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Mar 20, 2019 at 8:48 answer added Daniil Rudenko timeline score: 5
Aug 20, 2016 at 12:09 answer added Clément de Seguins Pazzis timeline score: 6
Aug 20, 2016 at 11:53 vote accept Clément de Seguins Pazzis
Aug 20, 2016 at 11:29 comment added Andreas Rüdinger If you generalize the element 2 in the diagonal to $c$, you get $(x^2-abx+a^2+b^2-D_1)(x^2-abx+a^2+b^2-D_2)$ with $D_{1,2}=c^2 \pm (2-c)ab$.
Aug 20, 2016 at 11:09 answer added Colin McQuillan timeline score: 12
S Aug 20, 2016 at 9:57 history suggested Rodrigo de Azevedo CC BY-SA 3.0
Added tag to this question, edited the title
Aug 20, 2016 at 9:49 review Suggested edits
S Aug 20, 2016 at 9:57
Aug 20, 2016 at 9:45 history asked Clément de Seguins Pazzis CC BY-SA 3.0