# Looking for a general and complete solution to the Diophantine $a^2 -2b^4 = -1$

The general and probably complete solution to $a^2+(b-1)^2 = (b)^2 is (2v+1)^2+(2v(v+1))^2 = (v^2+(v+1)^2)^2$ We get the triples $(a,b-1,b) = (3,4,5), (5,12,13), (7,24,5^2),...,(41,840,29^2),...,(239,28560,13^4),...$

If we only want those b that are also squares, we must solve the Pell like equation $a^2 -2b^2 =-1$ whose $(a,b)$ solutions beginning with the trivial $(1,1)$ are $(7,5), (41,29), (239,13^2),...$ given by the recurrent relation :$a(n+1) = 3a(n) + 4b(n); b(n+1) = 2a(n) + 3b(n)$. What is the general solution, if known, to $a^2 -2b^4 = -1$?

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The general theory says it will have only finitely many solutions. – Gerry Myerson Jan 10 '13 at 17:33
Why did you edit me , mister Gerry Myerson, so shortly after i published my question? What are you doing without my consentment for such a insufficient answer? – user55514 Jan 10 '13 at 21:30
I edited you because you don't know how to spell "diophantine". If you don't like being edited without consent, you've come to the wrong website. And that's Doctor Myerson to you. – Gerry Myerson Jan 11 '13 at 1:48
Always be scared by strong personalities, Mister Myerson. I am not suffering though from a lack of knowledge but from some kind of dislexia when i write fast; sorry Doctor. – user55514 Jan 11 '13 at 4:00
Apology accepted. Anyway, you can see the edit history by clicking on the time/date/whatever next to the word, "edited"; you'll find someone else did most of the editing, and it was done to improve the formatting (to make a^2 look like $a^2$, for example). Regular users are encouraged to do this type of editing. – Gerry Myerson Jan 11 '13 at 23:40

Mordell, Diophantine Equations, page 271, writes, "...for the special case $$y^2=2x^4-1$$ it has been known for two centuries that solutions are given by $(x,y)=(1,1)$ and $(13,239)$. It was proved by Ljunggren that these are the only positive integer solutions. The proof is exceedingly complicated." Mordell gives the citation, Zur Theorie der Gleichung $x^2+1=Dy^4$, Avh. Norske Vid. Akad. Oslo No. 5 1 (1942).

Guy, Unsolved Problems In Number Theory, 3rd edition, Problem D6 is "An elementary solution of $x^2=2y^4-1$." Guy cites Steiner and Tzanakis, Simplifying the solution of Ljunggren's equation $X^2+1=2Y^4$, J Number Theory 37 (1991) 123-132, and writes, "Whether Steiner & Tzanakis have simplified the solution may be a matter of taste; they use the theory of linear forms in logarithms of algebraic numbers."

Guy also cites a proof by Chen Jian-Hua, which he calls "unconventional." The bibliographic details are, A new solution of the Diophantine equation $X^2+1=2Y^4$, J Number Theory 48 (1994) 62-74 and A note on the Diophantine equation $x^2+1=dy^4$, Abh. Math. Sem. Univ. Hamburg 64 (1994) 1-10.

I'd suggest also looking at Wikipedia on Ljunggren.

And there's more: Konstantinos A. Draziotis, The Ljunggren equation revisited, Colloq. Math. 109 (2007), no. 1, 9–11.

Michael A. Bennett, Irrationality via the hypergeometric method, Diophantine analysis and related fields—DARF 2007/2008, 7–18, AIP Conf. Proc., 976, Amer. Inst. Phys., Melville, NY, 2008.

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Thanks for the good information, Gerry. I did not know it. We can then deduce almost immediately (am i wrong?) hat the equation a^4 -2b^4 = -1 has no integer non trivial solutions since 239 is not a square. I would be interested in knowing all the solutions, if there are, to a^2 - 2b^2n = -1 or alternatively to know all the solutions to a^2n - 2b^2 = -1 with the intention of proving a^2n - 2b^2n = -1 has no non trivial solutions. I believe, but could be wrong, proving a^(2n+1) - 2b^(2n+1) = -1 has no non trivial solutions might be harder. – user55514 Jan 15 '13 at 18:01
Now here's where aesthetics becomes mathematics. You write, b^2n. How is one to know whether you mean $b^{2n}$ or $b^2n$? If you plan to use this site regularly, please do yourself and others a favor, and learn a little about how to use TeX to format your mathematics here. – Gerry Myerson Jan 15 '13 at 23:35
You are half right, i am half wrong here. I should have put parenthesis; or brackets, i am not sure of the sense you give north by northwest (seen from where i´m living and belong to, this old and southerener Europe we love) to parenthesis: b^(2n). Furthermore i should have written b^(2*n) and 2*b^(2*n) to avoid any confusion (2b and 2n could be variables distinct from b and from n), but i was too lazy. I am an amateur and never used TeX, but i´ll try next time. – user55514 Jan 16 '13 at 9:54
Yet if you edit somebody to change the meaning of what he/she wrote, this seems to me less acceptable and near to censorship, even though the intention was good, to put it better and with a better style. But i think all is OK now. – user55514 Jan 16 '13 at 9:55
for(v=1,10^10,if(issquare((v^2+(v+1)^2),&n)&ispower(n),print([v,n]))) [119, 169] time = 3h, 7min, 887 ms. This maldroit very short program in Pari gp – user55514 Jan 16 '13 at 14:12