Talk:Ramanujan–Nagell equation

The article says that equation x² + D = A Bⁿ with A = 1, B = 2 has at most two solutions except in the case D = 7 already mentioned. But if D = 28, then x² + D = 2ⁿ has at least 3 solutions: (x, n) = (2, 5), (6, 6), (10, 7). So the statement in this article is false, or did I miss something? -- Spencer m67 (talk) 08:32, 22 October 2015 (UTC)[reply]

@Spencer m67: In fact there are five solutions for D=28: (x,n) = (2,5), (6,6), (10,7), (22,9), (362,17). The reason is that 28=4×7, so "x² + 28 = 2ⁿ" is equivalent to "(2y)² + 4×7 = 2^m+2"; divide by 4 to get "y² + 7 = 2^m".

So how does this fit in with what the article says? Well, the reference is to Saradha, N.; Srinivasan, Anitha (2008). "Generalized Lebesgue–Ramanujan–Nagell equations". There's a copy of that paper here. And it does indeed state that, "In 1960, Apéry considered (2.1) with λ = 4 and k = 2. He showed that x² + D = 2ⁿ⁺², D≠7, has at most two solutions." And (incorrectly for the statement as written) there is no requirement by that paper for D to be coprime with 4.

At this stage, I would turn then to the original source: Roger Apéry, Sur une équation diophantienne (1960). Actually Apery wrote two short papers with that exact same name. Sadly I haven't been able to source either of them. If you can look those up, I imagine that the terms should be more strictly defined.

-Stelio (talk) 15:07, 27 April 2020 (UTC)[reply]

(Redirected from Ramanujan–Skolem theorem)?

There is no Skolem in the article. МетаСкептик12 (talk) 09:59, 4 June 2019 (UTC)[reply]