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20h
comment Visual proof of $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$?
the only visualization I can think of is in 4 dimensions, finishing Robert Israel's proof. Is that okay? I could try to project the darn thing myself.
21h
comment How does one prove that $\mathbb{Z}\left[\frac{1 + \sqrt{-43}}{2}\right]$ is a unique factorization domain.
@AlonsodelArte Do you mean $\mathbb{Z}[\frac{1 + \sqrt{-41}}{2}]$ or $\mathbb{Z}[\sqrt{-41}]$ ? I meant $\sqrt{-43}$.
1d
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
May
24
comment Certain local inequality for volume and surface measures
Locally isnt a smooth curve just a line segment? Since you are integrating near $x$?
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
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
21
comment Visualizing Euclidean Algorithm in $\mathbb{Q}(\sqrt{-7})$ and $\mathbb{Q}(\sqrt{-11})$ with Convex Geometry
@mercio over $\mathbb{Z}$ we say $a = bq + r$ with $0 < r < b$. The interval $[0,b]$ is the "circle" with diameter $b$. We could just as easily say $|r| < \tfrac{1}{2}b$ and get a slightly different Euclidean algorithm.
May
18
comment Prove that $p\mid a^2+b^2\,\Rightarrow\, p\equiv 1\pmod{\! 4}$
how interesting! if $\mathrm{ord}_p (ab^{-1}) = 4$ then it's as if $ab^{-1} = \sqrt{-1}$
May
18
comment Bing's House and homotopies
Just tiny changes in the Spanish. It could just be the dialect of where I am from.
May
18
comment Prove that $p\mid a^2+b^2\,\Rightarrow\, p\equiv 1\pmod{\! 4}$
typically the result is $p \equiv 1 \mod 4 \leftrightarrow p = a^2 + b^2$ but if you replace the right side with $p | a^2 + b^2 $ there could be some more examples. The solutions seem to indicate not.
May
17
comment Enumerating Bianchi circles
This is also $K = \mathbb{Q}[\sqrt{-7}]$, right?
May
17
comment Enumerating Bianchi circles
I wonder if the naïve approach works, $\tfrac{a}{b}, \tfrac{c}{d} $ are neighbors iff $ad - bc = 1$. Unfortunately we have no analogue of "neighbors", but I mean you try something... The group structure of Bianchi groups seems to be open but I did find [1] and [2].
May
17
comment Enumerating Bianchi circles
Stange leaves many artifacts of her algorithm in the image gallary - especially those rhombus shapes. Sage implicitly uses Python, so it makes sense to define various ways to iterate over the Bianchi group $SL(2, \mathcal{O}_K)$.
May
11
comment proof that $\frac{x^p - 1}{x-1} = 1 + x + \dots + x^{p-1}$ is irreducible
@GregoryGrant Eisenstein's criterion is not a total free lunch as you have to invoke Gauss' Lemma (also here) or detour into notions of total ramification of primes.
May
2
comment Which pairs of matrices generate $SL(2, \mathbb{Z})$?
@HagenvonEitzen thank you. I know $a = b = 1$ is enough to generate all of $SL(2, \mathbb{Z})$ and hopefully there are many others.