@BalarkaSen do pythagorean triples arise in this at all?

@MeowMix Not in my solution, but try it out!

I think there are many ways to do it, of which I only know 2

my pythagorean theorem solution only works for multiples of 8

so i'm missing 7/8 of the naturals

:P

well, that's certainly not the way to go

Oh really? How did you do it

Oh wait

Does the center have to have natural radius / center?

Note that I wanted points to be inside the circle and not on the circle

OH

LOL

I read as "on". Sorry!

"On" is still an interesting question but I haven't solved it.

I doubt "On" is true

It is.

Once again, my intuition sucks

Actually

All you'd need to do to is find a circle containing $n-1$ points, such that when you expand it so that it contains another point, there's a radius such that it will only contain that point

Correct. You have to find appropriate centers and radii to do that.

That is, a circle containing $n-1$ points such that there is a unique point closest to the center

Not enough, but yes.

Anyways

So assume there is a circle containing $n-1$ points

If there is a unique point, QED

If there isn't, smidge it only a tiny bit to the left, right, up, or down

And re-expand the circle so that it contains some unique point

I suppose I should put this in formal terms

Sounds like an idea I initially had :) I couldn't make it work, but let's see if you can

Oh, I think I see how to do this

Let me bring out geogebra

So, suppose we have $k$ points which are equidistant from our circle's center (and outside of our current circle, and are all the minimal distance away)

Consider each pair of points

Let's call these points $B,C$ WLOG

and let's call the center $A$

The set of points equidistant to $B$ and $C$ is a line through $A$

$A$ cannot lie on this line, since then we wouldn't have a unique point

Construct all lines for each pair of points, and you have a bunch of lines through $A$ which $A$ can't intersect

Very astute.

Since there must only be finitely many lines, there MUST be some point in the circle not contained in any of these lines

In fact, we could get arbitrarily close to $A$ if we wanted

Just take a line that's not in the set of "no-go lines", and approach $A$ along it.

And we're guaranteed a unique point

All good ideas. Can you fix the center? Your current solution-sketch moves the center around.

The reason I think that I'm allowed to move it around is because I could make it arbitrarily close to $A$

So, if there were any points very close to the boundary of the circle (that might be excluded), we can find some point very close to $A$ that will account for it.

Yeah, that has to do with the disk being compact I think

I think you're allowed to move it too. But I'm asking if you can modify your proof so that it works for a fixed center.

@Astyx Nah.

I mean, sure, but it's the countability of Z^2 which is more important.

Oh, like there exists a center where there is always a unique closest point?

@MeowMix Err, unique closest point from the center doesn't guarantee there would be a circle containing for every $n$ exactly $n$ lattice points

What do you mean?

I mean that for any circle, that there is a unique closest point

Oh. Yes.

So that when you expand the circle to contain that point, there is also another unique closest point

Correct. There is such a fixed center.

In other words, no two points on the lattice have the same distance from Balarka's point

Hm... I'm thinking

You've already solved it in some sense

With what you stated earlier

it's very close to the set of ideas you're hovering about

Yeah, take any two points on the lattice

the set of points equidistant from those two points must be a line

however, there are countably many lines

so that can't cover $\Bbb R^2$

yeah man. great job

so there exists a point where no line is intersecting it.

Now, I'm wondering. What point is that? :P

An explicit example of such a point, btw, is (sqrt(2), sqrt(3))

oops

Oh, happy St. Patrick's day to those who celebrate it.

Hi @Mike

hi

Today I had a benchmark in geometry, and so for the last question, I was finished and bored, so I wrote a linear algebra proof of it as a joke.

(It's a - sort of not very good - translation of a riddle I like a lot)

So apparently someone needs a positive integer solution to $\frac x{y+z}+\frac y{z+x}+\frac z{x+y}=4$

and also apparently the solutions are all really large

did you try making it a giant big fraction?

No

and I kind of don't want to

ill take the liberty of doing so and then doing nothing

$\frac{x(x+z)(x+y) + y(y+x)(y+x) + z(z+x)(z+y)}{(x+y)(x+z)(y+z)}$

be right back.

good evening chat

how's everyone?

(i followed through on my promise to do nothing)

@AlessandroCodenotti still disappointed

By what?

loss of data and code

How did it happen?

Overwriting a partition with windows 10

@AkivaWeinberger Maybe one can do a greedy expansion on 7/(x+y+z) to estimate about how large x, y and z needs to be

That is, the partition had stuff on it, and it was overwritten with the windows 10 data, the deed wasn't done by windows 10

Ah, I see

I started learning standard ML, I'm doing a functional programming course in uni, pretty interesting stuff

have you done assembly yet

also ML = markup?

Not in school, but yes

it's the name of the language, I din't know that ML stands for

oh

I thought you meant general markup

like XML, HTML, YAML

Nope, nothing of this sort

anyways, for which processors?

Only intel stuff, x86 and higher, up to x64, with all the SSE and AVX instruction sets

oh

i don't like x86 that much

the instruction set i mean

i'm bad at combinatorics

should work on that for exams

@MeowMix it has been expanded a lot though

I mean compared to the 386 or what was the original instruction set

When will you have exams?@Balarka

monday

Courtesy of Wolfram Alpha:$$\frac{3 x y z + (x + y + z)^3 - 2 (x + y + z) (x y + x z + y z)}{-x y z + (x + y + z) (x y + x z + y z)}$$

is the fraction above in terms of symmetric polynomials

@BalarkaSen exams for what?

school

yeah, but you have combinatorics in school?

or $(3s_3+{s_1}^2-2s_1s_2)/(-s_3+s_1s_2)$

a thing or two in high school, yes

interesting..

i'm going to be taking algebra 2 next year

(i suck at it)

sighs

What kind of combinatorics?

Because I can't imagine you having troubles with the combinatorics done in high schools here

basic counting problems, permutations/combinations, binomial formula

It confuses the hell out of me

you're not alone

Lol, you should totally take a combinatorics class when you get the chance though

@Daminark Indeed, I plan to fill in a lot of what I don't know in my undergrad, even though I am constantly told to just jump into a PhD or something.

Woo!

@Balarka Good idea.

There's no undergrad course in higher cathegory theory :P