Timeline for Elementary description to count of perfect squares - II

11 events

when toggle format	what		by	license	comment
Dec 28, 2017 at 9:08	vote	accept	Turbo
Dec 28, 2017 at 9:08	vote	accept	Turbo
Dec 28, 2017 at 9:08
Dec 28, 2017 at 9:08	vote	accept	Turbo
Dec 28, 2017 at 9:08
Dec 28, 2017 at 9:08	vote	accept	Turbo
Dec 28, 2017 at 9:08
Dec 28, 2017 at 9:08	vote	accept	Turbo
Dec 28, 2017 at 9:08
Dec 28, 2017 at 8:51	answer	added	Fedor Petrov		timeline score: 1
Dec 28, 2017 at 8:25	history	edited	Turbo	CC BY-SA 3.0	deleted 35 characters in body
Dec 28, 2017 at 7:57	history	edited	Turbo	CC BY-SA 3.0	added 92 characters in body
Dec 28, 2017 at 7:33	comment	added	Turbo		@GHfromMO If $t=b^2+1$ then at $x=0$ and $x=1$ we have perfect squares and so is $g(b^2+1)=1$. Conversely if $g(a)=1$ then does $a=b^2+1$ hold? By this I mean can there be other $x$ and $x+1$ for some $t$ such that both $x(t-x)$ and $(x+1)(t-(x+1))$ are perfect squares? I think is impossible.
Dec 27, 2017 at 21:15	comment	added	GH from MO		This question is very similar to the question on gaps between sums of two squares, which has been studied extensively. See the recent arXiv preprint arxiv.org/pdf/1712.07243.pdf, and see also my response to your previous question mathoverflow.net/questions/289387/… to make the connection.
Dec 27, 2017 at 19:05	history	asked	Turbo	CC BY-SA 3.0