900 reputation
318
bio website
location Prague, Czech Republic
age 21
visits member for 1 year, 10 months
seen Dec 14 at 19:46

Mathematics student at the Faculty of Mathematics and Physics, Charles University in Prague.


Nov
18
comment Integer solutions for $a$ and $b$ such that $a^2 - 3b^2 = 2$
The trick of writing $a = 3m$ is actually ingenious.
Nov
18
accepted Integer solutions for $a$ and $b$ such that $a^2 - 3b^2 = 2$
Nov
18
comment Integer solutions for $a$ and $b$ such that $a^2 - 3b^2 = 2$
That's nice, thanks!
Nov
18
asked Integer solutions for $a$ and $b$ such that $a^2 - 3b^2 = 2$
Nov
4
awarded  Tumbleweed
Nov
2
accepted Find polynomial over a ring with nine roots
Nov
2
comment Find polynomial over a ring with nine roots
Thank you, @ajotatxe, how could I have not seen this?
Nov
2
comment Find polynomial over a ring with nine roots
I see! Thanks, I will try that now.
Nov
2
asked Find polynomial over a ring with nine roots
Oct
28
asked Reducing modulo random number, what is the number of matches?
Oct
25
awarded  Notable Question
Oct
25
comment Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
Thanks for all help, @egreg!
Oct
25
accepted Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
Oct
25
comment Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
One more thing: how can I be sure that such a set is contained in every subring of $\mathbf{C}$? Especially, how do I know that $\mathbf{Z}$ is contained in it? Is it because $1$ is in every subring, so that all natural numbers are in it, all opposites are in it and $0$ is in it?
Oct
24
suggested rejected edit on Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
Oct
24
comment Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
Right... Too late. Will fix it now.
Oct
24
suggested rejected edit on Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
Oct
24
comment Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
Well, that's how far I got (except I got $\alpha^2 = \alpha + 1$). But how can I tell that for nonzero $a,b$, $ab$ is nonzero? What is the essence of "subring of field is automatically integral domain"?
Oct
24
revised Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
added 13 characters in body
Oct
24
comment Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
Oh, sorry, that's what I meant.