@mr.FS well, you have blackboard bold in your avatar :)

@Charlie youtube.com/watch?v=M9LgEW8ikfw

can't beat Springsteen 8)

@AlexeiAverchenko o haha i forgot about that lol, I actually prefer the sans serif computer modern to the serif one, but I like the blackboard bold letters yes

@Argon So so cool!

@Charlie I like that song

@Argon What does it mean?

@WillHunting I got my results today :D

@Charlie The title means "You are beautiful"

@Argon hmm what about the rest?

@Charlie Some of it is in Amharic...

@WillHunting I know

it is midnight should i drink coffee?

@WillHunting you always make me feel better by saying these things :)

@Argon oh

I am the one who is always proselytizing that exams are trivial @WillHunting

@Charlie 1 sec, I will get lyrics!

@Charlie depends if you are planning to sleep anytime soon :)

@Argon ok!

@OldJohn haha

@Charlie sorry - moving target again!

that was meant for n3b

@WillHunting Good breakfast!

@OldJohn i realized

@WillHunting See you!

@Charlie good :)))

@Charlie #5 here: idanraichelproject.com/en/the-idan-raichel-project

@N3buchadnezzar it's 6 am, and I'm still good :D

@Argon WOW

@AlexeiAverchenko Why up so late?

@N3buchadnezzar poor regiment :)

@N3buchadnezzar double espresso ... and some snus - should keep you awake :)

and I'm trying to wade through technical theorems about CW complexes

proving every detail from an appendix in Hatcher

@OldJohn snus? Kill it with fire. And I am trying to sleep, I just felt like some coffee

@Argon Really nice

@N3buchadnezzar :)))

@Charlie I like his stuff

@OldJohn Do they have snus where you live?

@N3buchadnezzar no - but I have my sources :)

@OldJohn isn't snus an addictive drug? <_<

@AlexeiAverchenko yep

@OldJohn you junkie :P

@AlexeiAverchenko :P

@AlexeiAverchenko No, it is unicorns and rainbows contained in a small box, which eventually gives you cancer, then you die. Horribly.

@Argon Seems nice thing :)

hello again

@mr.FS FS

@mr.FS what does mean FS?

my initials

@AlexeiAverchenko вы русский?

Is there a way to check if a functor has left or right adjoint without trying to explicitly construct it?

@OldJohn угу

@mr.FS Fascinating

@Argon what other song do you like?

@mr.FS Francis?

nope

@Charlie Hmmm....

you're not going to guess my first name i can promise you

@Argon never listrned to this one...

@AlexeiAverchenko Украина?

@Charlie Which?

@mr.FS smelly bottocks!

@mr.FS Fraser

@OldJohn Сибирь :)

lol

@WillHunting nice blue square :)

@mr.FS Frumplestiltskin

a question here that shouldn't be difficult but i can't seem to find the answer, the problem is to find gcd(2^1120-2^1000,2^175-2^100)

@AlexeiAverchenko "cool" :))))

oh shoot

you got it

@OldJohn indeed :)

@Charlie :)

@mr.FS Frumplestiltskin?

yes

nice

@AlexeiAverchenko :))

thanks

@WillHunting I just thought this red was a handsome color

@WillHunting haha blue is my favorite, but it was taken

how do I get 15 rep on mathoverflow? :D

so many cool questions and answers, and I can't upvote them

@Argon excuse me for a few moments :)

@Charlie Ok

6 hours of math and I have almost worked through my first page

anyone that can help me with the question i pose a few minutes ago?

If someone has time, I'm struggling to understand steps in Mike Spivey's answer to a functional equation question.

@mr.FS Do you know the rule that gcd(ab, ac) = a*gcd(b, c)?

yes

I'm still working through it myself, but I'd start by trying to pull a term out.

ok

Any luck getting anything out?

no i tried some things but didn't get any further

i also tried applying euclid's algorithm but hit the wall immediately

If it helps, I started by noticing that both $2^{1120}-2^{1000}$ and $2^{175}-2^{100}$ were even, so they should have some factors of two in common.

yes

i know the answer but don't know how i would get to it

@mr.FS is the answer anything like $2^{75}$?

it's $2^{115}-2^{100}$

Then since I wanted to pull out as many factors of two as possible, the first term became $2^{1120}-2^{1000} = 2^{1000}(2^{120}-1)$, and the second term became $2^{175}-2^{100} = 2^{100}(2^{75}-1)$.

Is there no way to get to the main site from here?

Seems pretty poorly designed

@Jordan The bottom right corner has a logo, I think.

oh that was well hidden

Usually the bottom of the web page is where you hide the customer support links

@lewist I have a method, I think ...

@OldJohn Please feel free to step in, I'm a lousy teacher.

$\gcd(2^{1120}-2^{1000},2^{175-2^{100}}) = \gcd(2^{1000}(2^{120}-1),2^{100}(2^{75}-1))$

$ = 2^{100}\gcd(2^{900}(2^{120}-1),2^{75}-1)$

sorry for lack of reactions i was watching a video

i'm sorry if this shows lack of effort

ok and that one i can do

thanks

i understand it now

(the $2^{900}$ term disappears since the other factor is an odd number - so cannot have a commaon factor with a power of 2

yes

@mr.FS How? =)

Euclid

s algorithm

then you need to factorise $2^{120}-1$ and $2^{75}-1$

$2^{120}-1$=2^{45}(2^{75}-1)+2^{45}-1$

$2^{75}-1=2^{30}(2^{45}-1)+2^{30}-1$

Ahh, thank you folks, that last step is one I'm not good with.

$2^{120}-1 = (2^{15}-1)(2^{105}+\dots + 1)$

$2^{75}-1 = (2^{15}-1)(2^{60}+\dots + 1)$

so they have a common factor of $(2^{15}-1)$

yes

is it finished?

I just got FWD:FWD:FWDed this video about math

youtube.com/…

Guys, look this comment. Is it plausible? I don't know what he meant with that, because it seems there are polynomials in some unusual mathematical structures. At least I still don't see these linear combination of monomials in $n$ on these Chebyshev Polynomials.

@GustavoBandeira I think he is actually right ...

I have an interesting question!

That video is bad I think I got cancer from it

@OldJohn i think so, i got $\textrm{gcd}(2^{120}-1,2^{75}-1)=2^{15}-1$ by Euclid's algorithm so we just multiply this by $2^{100}$ to get $\textrm{gcd}(2^{1120}-2^{1000},2^{175}-2^{100})=2^{115}-2^{100}$

I don't know what he meant wit that but I can't see the linear combination of monomials on these Chebyshev Polynomials.

@mr.FS yep

You and a friend are walking along. Then a woman approaches you with 1000$, which she is willing to you. With the only condition that you both agree on how to split the money-

You want to split it fair 50/50, but your friend want to do it 90/10. And seems willing to lose all of it. What should you do?