@anon which answer?

@anon I see the divisibility answer :-)

@robjohn Here, I suppose.

@HenningMakholm that is what I figured, after doing some search.

totally regret CWing that one

CW?

made it community wiki

because the first version of my answer was too close to being a copy of what Arturo had already said in his answer

@anon I answered a question this morning long after the first answer. I didn't CW, and of course, it gets no votes :-) Perhaps it will get votes if I do :-)

Did Arturo CW his as a show of solidarity?

I feel burned out

and I don't know how to approach this problem anymore

one person says this, another says that

@HenningMakholm he did it because he was mainly elaborating on an idea I hit upon in the comment thread

actually, mostly people have been saying the same things, just with different numbers and letters :)

@Against, you wouldn't happen to have a cousin named Jordan, would you?

snap

@anon I'd just say when $x+1\,|\,4n^2+1$ :-)

@HenningMakholm Maybe he is a second years PhD student?

Jordan? No

That seems to be a common feeling among those. "I am past half of my time. I have done nothing. I don't know how to do it. I fail. I am stupid. Nobody knows how to solve this problem and I am the one they expect to solve it."

No I've just been hitting my head against this problem for like two days

Just burned out for the moment

Only two days? Oh joy.

I had the same thing for like six months, then I figured out it will never work and tossed it in the bin 8-).

@anon The smallest $n$ for a non-trivial $x$ would be $n=4$ since $65$ is the smallest, non-prime $4n^2+1$.

@AgainstASicilian which problem?

But at least I do know now that it will not work! 8-).

@robjohn the dividing problem. but it has a step which involves finding m^2 = -1 modulo 2k+1

@JonasTeuwen Impostor syndrome?

@HenningMakholm That is related I think.

@HenningMakholm That's that you think that you have been fooling people because they appear to think you are quite good and you actually think you know nothing?

That's a comorbidity.

@AgainstASicilian Your problem would be a very nice one for beginning students in number theory, because it covers so much - CRT, HL, QR - in just one problem. If you're only just familiar with modular arithmetic it's no wonder you couldn't get it in just two days.

@robjohn after thinking for awhile about the square root of the binomial coefficient problem, i can't see why it should converge. do you believe it does?

@anon It's just that it feels like there are too many pieces and I don't understand any of them

@JonasTeuwen Yup.

They often come together. 8-).

is there a way to solve this without the factorial thing? as per arturo's advice

wow it's been two hours. are we still talking about being stuck about the problem?

it's a hard problem

@AgainstASicilian I think Arturo's advice is to basically do a brute-force search, although first by trimming the search space a bit.

If I understand correctly.

For small primes $p$ this seems efficient.

@AgainstASicilian The basic square root of -1 modulo a prime? My preferred approach would be to try to guess a generator of the multiplicative group. There's lots of them, so finding one should be quick.

(Actually it might be more efficient across the board, I dunno.)

there are about 4 answers though.

@HenningMakholm The sad thing is that the complement of the imposter syndrome is much more common...

and it's been 2 hours we've been talking about it

@HenningMakholm hmm, checking if a number is a generator is more costly, but there are going to be way more generators. tough call for someone who doesn't know anything about computational complexity :)

@Eugene To be fair, I had asked for a particular form of explanation and had met resistence the entire way through

What's obvious to hardcore mathematicians isn't obvious to someone who's learning it

@anon Hmmm. I may have underestimated what it takes to check for a generator. My real idea was just to choose a random $x$ and compute $x^{(p-1)/2$ using exponentiation-by-squaring. The result is always ±1 by Fermat's little theorem, and if it happens to be $-1$, then $x^{(p-1)/4$ is a square root.

@JonasTeuwen The sufferers from that are probably happier, though.

@HenningMakholm I know.

ah, you're right, that doesn't sound too bad. expo-by-squaring I forgot about.

Could you guys be any more arrogant? Just because I said I was tired/exhausted/burned out doesn't mean I have Imposter Syndrome

Cut it out

Why do you think we are talking about you? Talking about arrogant...

not going to play pedantic wordgames with you -- cut it out.

Yes. That applied to you, why do you think the rest of it does?

What you said just reminded me of something related. The only connection with you is the reminder. Henning replied to that.

i just think maybe this problem is too big to solve

there are too many pieces

this m^2 = -1 thing is just one tiny pieces and even that is gargantuan

Just don't worry too much about it if you cannot solve a problem in two days, that is quite normal.

@AgainstASicilian If I understood your original problem, there's barely any more than finding the square root of -1 modulo 2k+1. Was what you posted actually part of an even bigger problem?

i've gone through this m^2 shit all day to the point where i don't even remember why i need it to begin with

@anon No, the dividing thing is the problem

finding solutions to that equation with n bound under some limit

as there are infinitely many solutions

I just want to warn you that some of us have been getting our comments flagged as offensive (not sure who's doing the flags) for expletives and other things. The users in this room will try to invalidate the flags, but people from other rooms might be unfamiliar with context and mark flags valid (the flags are 'outsourced' to third parties outside of the room). That can result in automatic 30 minute suspensions.

ok

(So: what happened to a few of us may potentially happen to you.)

@anon yeah. i wonder who's been so trigger happy.

thanks

Seeing how people misunderstand me here (and elsewhere) I am quite surprised that I did not get an automatic suspension yet 8-).

Would be quite ironic if somebody flagged ^.

i dont see any flags

I can flag

what the hell does this even mean

"One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard th eremor and the fraction who have not hear the rumor"

The wording is so ambiguous

@Jordan I can't see more than one reasonable thing it could mean, but that might be a failure of my imagination. Perhaps you could suggest a few plausible readings to help us get going?

?

Out of that I get $\frac{dy}{dt} = \frac{c}{y} \cdot \frac{y}{c}$

@Eugene Yes I do and I also believe that $\displaystyle\lim_{n\to\infty}\sum_{k=0}^{n}(-1)^k\binom{n}{k}^\alpha=0$ for $0<\alpha\le1$

@robjohn i see that for $n$ odd but how about $n$ even?

@Eugene For odd $n$, it is trivial. But it seems to be true for even $n$ as well.

@robjohn hm. i can't see why but then again i'm no analyst. =)

There is some deep truth here.

@Jordan The "fraction of the population" who have heard the rumor is just a different way to say the percentage who have heard it. It doesn't mean you should divide anything by $y$.

@Eugene I don't see why either, not exactly, but it seems to be so. The hard part is the cancellation of all these HUGE numbers.

@robjohn well i await for when you find the answer.

@Hans I can't really follow what is being said, it seems too ambiguous.

they mention a product, so I know that something is being multiplied

and that y is a fraction

which belongs to the population who have seen it

so I think that "have not" heard is a constant, and that y is a fraction taht represents who have

@robjohn That is peculiar I think. How do you show that?

Right for $\alpha = 1$ we just have the expansion of $(1 - 1)^n$.

That is like the Sherlock remark of today, but for smaller $\alpha$. Maybe it is quite easy.

@JonasTeuwen If I knew how to show it, I would have slept a lot better last week :-)

Oh you also think in your bed about funny identities with combinatorical coefficients? 8-).

this is interesting

At least we can show that it is bounded! 8-).

@JonasTeuwen for odd $n$, it is very simple by cancellation of equal terms. For even $n$, it is not so easy

Yes.

@robjohn Sounds like something Mike Spivey might know? Or JM?

Maybe it is just some special function.

Some kind of Bernstein polynomial perhaps.

@JonasTeuwen I have not heard of any research into something like this. I will ask JM next time he's here

@robjohn wait. are you back in ucla? i remember you saying you had to proctor an exam or something once before.

@Eugene I work for UCLA, if that is what you're asking.

@robjohn ah i see. that's awesome.

@robjohn That is pretty cool. 8-).

i will never understand why sometimes a question is voted much higher than the answers.

@JonasTeuwen I had not looked at it using Bernstein Polys. I will have to see what that brings. It looks close to something I was looking at, but perhaps some harder result related to them might help.

@robjohn It was just something it reminded me off... Don't take it too seriously, it is just something I would look into at first if I wanted to solve this and easy attacks fail.

Which they obviously do if they keep you awake for days 8-).

Hi ... can you check [this question]{)math.stackexchange.com/questions/162595/…)

@JonasTeuwen I worked on it most of one night and about half of another, so I would be surprised if there is an easy solution.

shouldn't you expect complex here

@robjohn If it would be me, that would be very much still possible... 8-).

wolframalpha.com/input/?i=%28-1%2F243%29^%28-2%2F3%29

Something like oh holy cow, the recursion is like soooo easy.

output on wolfram

@experimentX dude your expression is wrong

Hmm ...

@DylanMoreland did you see the split quaternions question? interesting...

@experimentX I think you should just delete that answer and avoid some downvotes...

okay ...

@experimentX when you put a negative number to a non-integer power, there are a lot of unspecified operations happening. For positive numbers, you just take the log and multiply then exponentiate. However, when you put a negative number in there, taking logs is a lot more complex. :-)

but why wolfram is giving complex??

i mean complex number ... i had a debate with someone and i told him if you have negative number inside and fractional power ... you will be getting complex number

@experimentX because there are three possible complex answers, depending on how you take the log of -243

@DylanMoreland also do you have any number theory/arithmetic geometry papers to recommend?

in advance peter just shut it about landau

@robjohn but everyone is getting real value on that problem

I hope i didn't make fool out of myself for that debate ..

@Eugene Hahahahha f*** off =)

@experimentX because there is a real answer, but that answer can be multiplied by $(-1)^{1/3}=e^{2\pi i/3}$ and $(-1)^{2/3}=e^{4\pi i/3}$ to get the other two

Hmm ... does that mean it has three solutions??

@experimentX But people tend to go for the real answer, if it's there.

@robjohn very punny. ;)

@experimentX there are always three numbers so that $x^3=a$ for any non-zero $a$

isn't it like saying $ \sqrt{a} $ as two values??

it's amazing how much number theory goes on at berkeley seeing that ribet is the one only left there.

I mean solving $ x^2 = a $ instead of calculating $ \sqrt a $ directly?? I mean $ \sqrt 9 = 3 \neq -3 $

@experimentX the square root is a function that is defined from the positive reals to the positive reals. $\sqrt{x}$ is a different thing than $x^{1/2}$. The latter is dependent on which branch of the $\log$ function you use.

Oo ... can you refer me something to read on this??

@experimentX it's just complex analysis

i have a question. why do people say maths?

thank you .. i'll see that later ... haven't taken complex analysis yet.

math is already an abbreviation. no need to put the s at the end.

and mathematics is always used with the s.

isn't it like saying "there's an appn for that?"

@Eugene But maths doesn't sound as strange.

yes but it's just as pointless and stupid

@Eugene Hhehe I dunno. In spanish now the singular "Matematica" is being prefered instead of "Matematicas".

@PeterTamaroff but isn't argentinian spanish different though?

@Eugene Not that much.

@PeterTamaroff i see

@Eugene My career is called "Ciencias Matematicas" which means "Mathematical Sciences".

but I don't know what the consensus is on the plural/singular word.

@PeterTamaroff does spanish put an s at the end?

@Eugene I think it does.

@PeterTamaroff don't you speak the language?

Actually both are accepted.

@Eugene Hahah but I don't know precisely.

Maybe one took over the other.

@PeterTamaroff i see

I have to check the RAE dictionary.

@Eugene According to the RAE dictionary, both are cool.

@PeterTamaroff i see

@Eugene Though I really don't give a damn....

@PeterTamaroff i just don't understand why the redundancy in the abbreviation

anyway i'm heading off now.

bye

I am so bad at math is is just discouraging to even try

I wish I could just transfer to a university instead of wasting so much time and money at a community college

it is so disheartening, there is absolutely no motivation to try anymore

@anon Are you around?

sorta

it's my brother's bday party

but I'm a pretty asocial guy irl

what's up?

@anon I'm with Algebra. And I'm trying to understand a proof.

The theorem is

Let $H$ be a linear hom sistem of $n$ equations by $n$ unknowns. Let $H'$ be an equivalent system to $H$ whose asociated matrix $B$ is triangular above. Then $H$ has an unique solution if and only if $B_{ii}\neq 0$ $\forall 1 \leq i \leq n$.

Now, I get the $\Leftarrow$ part.

equivalent system - does that mean the associated matrix is similar to the original?

@anon Yes. Transformed.

How did one code a matrix here?

\begin{pmatrix}a&b\\c&d\end{pmatrix}. Or bmatrix if you want brackets.

Proving $A\implies B$ is equivalent to proving $\neg B\implies \neg A$.

$$\begin{pmatrix}B_{11}&\cdots &B_{1n}\\0&\ddots & \cdots \\ \vdots & 0& B_{nn}\end{pmatrix}$$

Let that be the associated matrix to $H'$

show that if one of the diagonals is zero, there is more than one solution.

Then one gets $B_{nn} x_n=0$

anyway I'm going back up to the party, ping me for anything else in the meantime

@anon Good for you!

Have fun.

It's family, not friends. I just sit around and listen to people tell lame stories.

@anon Actually it is the other part I don't get ! hhehe

@anon Isn't there good food?

@PeterTamaroff youtube.com/watch?v=FywMOuMqNuI

@PeterTamaroff I already ate

@PeterTamaroff Okay, so you it's the $\implies$ part you get, not the $\Longleftarrow$?

@anon The thing is like this

@anon For $\Leftarrow$ the author states that if we triangulate the original matrix to get

$$\begin{pmatrix}B_{11}&\cdots &B_{1n}\\0&\ddots & \cdots \\ \vdots & 0& B_{nn}\end{pmatrix}$$

then the solution is $(0,\cdots,0)$ uniquely (quite boring, IMO)

For $\implies$, we assume that

$B_{11}\neq 0, \cdots, B_{ii}\neq 0$, but $B_{i+1,i+1}=0$

Then the author builds up this mega matrix

Give me some time to code it =)

@PeterTamaroff i've seen this before.

OK. That's it.

@anon He builds that matrix.

Now he says $(1,0,0,\cdots,0)$ is solution to the system with matrix $(\bar 0, M)$, that is $x_{i+1}=1$ and $x_{i+2}=0=\cdots=x_n$. That's OK right?

@anon Suppose we look at $\Bbb{R}$

If I "identify" $\Bbb{N}$ to a single point

I am guessing the equivalence relation in question is $x \sim y $ iff $x$ and $y$ are both in $\Bbb{N}$

But this does not really make sense because what does $x \sim x$ mean? @anon