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 Civic Duty
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May
21
revised Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
deleted 7 characters in body
May
21
revised Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
added 35 characters in body
May
21
answered Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
May
21
revised Concatenating countably many homotopies
added 115 characters in body
May
21
revised Concatenating countably many homotopies
added 434 characters in body
May
21
answered Concatenating countably many homotopies
May
21
revised Cauchy-Ramanujan Formula $ \displaystyle \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} $
added 17 characters in body
May
21
asked Cauchy-Ramanujan Formula $ \displaystyle \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} $
May
21
revised Visualizing Euclidean Algorithm in $\mathbb{Q}(\sqrt{-7})$ and $\mathbb{Q}(\sqrt{-11})$ with Convex Geometry
deleted 20 characters in body
May
21
answered Visualizing Euclidean Algorithm in $\mathbb{Q}(\sqrt{-7})$ and $\mathbb{Q}(\sqrt{-11})$ with Convex Geometry
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
21
asked Visualizing Euclidean Algorithm in $\mathbb{Q}(\sqrt{-7})$ and $\mathbb{Q}(\sqrt{-11})$ with Convex Geometry
May
19
revised Efficiently finding two squares which sum to a prime
deleted 18 characters in body
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
revised Bing's House and homotopies
tiny Spanish grammar changes. maybe my own personal bias
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
revised Bing's House and homotopies
tiny Spanish grammar changes. maybe my own personal bias
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
18
awarded  Civic Duty
May
17
comment Enumerating Bianchi circles
This is also $K = \mathbb{Q}[\sqrt{-7}]$, right?