Original proof of Ljunggren's equation

The equation $$x^2=2y^4-1$$ was studied and solved by Ljunggren, who showed that $$(1,1)$$ and $$(293,13)$$ are the only integer solutions. However, his proof was very difficult and L. J. Mordell thought there must be an easier proof. Nowadays, this equation can be rewritten in Weierstrass form and solved by an algorithm for elliptic curves. However, I was wondering what the original proof really was. Here are some of the references on Wikipedia that I can't seem to be able to open.

• Draziotis, Konstantinos A. (2007), "The Ljunggren equation revisited", Colloquium Mathematicum 109 (1): 9–11, doi:10.4064/cm109-1-2, MR 2308822.

• Ljunggren, Wilhelm (1942), "Zur Theorie der Gleichung x2 + 1 = Dy4", Avh. Norske Vid. Akad. Oslo. I. 1942 (5): 27, MR 0016375.

I have found a proof without the use of elliptic curves by Ray Steiner and Nikos Tzanakis, but the proof also requires some computational power, which Ljunggren didn't have access to. So if anyone knows a proof to this, it would be appreciated if you post it.

• The second solution should be (x,y) = (239, 13), not what your wrote. – jbuddenh Nov 16 '15 at 23:14
• The original proof uses a very delicate variant of Skolem's $p$-adic method. It can probably be recast as an application of Chabauty-type arguments, for what that's worth. – Mike Bennett Nov 17 '15 at 1:55
• Just trying to find some examples, I find the property where we get at least $z^2$ instead of $y^4$ (being squares) I find, that beginning with $x=7$ and the iterating $x=\text{floor} (\sqrt8 \cdot x))+3x+1$ if seem to find immediately all that $x$ which admit $z^2$ in the formula. (being $x=[7,41,239, ...]$) That only $x=7$ and $x=239$ give $z^2=y^4$ is then directly visible with numbers $x$ up to hundred digits and only some milliseconds to compute... I never related such a recursion with those type of questions. Amazing! – Gottfried Helms Nov 18 '15 at 7:28
• The mentioned paper by Draziotis is available in the site of the journal (here). – Pedro 2 days ago
• Thank you very much, @Pedro! – Gottfried Helms 2 days ago