I've read in a number of places that, building on previous work of T. Nagell, W. Ljunggren proved in 1 that the Diophantine equation

$$\frac{x^{n}-1}{x-1} = y^{2}$$

doesn't admit solutions in integers $x>1, y>1, n>2$, except when $n=4, x=7$ and $n=5, x=3$.

Since neither I have been able to spot a copy of Ljunggren's paper online (and even if I had made every effort to acquire an electronic copy of it, it all would have been to no avail because I don't read Norwegian) nor I have access to Ribenboim's edition of the Collected Papers of these authors, would anybody here be so kind as to explain in some detail how it was that Ljunggren proved this notable result? Unfortunately, Nagell doesn't say much about Ljunggren's demonstration in his review of 1...

Please, let me thank you in advance for your learned replies!


[1] W. Ljunggren, Some theorems on indeterminate equations of the form $\frac{x^{n}-1}{x-1} = y^{q}$ (In Norwegian). Norsk. Mat. Tidsskr. 25 (1943), pp. 17--20.

  • $\begingroup$ Try Ribenboim's book on the Catalan conjecture, if you can get a hold of that. $\endgroup$ May 15, 2015 at 7:29
  • 1
    $\begingroup$ Methinks it would have to be an international interlibrary loan, if such a thing is possible... $\endgroup$ May 15, 2015 at 7:36
  • 1
    $\begingroup$ I have access to the Collected Works in question and I am willing to scan the paper for you; you must however figure out the Norwegian part... ;-) $\endgroup$
    – knsam
    May 15, 2015 at 8:17
  • 1
    $\begingroup$ Related: mathoverflow.net/questions/58697/… $\endgroup$
    – C.S.
    May 15, 2015 at 8:31
  • 1
    $\begingroup$ According to the Zentralblatt review of Ljunggren's article, the method is to reduce to a (Pell-)Fermat equation. zbmath.org/?q=an:0028.00901 $\endgroup$ May 15, 2015 at 13:45

2 Answers 2


I went to my office today and scanned the Ljunggren's paper that OP asked for. I provide some bibliographical information first:

Wilhelm Ljunggren, Noen setninger om ubestemte likninger av formen $\frac{x^n - 1}{x-1}=y^q$, Norsk. Mat. Tidsskrift, 25 (1943), 17 -- 20 ( = Collected Papers of W. Ljunggren edited by P. Ribenboim, Volume 1, #14, p. 363 -- 366).

I put the paper in the web space I have. EDIT: Here is another scan (Internet Archive)

Since the paper is so small and I have been thinking about Brahmagupta-Pell equations recently, I decided to translate the part of the paper relevant to the OP. This was surprisingly easy (Two Ronnies and some knowledge of German finally paid off!).

What follows is the functional translation of nearly half the paper; it goes without saying that what follows is due to Ljunggren while the errors in translation rest with me.

We owe to T. Nagell [N] the following theorem:

The diophantine equation $$ \frac{x^n - 1}{x - 1} = y^2 \qquad (n > 2)\tag{1}$$ has only a finite number of solutions in integers $x$ and $y$. The possible solutions are found at the end of this work. Specifically, (1) is impossible with $|x| > 1$ if $n$ does not have one of the following four forms: $1^\circ. n = 4$; $2^\circ. n = p$; $3^\circ. n = p^2$; $4^\circ. n = p^2 q$ with $p$ and $q$ distinct primes, $q \equiv 1 \bmod{24 p}$ and $q < p^2 - 3$.

I will first show how one can deduce the following theorem about (1) using the theorems of K. Mahler [M]:

Theorem 1. The diophantine equation (1) is impossible with $|x| > 1$ in all cases except $n = 4, x = 7$ and $n = 5, x = 3$.

We need the following theorem of K. Mahler:

Let $D$ be a natural number that is not a perfect square. Furthermore, let $A$ be a square-free divisor of $2D$ ($A \neq 0$). Then, the solutions of the equation $$ x^2 - Dy^2 = A\tag{2}$$ are given by the following formula where $m$ an odd positive integer: \begin{align*} \pm x_m &= \frac{(u+v\sqrt{D})^m+(u-v\sqrt{D})^m}{2|A|^{\frac{m-1}{2}}}\\ \pm y_m &= \frac{(u+v\sqrt{D})^m-(u-v\sqrt{D})^m}{2|A|^{\frac{m-1}{2}}\sqrt{D}} \end{align*}

Here $u$ and $v$ are natural numbers that satisfy (2) with $|y_{\min}| = v$.

Furthermore, if we let $\mathfrak{n}(D, A)$ denote the set of odd integers $m$ such that the pair $(x_m, y_m)$ solves (2) and the set of prime divisors of $y_m$ is contained in those of $D$, then, either $\mathfrak{n}(D, A)$ is empty or $\mathfrak{n}(D, A) = \{1\}$ or $\mathfrak{n}(D, A) = \{1, 3\}$.

According to Nagell, it is enough to look at odd $n$. We assume now that $x > 1$. The equation (1) can be written in the form: $$ [(x - 1)y]^2 - x(x - 1) \left[x^{\frac{n-1}{2}}\right]^2 = - (x - 1).$$ Here evidently Mahler’s theorem applies; we have $D = x(x - 1)$ and $A = - (x - 1)$ and further we have that $u = x - 1$ and $v = 1$. This gives $$x^{\frac{n-1}{2}} = 1 \qquad \text{(impossible)}$$ or $$ x^{\frac{n-1}{2}} = 4x - 3.$$ For $x > 1$, it follows from the last equation that $x = 3$ with $n = 5$.

Suppose secondly that $x < - 1$. We put $x_1 = -x$ with $x_1 > 1$. The equation (1) is of the form

\begin{align*} \frac{x_1^n + 1}{x_1 + 1} &= y^2 \qquad\text{ or } \\ [(x_1+1)y]^2 - x_1(x_1+1)&\left[x_1^\frac{n-1}{2}\right]^2 = x_1 + 1 \end{align*}

Here we have $x_1^{\frac{n-1}{2}} = 1$ (impossible) or $x_1^{\frac{n-1}{2}} = 4x_1 + 3$ which is also impossible for $x_1 > 1$.


References. [N] = Nagell: (1) in the above linked pdf, [M] = K. Mahler: (1) in the above linked pdf.

[M] K. Mahler: Über den grössten Primteiler spezieller Polynome zweiten Grades, Arch. Math. Naturvidensk., 41 (1935), pp. 3-26
[N] Nagell, T., Sur l'équation indéterminée $\frac{x^n−1}{x−1} = y^q$, Norsk. Mat. Forenings Skrifter, Serie I, nr. 3, (1921).


  • The paper of K. Mahler is available from here. Incidentally, this archive of K. Mahler’s collected papers deserves to be more well known.

  • There are some simple details that need to be put into the proof but I will leave that to you. Feel free to ask if necessary and I can add some calculations/clarifications. :-) I must note that Ljunggren’s and Mahler’s papers are both so clearly written that you must read them entirely!

  • 1
    $\begingroup$ Wow! Thanks a bunch for taking the time to scan the paper and leave your translation of it here on MO! Incidentally, I had just read about the digitization of K. Mahler's papers in the latest issue of the Notices of the AMS: ams.org/notices/201505/rnoti-p526.pdf $\endgroup$ May 17, 2015 at 16:01
  • 1
    $\begingroup$ Thank you for this! Very interesting result(s). Unfortunately, the linked paper appears to be unavailable, and there is no direct reference for the Mahler paper. Could you please provide a title or link? $\endgroup$ Mar 18, 2016 at 21:57
  • 4
    $\begingroup$ @knsam: Where did those scans go? :( $\endgroup$ May 20, 2016 at 18:21

K. Mahler's paper (in German) was this one: Über den grössten Primteiler spezieller Polynome zweiten Grades, Archiv. Math. og. Naturv. 41 (1935), nr. 6, 3-26; Zblatt 13, 389. This is the corrected link.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.