I wrote my proof here: latex.artofproblemsolving.com/miscpdf/…

Can you proofread it?

if they were in arithmetic progression, then both $q^{1/3}-p^{1/3}$ and $r^{1/3}-q^{1/3}$ would be integer multiples of some common $d$

i.e. $\frac{r^{1/3}-q^{1/3}}{q^{1/3}-p^{1/3}}$ would have to be a rational number...

which, uh, seems highly unlikely

yeah I wrote that, did you read my proof?

looking at it now

i hadn't actually opened the link before writing that...

oh ok good to know we're thinking the same

yeah. i'm not convinced that what you've got is a valid proof, though. have to think some more

I think I need a more rigorous explanation of how the equation can not be true

right.

I mean it's pretty obvious that you can't even combine the radicals so...?

what i have in mind is taking that last equation you obtained, and cubing both sides

the resulting RHS is definitely an integer, as are some of the terms on the LHS

oh, that makes sense

and because there are still radicals left in the LHS it can never be equal

unless they cancel out...

but no it's all positive ok

it's something like $l_1 (p^2 r)^{1/3}+l_2 (p r^2)^{1/3}=l_2$ where the $l_k$ would all have to be integers

ok yeah thanks for that

might need to do that once more to get it fully rigorous

I mean...

you know there are nonzero radicals on the LHS and no radicals on the RHS

hm, true

and the LHS radicals can't possibly cancel out because they're all positive

so yup no way

consider p q r k1 and k2 are integers

Hey, thanks so much for your help on these topics.

Do you use AOPS?

np. and no, i don't

actually, if you divide both sides of the above by $p r$

then the above becomes $l_1 (r/p)^{1/3}+l_2 (p/r)^{1/3}=l_2/pr$

ok...

which if you define $(r/p)^{1/3}$ to be $x$, gives $l_1 x+l_2/x=l_3/pr$...not sure where I'm going with this

man, order and primitive roots is really getting to me

I think I have that proof down

now

Do you want to talk more about order and primitive roots?

sure. actually, here's the conclusion: $x$ satisfies a quadratic equation with rational coefficients

ok so

so at most it must be of the form $a+b\sqrt{r}$ where $r$ isn't square. can't be the cubed root of a rational number, viola

ok, so I'm thinking d must be a divisor of 18

because umm

well, you'd need to have $x^d=1$ mod 19

yup

Ok so in the class the first step they concluded was that all possible orders mod 19 are divisors of 18

I don't know how they got that

WAit I know

i think you can use the primitive root approach from before to make it work

By FLT $a^18\eqwuiv 1 \pmod{19}$

whoops

$a^{18}\equiv 1 \pmod{19}$

i think you'll need to use that, yes

So then umm you know one possible value of d....

here's my thought. one of the primitive roots of 19 is 2, yes?

yes

that was mentioned in the convo too

haha

But I'm completely confused when they wrote powers of 2 from 1 to 19 around a circle

so each element is of the form $2^x$ for some $x$ from 0 to 18

do you want a screenshot?

nah

i think i can guess where they're going

now, suppose i want to see whether $2^x$ is of order $d$

what I do is look at $2^{x d}$

now, if $xd$ isn't a multiple of 18, what does FLT tell us?

wait a second

I think it'll be helpful for you to see where i'm going off of

alright

drive.google.com/file/d/0B8uAGajoMDYxQ3M2N2MzM0cybGc/…

need access, i've sent a request

The problem right now is on page 20

hmm wait a sec

ok link should work now

got it

anyways, do you see where my reasoning was going?

hmm ok

so 2^x represents every nonzero residue in mod 19

so uh why would you look at $2^{xd}$

because i want to know if it's an element of order $d$

if it is, then $2^{xd}=1$ mod 19

oh yes

haha ok hmm

So if it isn't a multiple of 18...

That's my question I guess

yeah. what does FLT tell you in that case?

so for instance, suppose xd=184 (picked at random)

then 184 = 18*10+4. what does FLT tell you about $2^{184}$?

darn umm

more to the point, what does FLT tell you about 2^{18} mod 19?

oh yeah oh yeah oh yeah

ok so umm

$2^{18*10+4}\equiv 1*2^{4}\equiv 16 \pmod{19}$

right. the main thing is, it's definitely not equal to 1 mod 19

So it can't possibly be 1 mod 19...

it can't be

yeah

umm

ok but that's for xd over 18

what about between 0 and 19

Also it's not necessarily definetly not equal to 1 mod 19

becuase what if

right. if it's less than that, though, you already know it doesn't equal 1. otherwise you wouldn't be able to represent every integer from 1 to 19 using 2^0 through 2^18

so

ok I have no idea where this going

well, what i'm getting at is the following

suppose i want to find all $p$ such that $2^p=1$ mod 19

now, if $p=18$, we know that's true by FLT. same for any multiple of it

does any other $p$ work?

yeah so the first thing in the transcript says

the divisors of 18 could work

but I don't know why

we would thnk that

hmmmm

this isn't an answer, alas, but here's an example of what goes wrong if it's not a divisor

suppose some element of order d=5 existed modulo 19

that shouldn't work, since 5 doesn't divide 18. let's see what goes wrong

representing this element as $2^x$, we'd have $(2^{x})^5=2^{5x}=1$ mod 19

what would tell us about 5x mod 18? @WidowMaven

darn is this the primitive root stuff

umm

think back to FLT

5x=18??

5x would have to be a multiple of 18, yes

wait what???

oh wow

otherwise division by 18 would leave some residue, and then taking 2^residue would give us some nonzero element

since each element 1 through 18 can be represented as such a power

@anon does the infinite alternating group have a faithful representation on any $\Bbb R^n$?

but if 5x is a multiple of 18, then x must be a multiple of 18. 5 doesn't help us

ok i guess

in which case, what's $2^x$ mod 19?

Heya @EricS

hi

by the way, @anon, your question from a few days ago cleared up something I was confused about independently today :)

i.e. if $x=18n$ for some $n$ then $2^{18n}=(2^{18})^n=?$

@MikeM: I hope anon's doing better. Haven't seen him in a few days.

I see him in the user list is all. Was he sick?

He had all his wisdom removed and recovery was slow and painful.

Oh gosh.

ok ok

so I iunderstand it's a divisor of 18 now

sorry was ordering pizza

np

ok good

so now we look at the divsors of 12 and see how we can write

$a^x \bmod{19}$

(strictly speaking, the example i gave above only showed that it couldn't be coprime to 18. but it's easy enough to reason that it has to be an out-and-out divisor)

in terms of $2^y$

okay. so then we take some $(2^y)$ and ask if it's an element of order, say, 6

i.e. $(2^y)^6=2^{6y}$.

what's the smallest $y$ such that that's true?

My best way of finding interesting questions so far: Look up the posts of a user with interesting answers. Somehow that can be full of recent posts to interesting questions in my tag set that I have never seen before. Puzzled.

hmm

such that $2^{6y}=1$, i mean, so that $2^y$ is of order 6 mod 19

y=0??

I haven't tried to keep such track, @ccorn.

...well, yes. but 2^0=1 isn't interesting, heh

what's the smallest nonzero example?

am I supposed to calculate these?

Or use euler's theorem guess and check?

to try to find phi n=6

well, reason from FLT

oh 3

right?

right.

so 2^3=8 works.

OH GEEZ

GOOD GOLLy

now I see why primitive roots are useful

now, so does 2^6=64 = 7 mod 19

so that we can express each residue thingy like that

right

ok so

so 8 has order 6 mod 19

and umm is that the only one

nope

like the only one order 6

ok then

2^(6y)=1, we need 6y to be a multiple of 18 i.e. y is a multiple of 3

...

oh yeah multiple of 18

ok so it could be 2^9

so y=3 works, but so does 6, 9, 12, 15. what happens when y=18?

which is 18 mod 19

it's just 1?

the same as

right. 2^18=1, so that's not an interesting example

y=0

and past that it just loops around

ok good

okay, i need to head off

yeah thanks for all of your help today

I can just repeat the procecss for the other divisors of 18

check my work to see if i did something silly

k bye

(i think 2^9=18 is a bit weird, since 18=-1 and so (2^9)^2=1 has least order 2

so, uh, might want to check that :/)

@MikeMiller if it did, we could complexify then induct to get a finite-dim rep of S(inf), which then restricts to a faithful finite-dim rep of S(n) for every n. but all irreps (besides sign and trivial) of S(n) have dimension n-1 or bigger (except n=4), so we can get a contradiction.

@MikeMiller oh?

really nice argument, thanks.

@anon If I have a flat connection on a G-bundle and fix a basepoint in the total space, as before this gives a holonomy representation. When I change the flat connection by a gauge transformation, this conjugates the holonomy.

I was forgetting why until I remembered the trick that if x parallel transports to gx, then ax parallel transports to agx (hit the old path with a!) I was doing the same thing in my head as you were in the question until I remembered it

cool

Hey guys, I coded a program that runs a martingale system with some starting amount and stops after a number of bets. I then added a monte carlo method to run the program a lot of times and conclude on the average winnings by calculating total_won/iterations(total_won is the sum of your current money in cases where you didn't lose money)

do you know off the top of your head what the irreps of SO(3) are (other than the obvious one?)

Is there a way to calculate the average winnings analytically using probability?

@MikeMiller I think reps of spin(n)s is indexed by half-integers, and integers correspond to ones on so(n)

or something like that

never really consolidated the stuff I read into an understanding on that question

weird

or maybe that was just so(3)=su(2)

that's all I need

the index labelling the irrep is what physicists call spin

do you know what dimensions they're in and how you "see", say, the simplest one?

nope

somehow they're related to spherical harmonics and laplace's equation

rather unpleasant

probably don't need any actual explicit understanding but that's a shame

hey

yo