1
$\begingroup$

I am looking for the following paper by Ljunggren, Wilhelm: "Zur Theorie der Gleichung $x^2 + 1 = Dy^4$", Avh. Norske Vid. Akad. Oslo. I., 1942 (5): 27

The main result of this paper which I am interested in is that the solutions in positive integers of the equation $x^2 + 1 = 2y^4$ are $(1,1), (239,13)$, so a reference summarizing Ljunggren's proof would be welcome as well.

$\endgroup$
5
  • 1
    $\begingroup$ This comment says "The original proof uses a very delicate variant of Skolem's p-adic method". $\endgroup$
    – Wojowu
    Jun 11, 2020 at 18:29
  • $\begingroup$ Also user whose name resembles Jose Stgo. has a post on MathOverflow which summarizes a related paper and may have a link to your paper or near to your paper of interest. So search MathOverflow for It. Gerhard "You're In The Right Place". Paseman, 2020.06.11. $\endgroup$ Jun 11, 2020 at 18:35
  • $\begingroup$ Thanks, hopefully I'll find it. $\endgroup$
    – Random
    Jun 11, 2020 at 19:32
  • $\begingroup$ An idea that has worked for me is to ask for a copy to someone who has recently cited the article in question. Many people (me included) keep scanned copies of articles which are impossible to find online. $\endgroup$
    – efs
    Jun 11, 2020 at 23:21
  • $\begingroup$ Thanks, I'll try that! $\endgroup$
    – Random
    Jun 12, 2020 at 14:04

1 Answer 1

1
$\begingroup$

Two papers that simplify Ljunggren's proof:

An Elementary Proof for Ljunggren Equation (2017)

Simplifying the Solution of Ljunggren's Equation (1991)

$\endgroup$
2
  • 1
    $\begingroup$ I believe that the first paper has an error, specifically the claim at the bottom of page 2 that if (c,b,13k) is a Pythagorean triple then b/c is either 5/12 or 12/5. In any case, I am aware of other proofs but I am interested specifically in the orginal proof. I should probably have emphasized that in the original question. $\endgroup$
    – Random
    Jun 11, 2020 at 19:29
  • $\begingroup$ $(16,63,65)$, $b/c=16/63$. $\endgroup$ Jun 11, 2020 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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