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11h
asked How does one prove that $\mathbb{Z}[\sqrt{-41}]$ is a unique factorization domain.
12h
comment Can we find the 18 imaginary quadratic ffields with class number 2 algorithmically?
@BillDubuque it is indeed; don't be surprised if I revise this question or post another one asking for more details
12h
revised Can we find the 18 imaginary quadratic ffields with class number 2 algorithmically?
added 356 characters in body; edited title
12h
revised Can we find the 18 imaginary quadratic ffields with class number 2 algorithmically?
added 356 characters in body; edited title
12h
asked Can we find the 18 imaginary quadratic ffields with class number 2 algorithmically?
2d
comment Certain local inequality for volume and surface measures
Locally isnt a smooth curve just a line segment? Since you are integrating near $x$?
2d
revised Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
Introductory text
May
23
comment Prove: $\gcd(n^a-1,n^b-1)=n^{\gcd(a,b)}-1$
You raise an interesting point. My argument is simple but unfortunatetly does not prove the result. I will ask a question on this site
May
23
comment Prove: $\gcd(n^a-1,n^b-1)=n^{\gcd(a,b)}-1$
I am not asking you for permission here. Then use cyclotomic polynomials if you are dissatisfied. I think we have shown that $\mathbb{Q}[x]/(x^a-1,x^b-1)$ is a very interesting ring.
May
23
comment Prove: $\gcd(n^a-1,n^b-1)=n^{\gcd(a,b)}-1$
Uh... If I hate zero divisors I can remove factors of $(x-1)$ and extend by $1+x+...+x^{a-1}$. And multiply it back each time. That way only adjoin primitive roots of unity.
May
23
comment Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$
Sort of like the Fibonacci sequence!
May
23
comment Prove: $\gcd(n^a-1,n^b-1)=n^{\gcd(a,b)}-1$
Okay its just a polynomial ring with a bunch of zero divisors. Like $\mathbb{Z}/8\mathbb{Z}$.
May
23
answered Prove: $\gcd(n^a-1,n^b-1)=n^{\gcd(a,b)}-1$
May
22
revised Enumerating Bianchi circles
added 375 characters in body
May
22
answered Enumerating Bianchi circles
May
22
comment Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
math.stackexchange.com/questions/957224/…
May
22
comment Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
See Master's Thesis of Abebe Simachew (Ethiopia / Boreaux) A Survey of Euclidean Number Fields
May
22
revised Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
demonstrate no longer true for Zsqrt19
May
22
revised how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$
added 522 characters in body
May
22
asked how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$