Timeline for Euler Bricks in High Dimensions

6 events

when toggle format	what		by	license	comment
Dec 6, 2016 at 18:38	comment	added	Will Sawin		I was hoping that there would be a local obstruction at some prime, making this problem easy. Unfortunately there is not - for each $p$ there are bricks where all diagonals are $p$-adic integers. For $p>2$ one can take the side lengths to be $1,p,p^2,\dots$, and then every diagonal will be of the form $p^n \sqrt{ 1 + m p}$ for integers $n,m$ and so will be a $p$-adic integer. For $p=2$ the same thing works but with powers of $4$.
Dec 1, 2016 at 18:30	comment	added	Pace Nielsen		@YemonChoi Yes, so this problem should be easier to solve (given it is likely that there is no Euler brick).
Dec 1, 2016 at 18:29	comment	added	Pace Nielsen		@EmilJeřábek That claim does throw some questionability on their work, doesn't it. :-)
Dec 1, 2016 at 16:59	comment	added	Emil Jeřábek		According to erpublication.org/admin/vol_issue1/upload%20Image/… , there are no 4-dimensional Euler bricks. (Also, the ABC conjecture is true.)
Dec 1, 2016 at 16:35	comment	added	Yemon Choi		Aren't the faces of an Euler 4-brick themselves Euler 3-bricks? (I may be being stupid here)
Dec 1, 2016 at 16:29	history	asked	Pace Nielsen	CC BY-SA 3.0