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 14h comment In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ Thank you for your most generous replies 18h comment In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ Dear Professor - Thank you for your kind answers. As you probably have guessed, I am a fervent but uneducated self-studier. I started in my late 60's and wish I was more proficient. If it is not too much of an imposition, maybe you would please add a remark with an intuitive explanation that goes beyond the "brute force" approach of cubing both sides of $a=1+b \lambda$ to get $a^3\equiv \pm 1\pmod 9$ in my previous question but does not entail ramification. With regards, 18h accepted In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ 1d comment In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ @AndréNicolas Sorry to trouble you. I was thinking about your comment. You could have $1+1+1\equiv 0\pmod {1-\omega}$, but $1-\omega \nmid a,b,c$ 1d comment In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ @HagenvonEitzen Please forgive me, but if $a\equiv b\equiv c \equiv 1\pmod{1-\omega}$ does that not imply that none of $a,b,c$ are divisible by $1-\omega$. Thanks 1d comment In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ @AndréNicolas Thanks. I was wondering if I could do that, since in $\mathbb{Z}/3\mathbb{Z}$ $1,2,3$ are their own cubes. But the hint in the text of the question suggests to use the second problem result I mentioned above. With regards, 1d revised In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ added 32 characters in body 1d asked In $\mathbb{Z}[\omega]$, if $a^3+b^3+c^3=0$ then $1-\omega$ divides at least one of $a,b,c$ 2d accepted Proving an equivalence relation, $a^3\equiv\pmod 9$, in $\mathbb{Z}[\omega]$ 2d comment Proving an equivalence relation, $a^3\equiv\pmod 9$, in $\mathbb{Z}[\omega]$ Does that change the remark that the residue class is then $\bar{b^3} -\bar b =0$ 2d comment Proving an equivalence relation, $a^3\equiv\pmod 9$, in $\mathbb{Z}[\omega]$ My apologies, but I think the expression is $(b^3 - \omega^{2} b)$ Thanks. With regards, Apr 25 asked Proving an equivalence relation, $a^3\equiv\pmod 9$, in $\mathbb{Z}[\omega]$ Apr 25 comment Show $a+bi \equiv 0,1 \pmod{1+i}$ Thanks for this. I was able to apply the concept to the next problem to show $a+b\omega \equiv -1,0,1 \pmod{1-\omega}$. With regards, Apr 24 accepted Show $a+bi \equiv 0,1 \pmod{1+i}$ Apr 24 asked Show $a+bi \equiv 0,1 \pmod{1+i}$ Mar 5 awarded Popular Question Mar 5 awarded Famous Question Mar 3 accepted Calculation of the limit of the difference of binomial coefficients Mar 3 comment Calculation of the limit of the difference of binomial coefficients Thanks, Joriki. The mistake and confusion are all mine. I'll start over. Previously I had put the binomial coefficients into factorial form and combined them over a common denominator. I'll go back and look at that with your original answer in mind. If I'm still stuck, maybe you wouldn't mind if I asked for further help. Mar 3 asked Calculation of the limit of the difference of binomial coefficients