now, if a power series is of the form $a_0 + a_1 n + a_2n^2$ with $|a_1| > |a_2|,|a_3|,|a_4|,\ldots$, then it is a bijection from $\Bbb Z_3$ to $a_0+a_1\Bbb Z_3$.

The $a_1$ coefficients modulo $81$ in our three sequences are $9-9=0, 9-9 \times (-5) = 54 \neq 0, 9-9 \times (-1) = 18 \neq 0$.

In the last two cases we deduce that the sequence is injective so the initial occurence of $1$ is the only one.

As for the first sequence, we have that both $a_1$ and $a_2$ are of the order of $3^4$. But then its derivative is of the right form to show that it has a single zero at some $n_0 \in \Bbb Z_3$. By setting $m = n-n_0$, we get a new power series of the form $b_0 + b_2m^2 + b_3m^3 + \ldots$ where $|b_2| > |b_3|,|b_4|,\ldots$. This shows that the power series is $2$-to-$1$ (with the exception of $0 \mapsto b_0$), so the two values of $1$ we already know are the only ones.

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  • $\begingroup$ (+1) The idea to use $\nu_3$ and the binomial theorem is pretty clever. $\endgroup$ Aug 21, 2015 at 18:03
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    $\begingroup$ @Jack i can't count the number of times i used the log power series, then the exp one only to realize sheepishly that i could have just used the binomial theorem $\endgroup$
    – mercio
    Aug 21, 2015 at 18:37
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    $\begingroup$ besides, this $p$-adic method is classic and known as Skolem's method. Though there is no guarantee that it works as you want it to. $\endgroup$
    – mercio
    Aug 21, 2015 at 18:38
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Sketch of My Idea:

Let $\omega:=\frac{-1+\sqrt{-3}}{2}$ and $\bar{\omega}:=\frac{-1-\sqrt{-3}}{2}$. Then, $7=(2-\omega)(2-\bar{\omega})$. Hence, $$(x-\omega)(x-\bar{\omega})=x^2+x+1=7^y=(2-\omega)^y(2-\bar{\omega})^y\,.$$ Since $7\nmid x-\omega$, we may assume without loss of generality that $x-\omega=u(2-\omega)^y$, where $u$ is a unit (i.e., $\pm1$, $\pm\omega$, and $\pm\bar{\omega}$). Then, $x-\bar{\omega}=\bar{u}(2-\bar{\omega})^y$, where $\bar{u}$ is the complex conjugate of $u$. Thus, $$-1=\frac{u(2-\omega)^y-\bar{u}(2-\bar{\omega})^y}{\omega-\bar{\omega}}\,.$$

Now, define $a_n(u)$ to be $\frac{u(2-\omega)^n-\bar{u}(2-\bar{\omega})^n}{\omega-\bar{\omega}}$ for $n\in\mathbb{Z}$. You have that $a_n(u)=5a_{n-1}(u)-7a_{n-2}(u)$ for all $n\in\mathbb{Z}$. Then, investigate each $u$. You should find that, only for $n=0$, $n=1$, or $n=3$, there exists a unit $u$ such that $a_{n}(u)=-1$. Now, since $a_n(-u)=-a_n(u)$, you may only look for solutions to $a_n(u)=\pm 1$ with $u \in \{1,\omega,\bar{\omega}\}$.

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  • $\begingroup$ Nice idea,+1.But How to prove $a_{n}(u)=1$ only $n=1,n=3?$ $\endgroup$
    – user246384
    Aug 20, 2015 at 8:47
  • $\begingroup$ @ConĐườngNghệ: that may be hard, since the sequence $\{a_n\}$ takes both positive and negative values infinitely often (the roots of the characteristic polynomial are conjugated complex numbers with norm $7$ and argument $\pm\arctan\frac{\sqrt{3}}{5}$, so no more than $9$ consecutive terms may have the same sign). $\endgroup$ Aug 20, 2015 at 14:52
  • $\begingroup$ If I made an impression that I solved the problem, I'm sorry. I meant to say that it's the only idea came to my mind. $\endgroup$ Aug 20, 2015 at 19:01
  • $\begingroup$ Beautiful idea, I am taking this home with me tonight! $\endgroup$
    – user284001
    Apr 29, 2016 at 14:42
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A sketch of my thoughts:

Let $\nu_7(n)=\max\{m\in\mathbb{N}:7^m\mid n\}$. A good idea may be to prove that if $\nu_7(x^2+x+1)=\nu_7(x^3-1)-\nu_7(x-1)=y>3$ then $x$ has to be large, say $x\geq 7^{y-1}$. In such a case, however, $x^2+x+1$ is too big to be just $7^y$ and it must have some other prime factor.

By this way, the problem boils down to finding (or, at least, lower-bounding) the two elements of order three in $G=\left(\mathbb{Z}_{/(7^y\mathbb{Z})}\right)^*$, that is a cyclic group with $o(G)=6\cdot 7^{y-1}$.

If $y=1$, that elements are $2$ and $4$. If $y=2$, that elements are $18$ and $30$. If $y=3$, that elements are $18$ and $18^2$. In general, we may compute such elements by solving $z^2+3\equiv 0\pmod{7^y}$. If $y=4$, such elements are $1047$ and $1353$. If $y=5$, such elements are $1353$ and $15453$.

We just need to find a pattern, or an alternative reason for which $4\cdot 7^y-3$ cannot be a square for $y>3$. Obviously if $y$ is even $4\cdot 7^y-3$ is too close to a square to be a square itself, so we may assume that $y$ is odd.

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    $\begingroup$ Would writing $x^2+x+1=\dfrac{(x+1)^3-x^3+2}3$ offer any particular insight ? $\endgroup$
    – Lucian
    Aug 21, 2015 at 17:38
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While not exactly solving your question, we can show that it is possible to find all solutions to a more general class of equations which includes the one asked for here. This shows that the equation asked for has only finitely many solutions and gives a method of finding them all - so, it shows that the question can be solved by performing some (rather tedious) calculations.

Letting $Q\in\mathbb Q[X]$ be a rational quadratic polynomial and $\mathcal P$ be a finite set of primes, then there are only finitely many solutions to \begin{align} Q(x)\in\mathcal P^*&&{\rm(1)} \end{align} for $x\in\mathbb Z$, where $\mathcal P^*$ is the set of integers whose prime factors are all in $\mathcal P$. Furthermore, the set of solutions is effectively computable.

In the question asked here, we have $Q=X^2+X+1$ and $\mathcal P=\{7\}$, and want to show that the only solutions are $x=-19,-3,-1,0,2,18$, for which $Q(x)\in\{1,7,7^3\}$. Now, this answer does not quite answer your question -- as I am going to apply an effective theorem on diophantine approximation to quadratic irrationals, although I haven't yet worked out how to compute the solutions. The theorem says that an algorithm exists, although a closer reading of the proof should give the algorithm (whether it is simple enough to be easily applied is another matter though).

To prove that (1) has finitely many solutions, we can complete the square to write $$ a^{-1}\left(L(x)^2-b\right)\in\mathcal P^* $$ for a degree one polynomial $L\in\mathbb Z[X]$ and fixed integers $a,b$. This can then be written as $$ L(x)^2-b=au^2v $$ for nonnegative $u\in\mathcal P^*$ and $v$ one of the finitely many squarefree elements of $\mathcal P^*$. In the example given, we have the case $v=1$ which is solved in the question, and $v=7$ which is asked for. We can rearrange as $$ \left(L(x)+u\sqrt{av}\right)\left(L(x)-u\sqrt{av}\right)=b. $$ From this, we obtain the inequality $$ 0 < \lvert u\sqrt{av}-w\rvert u\sqrt{av}\le\lvert b\rvert, $$ where I have set $w=\lvert L(x)\rvert$. Rearranging, we express this as a rational approximation to the algebraic number $\sqrt{av}$, \begin{align} 0 < \left\lvert\sqrt{av}-\frac wu\right\rvert\le \frac{\lvert b\rvert}{\sqrt{av}}u^{-2}.&&{\rm(2)} \end{align} Now, a result on diophantine approximation says that if $\xi$ is a real quadratic number then there exists effectively computable positive reals $C,\kappa > 0$ such that $$ \left\lvert\xi-\frac wu\right\rvert\ge Cu^{-2+\kappa} $$ for all integers $u,w$ with $u\in\mathcal P^*$. See the statement and proof of Theorem 1.2 in Effective Results For Restricted Rational Approximation To Quadratic Irrationals by Bennet and Bugeaud (link). Taking $\xi=\sqrt{av}$ and comparing with (2), $$ Cu^{-2+\kappa}\le\frac{\lvert b\rvert}{\sqrt{av}}u^{-2}. $$ Hence, $$ u\le\left(\frac{\lvert b\rvert}{C\sqrt{av}}\right)^{1/\kappa}. $$ So, we have a computable upper bound for $u$, and there are only finitely many solutions to (1).

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An elementary solution [albeit not using Pellian methods] is reasonably simple to obtain.

As you show, there is no solution when $y$ is even. Now assume $y$ is odd. Observe that $7 \mid (x^2+x+1)$ implies $x \equiv 2\!\pmod{7}$ or $x \equiv 4\!\pmod{7}$.

In the first case, we substitute $x=7z+2$ into the equation and simplify to obtain $$z(7z+5)=\bigl(7^\frac{y-1}{2}-1\bigr)\bigl(7^\frac{y-1}{2}+1\bigr).$$ Now our assumption of a positive solution with $y \ge 3$ implies there exist positive integers $a,b,c,d$ such that \begin{align} z &= ab, & 7^\frac{y-1}{2}-1 &= ac, \\ 7z+5 &= cd, & 7^\frac{y-1}{2}+1 &= bd. \end{align} The conclusion follows fairly directly via algebraic manipulations.

The proof in the second case is nearly identical.

EDIT:

Here's a hint to get most of the way through the proof… Writing $k=(y-1)/2$, we have $$ \frac{z}{7^k+1} = \frac{a}{d} = \frac{7^k-1}{7z+5}. $$ Solving each relation for $7^k$ gives $$ 7^k = \frac{dz-a}{a} = \frac{7az+5a+d}{d}, $$ and solving the second relation for $z$ yields $$ z = \frac{a(2d+5a)}{d^2-7a^2}. $$ Since $z$ is an integer by hypothesis, there must be a prime $p$ dividing $\gcd(d,a)$, or $(d^2-7a^2) \mid (2d+5a)$, from which we can quickly deduce $p \mid 3a$.

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  • $\begingroup$ Can you show me how to do those algebraic manipulations? $\endgroup$
    – S.C.B.
    Apr 29, 2016 at 5:48
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We work in the Eisenstein integers. The equation then becomes$$(x + 1 + \omega)(x - \omega) = 6^y.$$Suppose $d\, |\, (x + 1 + \omega)$ and $\text{ }d\,|\,(x - \omega)$.Then we have $d\,|\,(1 + 2\omega)$, so $N(d)\,|\,3$. However, we also have $N(d)\,|\,7^{2y}$, so $N(d) = 1$. On the other hand, $7 = (-2 - 3\omega)(1 + 3\omega)$ and $-2 - 3\omega$, $1 + 3\omega$ are relatively prime. Thus, it follows that $x + 1 + \omega$ and $x - \omega$ are $e(-2-3\omega)^y$ and $f(1 + 3\omega)^y$ in some order, where $e$ and $f$ are units $\pm1$, $\pm\omega$, $\pm1\pm\omega$ with $ef = 1$. We just need to show that if$$a + b\omega = f(1 + 3\omega)^y,$$then $|b|$ can never be $1$, which would solve the problem. Indeed, let$$A = 1 + 3^3\binom{y}{3} + \dots,\text{ }B = 3\binom{y}{1} + \dots,\text{ }C = 3^2\binom{y}{2} + \dots.$$We have$$a + b\omega = f(A - C + \omega(B - C)),$$so we want to show none of $|A - B|$, $|B- C|$, $|C - A|$ are one. However, this is a relatively easy task for $y > 3$, so we are done.

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    $\begingroup$ "However, this is a relatively easy task for y>3, so we are done" care to give some details ? $\endgroup$
    – mercio
    Aug 27, 2015 at 16:56

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