WLOG, assume $2k < n$.

well it has to be

$p^k(p^{n-2k} - 1) = 24$.

There are only finitely many such $p$ for a given $n$.

As $n$ grows, I'd bet $\#\, p$ gets smaller.

So almost surely finitely many $(m, n)$s.

You can rigorise my ideas to prove it thoroughly @Alizter. That's an exercise for you.

Note that $m$ is pretty much more or less irrelevant here.

m should be above twelve though

so any m above 12

Heh?

If $m > 12$ then $m - 12 < 0$

?

Are you assuming that $p$ can be a negative prime too?

Otherwise $p^n < 0$ hence $p < 0$, contradiction.

@BalarkaSen m>12 =/> m-12 < 0

Fair enough. I am sleepy.

=P

In any case, $m$ doesn't matter.

@BalarkaSen you have given me good insight though

thank you

Hmm, well, there might be infinitely many of them as well.

$p^{n-k} - p^{n} = 24$, $p^n \cdot ( p^{n - k} - 1) = 24$. Oh no, there ARE finitely many of them.

As $n$ grows $p^n$ gets huge. The equality holds only finitely many times.

Well, proved. QED.

So a largest $n$ would be $2^n<24$ n <4?

The largest $n$ would be $p^n < 24$. So both $p$ and $n$ are bounded.

smallest prime to largest power

2^4

no?

Brain farts on elementary mathematics. I dunno.

there are finitely many cases to check

so its fine

Yeah, I guess yes.

This was the hardest problem on a junior olympiad

Make sure if I didn't mess up.

Hmm. Guess I didn't. Well, enjoy.

@Alizter As I said these Olympiad problems are BS.

Tricks everywhere. Ugh.

@BalarkaSen $p^{n-k}-p^n\ne p^n(p^{n-k}-1)$

$p^n(p^{n - 2k} - 1)$

The argument still holds.

yes good

Oops. Still typoed.

'Twas $p^{n - k} - p^k$ I believe =P

So $p^k(p^{n - 2k} - 1)$. Now the argument gets void.

Well, expectancy is still finite. If $k$ is small, $n - 2k$ is huge.

If $n-2k$ is small, $k$ is as huge as $n$.

But the bound is void, though @Alizter

grabs a cup of coffee

Well $24=(p^{n-2k}-1)p^k$ therefore only 4 cases of k

because $24=2^3\cdot 3$

Duh. Yeah, well, yeah.

k=1, 2, 3 and p=2. k=1 and p =3

Man I am too sleepy.

@BalarkaSen don't sweat it :)

Full write-up : $(m - 12)(m + 12) = p^n$. Thus $m - 12 = p^k$ and $m + 12 = p^{n - k}$. Thus $p^k + 24 = p^{n - k}$, hence $p^{n - k} - p^k = 24 \Rightarrow p^k(p^{n-2k} - 1) = 24$. Thus $(k, p) = (1, 2), (2, 2), (3, 2)$ or $(1, 3)$. Subbing these respectively in $p^{n - 2k} - 1$ and solving for the $p$-free part gives back the corresponding $n$s. Hence there are only finitely many.

I have sent that to the NT room as feeds, @Alizter.

bloomingdales

@TomCruise!

Long time.

@BalarkaSen for the original problem. There is only a single solution

$(20, 8)$

If you check all four cases only two are valid. Then checking those with m makes only a single case valid

Ah. =)

Cute problem. Must show my friend the solution tomorrow. He did pose it after all

Thank you ever so much @BalarkaSen for your guidance :)

No problem.

And that $(20, 8)$ comes from the $(k, p) = (3, 2)$ case, I presume?

Hello @TedShifrin

hello, @Balarka

How was your day?

@studentmath: I've only graded the first problem so far, but, much to my dismay, only a handful of students got it remotely correct. No one read the problem. They all told me (or tried to ... I didn't grade this for partial credit) how many possible bridge hands there were with or without voids.

The night is going to be far worse, @Balarka. I'll be grading ...

Oh noes.

now you sound like @Pedro

I said it was not an answer

and should be a comment

but the audit system said no

Cleo did not attempt to answer the question in my personal opinion

Doesn't look like an answer to me ... but I haven't seen the question.

@TedShifrin It is one of vladamirs integrals

shrug ... I am ignoring most posts these days.

@TedShifrin I'm pretty sure I followed guidelines

Bleh it was NOT an answer

I really have no interest in this sort of thing ...

None of Cleo's are.

Me neither @TedShifrin

The one esoteric integral I got involved with involved some serious thinking, not technical pyrotechnics.

Oops miss-clicked sorry

@TedShifrin I bet it was about computation of curvature of some complicated surface over $\Bbb R^5$ or some sort...

no, no, not remotely.

hmmm?

Here you go.

Whoa yeah I have seen that

And a VERY different solution too

I still want to know where the OP got that question ...

@TedShifrin Are you interested in another solution?

Hi, I would like to prove that for all $x,y\in \Bbb{K}^n$ we have $\vert x^{}y\vert\le \vert x\vert\vert y\vert_{\infty}$, I don't see how can I start, I would like to use CS so I have to prove that $\vert x^{}y\vert$ define a dot product right?

What's another solution?

It was posted on a forum I was in years ago and someone came up with a solution.

@Marc: First we have to decipher what you're typing ...

hi @TedShifrin

It was not nearly similar to the one you posted, IIRC

heya @Mike

I think I passed today

@MikeMiller throws a table for a lack of hello

h

Yippeee ... everything, @Mike?

The topology test, @TedShifrin

Algebra is on friday

ah

I'm reservedly very proud of you, @Mike.

You won't be after I tell you what I missed :P

I said reservedly. I can retract.

The one problem I couldn't do in any form was showing that rank 1 matrices are a submanifold of the $2 \times 3$ matrices over $\Bbb R$

Oh, blah.

I kept trying to do it as a regular value problem... instead of just writing down charts

@TedShifrin Sorry. I hope this one is more clearer.

You should do it as a regular value problem ... but you need to work on a chart where you assume the matrix is of block form with ...

and $x^*$ is the adjoint.

@TedShifrin Right... I wanted to cut out the whole submanifold at once

What is $\Bbb K$? And what are $\|x\|$, $\|y\|_\infty$?

Can't do it, @Mike.

Besides, it's silly, since being a submanifold is a local question.

I know :P

But I can cut out $SL_n$ all at once... :P

Yeah, I can too.

To cut out rank $1$ matrices globally, you need a lot more equations than the codimension.

It's not a complete intersection.

Yeah, I believe you

Problem's that when you've got a finite amount of time and your mind set on an approach, that's all you try.

Well, I'm not yet totally stupid :D

Need to learn to try something else when you're stuck.

yeah

I think here $\Bbb{K}=\Bbb{R}$ or $\Bbb{K}=\Bbb{C}$, and $\|{\textbf{x}}\|_{\infty}=\max\left(|x_1|, \dots, |x_n|\right)$, $\|{\textbf{x}}\|_1 = |x_1| +\ldots+|x_n|$.

gonna take a break for a while before I start hitting the algebra

yeah, you need a break, @Mike

oh, there's a $_1$ on the $\|x\|$?

Yes! I am sorry @TedShifrin I am on my mobile!

Then this isn't Cauchy-Schwarz. It's Hölder. You only get $\|\cdot\|_2$ norms with Cauchy-Schwarz.

So, no, it's not a dot product. At least, I don't see how it is.

Ok, I will try Hölder, thanks.

Oh noes @TedShifrin it WAS your solution now that I compare the two. Apparently someone copy-pasted it without reference.

Oh lovely ... plagiarism wins.

Grumph. I don't usually encourage this kind of attitude but well, it has been a year.

Otherwise I would've PM'd him.

Nice solution @TedShifrin. I starred it.

Ted, for Holder I take $p=1$ and $p=\infty$, and I have to prove that $\vert x^{}y\vert\le \sum x^ y$?

sorry, on phone

no problem :)

@Sawarnik

@MarcGato: In general, if $1/p + 1/q = 1$, Hölder should say that $$|x^*y|\le \|x\|_p\|\|y\|_q.$$

In $\Bbb R^n$ or $\Bbb C^n$ you should be able to prove this by Lagrange multipliers.

@Ted but you even clarified it even further in the example! sigh I guess reading is the enemy of all students.

At least they will learn from this that in probability you really have to read the question thorough. (and in any subject, but here you can't ever get away with it)

you know i really do love this site

It's not everywhere that you can recommend somebody a book and the author will come out of the woodwork to recommend alternatives/supplemental reading

Here is a question for the room. Let $G$ be a digraph of rooms, and for each edge $i\rightarrow j$ in $G$, denote by $f_{ij}(t)$ the number of people who have walked from room $i$ to room $j$ by time $t$. Denote by $w_i(t)$ the average amount of time that a person has spent in room $i$ by time $t$, and denote by $A_i(t)$ the total number of people who have come into room $i$ by time $t$ (so, $A_i(t)=\sum_{j\in \text{In}(i)}f_{ji}(t)$, where $\text{In}(i)$ is the in-neighborhood of $i$).

if we suppose that $\lim_{t\to\infty}A_i(t)$ and $\lim_{t\to\infty}w_i(t)$ both exist and are finite for one $i$, can we conclude that $\lim_{t\to\infty}A_i(t)$ and $\lim_{t\to\infty}w_i(t)$ both exist and are finite for all $i$?

@AlexanderGruber who is "a person" in your definition of w(t)?