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Sep 24, 2020 at 17:48 vote accept tomos
Sep 23, 2020 at 9:35 vote accept tomos
Sep 24, 2020 at 17:48
Sep 20, 2020 at 18:45 comment added Servaes Also, for the differentiability condition for Hensel's lemma to hold, it is necessary that $p$ doesn't divide $n$, and that $p$ doesn't divide $R$.
Sep 20, 2020 at 18:37 comment added Servaes So when you say that the number $N(n,p^{\alpha},R)$ of solutions to the congruence $$x^n\equiv R\pmod{p^\alpha},$$ satisfies $N(n,p^{\alpha},R)\ll_n1$, you mean to say that $$N(n,p^{\alpha},R)\leq C(n),$$ for some function depending only on $n$?
Sep 20, 2020 at 18:34 answer added Servaes timeline score: -1
Sep 20, 2020 at 18:25 comment added tomos sorry, i mean: If in context we have variables $v_1,...,v_n,u_1,...,u_m$ and functions $f,g$ with $g$ positive, then $f(v_1,...,v_n)\ll _{u_1,...,u_m}g(v_1,...,v_n)$ means $|f(v_1,...,v_n)|\leq C_{u_1,...,u_m}g(v_1,...,v_n)$ for some positive constant $C$ independent of $v_1,...,v_n$ but dependent possibly on $u_1,...,u_m$
Sep 20, 2020 at 18:20 comment added Servaes I still don't understand what you mean by that symbol, or that sentence.
Sep 20, 2020 at 18:19 comment added tomos independent of $p,\alpha $
Sep 20, 2020 at 18:16 comment added Servaes What do you mean by "...has $\ll _n1$ solutions..."?
Sep 20, 2020 at 17:55 history asked tomos CC BY-SA 4.0