In this post I present my variations of the problem involving Nagell-Ljunggren equation, that is explained in pages 10 and 11 of Highlights in the Research Work of T. N. Shorey by R. Tijdeman, from Diophantine Equations, (Editor) N. Saradha, Tata Institute of Fundamental Research, Narosa Publishing House (2008).

Consider for integers $x>1$, $1<y<z$ and integers $1<p$ and $2<n$ the equations $$y^p+z^p=(y^{p-1}+z^{p-1})\cdot\frac{x^n-1}{x-1},\tag{1}$$ and $$2yz=(y+z)\cdot\frac{x^n-1}{x-1}.\tag{2}$$

Conjecture 1. Consider the equation $(2)$ over integers $x>1$, $1<y<z$ and $2<n$, then there exists a positive integer $n_0\geq 8$ such that $\forall n>n_0$ this equation has no solutions.

Conjecture 2. We consider the equation $(1)$ over integers $x>1$, $1<y<z$, $1<p$, and $2<n$ then there exists a positive constant $C$ such that the equation has no solutions whenever $n\cdot p>C$.

Both equations ask for Lehmer means $L_p(y,z)$ with all the digits equal to $1$ in the corresponding $x-$adic expansion


Wikipedia has an article for Lehmer mean, and their relationship to the harmonic mean.

Question. I would like to know if it is possible (what work can be done about it) to prove or disprove Conjecture 1 and/or Conjecture 2. Many thanks.

I don't know if preveious problems are in the literature, I hope that these problems have a good mathematical content and are interesting for you. I'm asking about what work can be done about the Conjecture 1 or well about the Conjecture 2. Finally I add two solutions from the following examples.

Example 1. One has that $(x,y,z;n,p)=(16,225,300;3,3)$ solves $(1)$ because $$\frac{225^3+300^3}{225^2+300^2}=272=1+1\cdot 16+1\cdot 16^2.$$

Example 2. Similarly $(x,y,z;n)=(11,91,247;3)$ solves $(2)$ because $$\frac{2}{\frac{1}{91}+\frac{1}{247}}=133=1+1\cdot 11+1\cdot 11^2.$$


1 Answer 1


Conjecture 1 does not hold as for any even $n$, (2) has a solution: $$(x,y,z)=(2,\frac23(2^n-1),2(2^n-1))$$

Likewise, Conjecture 2 fails as for $p=2$ and any even $n$, (1) has a solution: $$(x,y,z)=(4,\frac2{15}(4^n-1),\frac25(4^n-1))$$

  • $\begingroup$ Many thanks, what good counterexamples! I'm accepting your answer. Feel free to study the (veracity of conjectures) case $n$ odd for these equations, if it is interesting for your work. $\endgroup$
    – user142929
    Sep 22, 2021 at 15:09
  • $\begingroup$ While my conjectures are both false, your examples of families of solutions provide more value to the equations/problems $(1)$ and $(2)$. $\endgroup$
    – user142929
    Sep 22, 2021 at 15:15

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