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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