Timeline for A modification of the Ljunggren-Nagell equation

9 events

when toggle format	what		by	license	comment
Nov 8, 2017 at 19:47	vote	accept	Pace Nielsen
Nov 8, 2017 at 17:28	comment	added	Pace Nielsen		Here is an even easier solution in the latter case: If $k$ is odd, we cannot have $(a^k-1)/(a-1)$ twice a square. But if $k$ is even, then $a^k+1$ cannot be a square, since integer squares are more than one apart (for $a>1$).
Nov 8, 2017 at 16:24	comment	added	Fedor Petrov		$a^k+1=c^2$ was done long before Mihailescu in elementary and short way, see E. Z. Chein, Proc. AMS 56 (1976), pp. 83-84.
Nov 8, 2017 at 16:15	comment	added	Fedor Petrov		If $a$ is odd, than $m$ and $(a^m-1)/(a-1)=1+a+\dots+a^{m-1}$ have the the same parity, we do not need theorems for seeing this.
Nov 8, 2017 at 14:59	history	edited	Max Alekseyev	CC BY-SA 3.0	deleted 11 characters in body
Nov 8, 2017 at 14:57	comment	added	Max Alekseyev		@PaceNielsen: Mihailescu's theorem may be an overkill here, but it's the most straightforward approach. Alternatively, one can get the same result by factoring $a^k=(c-1)(c+1)$.
Nov 8, 2017 at 14:53	history	edited	Max Alekseyev	CC BY-SA 3.0	links
Nov 8, 2017 at 14:51	comment	added	Pace Nielsen		This is really nice, although it pulls out the big guns by appealing to Mihailescu's theorem! In my work I also need the cases when the right-hand side is $3b^2$ or $6b^2$. It looks like the $6b^2$ case should be similar, but I'll give them a try using your ideas.
Nov 8, 2017 at 14:43	history	answered	Max Alekseyev	CC BY-SA 3.0