On a result attributed to W. Ljunggren and T. Nagell

Question

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!

References

[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.

Answer 1

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*}

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

  <p>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 $\mathfrak{n}(D, A) = \{1, 3\}$.  </p>
</blockquote>

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).

Remarks.

• 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!

Answer 2

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.