Skip to main content
11 events
when toggle format what by license comment
Mar 25, 2010 at 16:33 vote accept Aeryk
Mar 24, 2010 at 21:01 history edited darij grinberg CC BY-SA 2.5
added 1 characters in body
Mar 24, 2010 at 20:14 comment added darij grinberg Thanks, I found the flaw in my proof. It is now fixed to work for the case when $f$ has leading coefficient $1$. The assumptions on $g$ are pretty generic.
Mar 24, 2010 at 20:13 history edited darij grinberg CC BY-SA 2.5
deleted 1488 characters in body; deleted 14 characters in body; added 16 characters in body; added 4 characters in body
Mar 24, 2010 at 19:41 history edited darij grinberg CC BY-SA 2.5
added 81 characters in body
Mar 24, 2010 at 19:35 history edited darij grinberg CC BY-SA 2.5
added 863 characters in body; added 181 characters in body
Mar 24, 2010 at 19:22 history edited darij grinberg CC BY-SA 2.5
added 146 characters in body
Mar 24, 2010 at 19:15 history edited darij grinberg CC BY-SA 2.5
added 2193 characters in body; added 32 characters in body
Mar 24, 2010 at 18:37 comment added Sergei Ivanov This argument works for $f(x)=x+const$, the case $f(x)=ax+b$ where $a\ne 1$ is different.
Mar 24, 2010 at 18:33 comment added Aeryk This seems to do the trick for my case: Thanks! I don't know if I'm convinced this will work for "uglier" polynomials in $Z[x]$, like say $f(x)=x+7$ and $g(x)=x^2+3x+9$ ...
Mar 24, 2010 at 18:19 history answered darij grinberg CC BY-SA 2.5