LaTeX
Euler
function
MATHCOUNTS
inequalities
analytic geometry
geometry

I aint the moderator, or adiministrator. but i want to say one thing

Make a good title to your post.

A lot of people make titles using latex, or just saying "Hard one" etc...

it is so annoying...

We wast our bandwith clicking on the link and seeing the post, to know what it is.

It's better make good titles for 2 reasons

1 - More people interested on it will click.
2 - For the "salute" of AoPS

Thanks

I agree. Titles like "Question: help please" or "On a theorem of Euler" give readers no information whatsoever.

I also remember the time there were lots of interesting discussions here, it's like in one and a half year or something, the level really dropped.  Imean there are interesting topics (about sarkozy, about cultural diversity, that could become interesting, but they don't) :|

This topic isnt to be interesting.

Is just to make others topics interestings.

people can learn how to make problems.

A tipical problems in AoPS are

[quote]
$y = x^{3}+7 x$
$x = y^{3}+7 y$

solve
[/quote]

Is this a problem? have You ever seen a test like that?!
This isnt a problem, this isnt anything.

People can make the problem more real.
Try to make inteligent questions. etc.

And ask something

1 - Why are you making the problem?
2 - What you want of the problem?

Because If you post a problem is for 2 reasons

1 - You dont know the answer
2 - You want to challeng others people

if you want 2. so. try to make short tips. or after all solve the problem.

if you want 1. so type what are the problem, what you dont know. etc...

Let's make a better AoPS and stop post unecessary things. or trying to post intelignt things...

I agree but I find some problems:

Well like you said there are only 2 reasons why you want to post a question.

1. You need help
2. You want to challenge others.

Without the second option, AoPS would most likely be a dead forum with inactive users. Why would you want to post hints to a question? If they can't get it, then they can't get it. If the problem is too hard, hints can be offered upon request.

Also, AoPS stands for Art of [b]Problem Solving[/b]. Questions like

$y= x^{3}+7x$
$x=y^{3}+7y$

is based on text and book solutions. AoPS primarily functions to help other people with competition problems (ie MATHCOUNTS, AMC, USAMO). Questions that involves tricks that haven't been taught or logical solutions that haven't been seen.

On your subject of intelligent questions, and making them more real, you would have more success if you asked your math book to base each and every excersise on something in real life.

I don't know what forum you are basing this on, but I find HSB quite allright in terms of subject title.

[quote="yagaron"][quote]
$y = x^{3}+7 x$
$x = y^{3}+7 y$

solve
[/quote]

Is this a problem? have You ever seen a test like that?!
This isnt a problem, this isnt anything.

People can make the problem more real.
Try to make inteligent questions. etc.[/quote]

$y = x^{3}+7 x$
$x = y^{3}+7 y$

solve

That is [b]definitely[/b] a good math problem. What you want is word problems? Well, those are given too.

And I don't think any problem can be classified as not "intelligent" to use your words because intelligence is a human characteristic and in my opinion, no math problems are dumb math problems. It all depends on your capability.

[quote="yagaron"]
Let's make a better AoPS and stop post unecessary things. or trying to post intelignt things...[/quote]

Um...I have seen very few "unecessary" posts in the purely problem-solving forums of this website (i.e. basic,intermediate,etc.)

What he's saying I think is that there are way too many problems that are posted with descriptive titles like "inequality" and "hard" (source: I think so) that pretty much make you click through each topic to find the problems you like. Also many times the problems that are proposed contain typos or are otherwise unclear, so people have to waste time asking for clarification, etc.

what I'm saying is:

[quote]
$y = x^{3}+7 x$
$x = y^{3}+7 y$

solve[/quote]

people can be more friendly in the posts.

For me posts like that is just like "TAKE THIS AND SOLVE, NOW!!"

and the trics, information about the problem is:

"I tryed to solve this, but I cannot simplify..."

or "there is a easy way to..."

I'm saying that posts like that dont offer good information for the reader.
And it dont give a lot of motivation for the reader to solve it.

The "poster" need to make a problem friendly and too good that people will want to solve or help that person.

Posts like

[quote]
I'm having problem on y = x^3 + 3 y x + 3
[/quote]

We will ask "what kind of problem?"  "what you have tryed to solve?"

because if your problem is 1- dont know how to solve.
you need to say what kind problem, people learn with theirs mistakes.

If you dont give the information of what you want of the problem you cannot be helped.

and if 2- you want to challange people.
so, make a small text saying someting about the problem.
"You make the problem?" So, how you make, why?
what kind of solution you want?

and the most important thing

If you make a problem, solve it first, if you are unable to solve, try to solve more.
people give up to early.
If you dont solve the problem in 1 hour, try more 3 hours, with this you will get better on solving  problems.

and another most important thing: Learn to use Google and Wikipedia.

I recentelly reply a post talking about of polar coordinates, the area inside 2 equations.
ok, the person dont know how to solve.

but, if you put on google "polar equations" and search 10 pages, you probabily will find what you want.

these are what I call Unecessary posts.
they can be solved fast with google and wikipedia.
or searching on AoPS forum databasis. etc...

So, If you want to make a interesting problem, that people will "wanna" try to solve, give information.

If you see the list of posts without repply you will understand what I'm saying.

And Dont give up early.

In a recent post I type "mon", and a person ask me "what is a mon?"
but, if that person read the text carefully will see that I'm talking about the moon.

It seens me that people just want to post. But, people, we dont gain Points making replies, so try to understand, try to learn. and dont make unecessarily posts.

I know a lot of people agree with me.

Good posts, with good formulation make we try to solve it, and if we dont know what the post is saying about, the post make we learn, seach etc...

and to the replies, they can make more informative solutions.
like

"I used the inequality of ..."

Because a lot of people loking the article dont know the solution, so they will ask "What is this? what did you make?" etc...

Talk Is good. Talk necessary things. usefull and informative things.

Think about this. We are mathematicians, we need to be clear and usefull.

[quote="yagaron"]
and another most important thing: Learn to use Google and Wikipedia.
[/quote]

Psh. Google and Wikipedia suck in terms of math researching. A better site would be [url]http://mathworld.wolfram.com/[/url].

Also, sites like Wiki and Google can become corrupted, and sometimes the information can be misleading.

-------------------------------

[quote="yagaron"]what I'm saying is:

[quote]
$y = x^{3}+7 x$
$x = y^{3}+7 y$

solve[/quote]

people can be more friendly in the posts.

For me posts like that is just like "TAKE THIS AND SOLVE, NOW!!" [/quote]

Well it does keep AoPSers pre-occupied. Is it any different from problems in your math textbook?

Questions like that are essentially neutral. Your math book isn't mean. The AMC 12 questions aren't mean. Get used to questions being asked the way they are.

-------------------------------

[quote="yagaron"]
and the trics, information about the problem is:

"I tryed to solve this, but I cannot simplify..."

or "there is a easy way to..."

I'm saying that posts like that dont offer good information for the reader.
And it dont give a lot of motivation for the reader to solve it.
[/quote]

It doesn't give good information to the reader, but in terms of learning a new concept, which is better, knowing the answer, or knowing the solution? The actual steps in which someone responds is better than a post just saying a random (but sometimes correct) answer. Leaving an unfinished solution is perfectly fine, as it can invite more people to solve a particular thread.

-------------------------------

[quote="yagaron"]
If you make a problem, solve it first, if you are unable to solve, try to solve more.
people give up to early.
If you dont solve the problem in 1 hour, try more 3 hours, with this you will get better on solving  problems.
[/quote]

I agree. But posting problems that you already know how to do can spark several interesting options:

1. a completely new method that is quicker, faster, and easier that you didn't know

or

2. a completely new point of view that can change how you tackle the next problem

-------------------------------

[quote]A truly wise man has more questions than answers.[/quote]

[quote]It doesn't give good information to the reader, but in terms of learning a new concept, which is better, knowing the answer, or knowing the solution? The actual steps in which someone responds is better than a post just saying a random (but sometimes correct) answer. Leaving an unfinished solution is perfectly fine, as it can invite more people to solve a particular thread. [/quote]

The question is not the problem itself.

Some problems posted here can be wonderfull, but If it is bad-written with a "Solve it" we lose our will to solve it.

And last be reallist, if the posts and repplies were more well-written we will learn much.

the "replier" need to know that:
He solve a hard problem, hm... ok. But the problem is not only for he. others persons will learn by the solution of this "replier".

I just began with this BIG post because have a lot of interesting problems and solution in AoPS, but it is too difficult to understand, the WWW (why, what, woow)

Let's see a pratical example.

[quote="mli7"]problem:  Verify that Stokes' Theorem is true for the given vector field F and surface S.

F (x, y, z) = yi + zj + xk,

S is the hemisphere x^2+ y^2 + z^2 = 1, y>=0, oriented in the direction of the positive y-axis.[/quote]

1 - Why I will Try to solve this? Stokes theorem is true. Have a proof of this.
2 - People should learn latex. The problem in this form is too ugly that we lose our motivation
3 - It is not a problem. If that person had said someting like "I cannot proof this, I'm having trouble" etc...
But not!! People just like to make problems, but 60% of the problems made by AoPSers are wrong, have a unestructural error, or others errors, or others things.

I tryed once make a problem here, And... I fail. Why? Because I did not thougth more in the problem.
I made anothers, but they I solved myself, I post because I saw a intereting solution.

We and others normal persons see a lot of interting problems here, but we cannot read because people "like" to think "this is so trivial that I will not about this"

they make problems so hard, and confused that no one reply this.

If they make the problem better. with explanation, just saying what he used, what the name of the theorm? etc?

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145184

Ok, somebody will answer it. etc.

but it is not for he understand! not only for he! is for a lot of people!

Think, what is better, only 2 persons know the solution, so they write in a very difficult way, because they know, they dont give information, because is so trivial. etc..

or everybody in AoPS know the solution, a good one, beautifull, with explanation, links, etc etc?

If I can undestand, and you can understand, ok.

but let's think about of the people that dont understand.

I agree that all people should learn LaTeX. But as for the will, you said it yourself that people get practice from solving questions like these.

well you don't need to learn latex until it becomes ambiguous because of your problems with expressing yourself...although it would be nice.

The complaints raised here aren't anything new.  They've been known for years.  However, if you stick around the forums long enough you'll see that one of the benefits of the message boards is that people do indeed become better at making nice, logical, posts.  As annoying as it might be, you'll always have to live with n00bs that don't know any better.  We were all n00bs once and pretty much still are.

The only solution seems to help less practiced posters out here and there.  But be courteous about it.  That's the best solution unless you can come up with a better one.  And, no, flaming about it in this thread isn't a solution...since, you know, solutions actually have to solve problems.  Others ideas have been tossed around before such as making users view a netiquette page once they join and then possibly making them pass a test or something.  But (a) directing them to a page of neitquette won't be successful since most people wont actually read it and if they do they probably wont understand immediately.  And (b) making people pass a test is unwelcoming and not at all in the spirit of AoPS.

[quote]In a recent post I type "mon", and a person ask me "what is a mon?"
but, if that person read the text carefully will see that I'm talking about the moon.[/quote]

Ah, you don't understand; it's designed to be humorous. If one makes a tpyo and someone else points out his typo, they aren't asking it because they don't know what you mean, they are asking it to point out the typo and to laugh at. It's nothing personal, but typos make posts hard to read and are amusing.

Some comments about the Stokes' Theorem post by mli7 that yagaron was complaining about:

First of all, it was quite easy to understand without $\text{\LaTeX}.$ I can see complaining about that in some cases, but not in this case.

It's a textbook exercise, typical of an American "Calculus III" or "Multivariable Calculus" course - and it's a textbook exercise from the most difficult portion of that course. The statement "Verify that Stokes' Theorem is true" is instructions to perform some specific calculations. In teaching the course, I would probably assign that exercise. I don't particularly object to textbook exercises being posted in the "Calculus Computations and Tutorials" section, although if I find them posted elsewhere (such as the "Calculus-Analysis Unsolved and Proposed Problems" I will move them.

The big issue: several people, including me, thought this was interesting enough to be worth answering. So don't go slamming that post. Yes, the poster is a relative newbie; yes, he's posting his calculus class homework. Those considerations don't automatically disqualify the post - it can be interesting enough anyway.

To Kent Merryfield
I will be short.

I didnt "verificate" that calculation, It can be a good exercise of algebra and calculus. etc...

But for me it seens like "Verify if $f'$ is really the derivative of $f$"
We can proove or disproove this. But is not a true problem.
It's a paradox "verificate if these Theorem is true"
Theorems are every true. Conjectures that can be false.
I really dont get this...

To solafidefarms

I really enjoy jokes like that, but... Waste time in No-Fun post reading this is annoying.
Suppose everytime I write someting wrong somebody make a joke.
Dude, we will have 357 posts just with jokes!!! uhauhau

To joml88

I dont understand nothing you said! Sorry...
I know this topic is a pretty old thing and it is not well "formalized".

It's just for the users like yours with 3333...333 posts think more about in the neewbies like me and others on the solutions, because we are not god like yours...

and for the newbies like me, to think more before posts a problem.
we can "set up" thing on our mind and forget to write this. he...

I think It ends. It's not necessary continue.

The topic itself didnt help a lot... sorry.

buts lets try our best.
hm...

adie

[quote="7h3.D3m0n.117"]Psh. Google and Wikipedia suck in terms of math researching. A better site would be [url]http://mathworld.wolfram.com/[/url].[/quote]
Some topics are better covered in Mathworld, others in Wikipedia; Google finds me both so I can choose the one I like.

To counter [b]yagaron[/b]'s point: all a website can give is information. AoPS forums provide guidance.

[quote]To counter yagaron's[/quote]

why? I aind saying the opposite...

And this discution had already finished. IT's annoying.

we want AoPS be the best site in internet, so, let's make everything possible, in our manners.

The End


After rationalizing the numerator of $ \frac {\sqrt{3}\minus{}\sqrt{2}}{\sqrt{3}}$, the denominator in simplest form is:

$\textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2}) \qquad
\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2}) \qquad
\textbf{(C)}\ 3-\sqrt{3}\sqrt{2} \qquad\\
\textbf{(D)}\ 3+\sqrt6 \qquad
\textbf{(E)}\ \text{None of these answers}$

[hide="solution"]$ \frac {\sqrt {3} \minus{} \sqrt {2}}{\sqrt {3}} \equal{} \frac {(\sqrt {3} \minus{} \sqrt {2})(\sqrt {3} \plus{} \sqrt {2})}{\sqrt {3}(\sqrt {3} \plus{} \sqrt {2})} \equal{} \frac {1}{3 \plus{} \sqrt {6}}$

So the answer is D. [/hide]

You should put these all in one post, to prevent flooding the forum.

No, this is fine because these problems are to be used to for Resource section.  :)


Assume that $x,y,z$ are the positive numbers such that $xy^2z^2+x^2z+y=3z^2$.
Find the max of \[ P=\frac{z^4}{1+z^4(x^4+y^4)}.  \]

If we put $z=\frac1t$, then the condition imposed gives $xy^2+x^2t+t^2y=3$, and the epression $P$ reads
\[ P=\frac1{x^4+y^4+t^4}.  \]
So that to maximise $P$, we need to prove that $x^4+y^4+t^4\geq 3$. Please do it yourself for the rest, DUC.

Thanks to sir Pham Van Thuan, I'll post the complete solution to the posted problem:
Put $z=\frac{1}{t}$, the condition imposed gives $xy^2+yt^2+tx^2=3$
Next we'll prove that $x^4+y^4+z^4 \geq 3$.
Using the AM-GM inequality, we have the followings:
$x^4+x^4+x^4+y^4+y^4+y^4+y^4+y^4+y^4+1+1+1 \geq 12xy^2$
i.e $3x^4+6y^4+3 \geq 12xy^2$
Analogously, we have $3y^4+6t^4+3 \geq 12yt^2$ and $3t^4+6x^4+3 \geq 12tx^2$
Summing up three above inequalities, we have the result as claimed. And it's clear that $P_{max}=\frac{1}{3}$


Let $ n$ denote the smallest positive integer that is divisible by both 4 and 9, and whose base-10 representation consists of only 4's and 9's, with at least one of each.  What are the last four digits of $ n$?
(source: AMC 10B 2007)

Thanks

[hide="Solution"]If it's divisible by 4 the last two digits are divisible by 4, which means they must be 44. We're not going to get any help adding 9s because they contribute nothing to the digital sum mod 9, so we're going to end up with some permutation of one 9 and nine 4s. The smallest of these permutations will be the one where $ 9$ is as far right as possible, which would end with $ \boxed{4944}$.[/hide]

You need to start titling these something more specific than "AMC Question," and you need to stop reposting problems that are already in the resources section


[quote="binomial_4eva"]A shark weighs 600 kilos plus $\frac{1}{3}$ of its weight.
what does the shark weight ?[/quote]

???

$600 \times \frac{1}{3}= 200+600 = 800$? it seems to easy  :maybe:

[quote="binomial_4eva"]A shark weighs 600 kilos plus $\frac{1}{3}$ of its weight.
what does the shark weight ?[/quote]

[hide]$600 \cdot \frac{1}{1-\frac{1}{3}}=900$ kilograms.[/hide]

I agree with the second answer (cognos599) anyone disagrees?

Ah sorry. You worded the problem weird and i misinterpreted it  :blush:


probability
analytic geometry
geometry
integration
calculus
AMC

Two circles of radius 1 are to be constructed as follows. The center of circle $ A$ is chosen uniformly and at random from the line segment joining $ (0,0)$ and $ (2,0)$. The center of circle $ B$ is chosen uniformly and at random, and independently of the first choice, from the line segment joining $ (0,1)$ to $ (2,1)$. What is the probability that circles $ A$ and $ B$ intersect?

$ \textbf{(A)} \; \frac{2\plus{}\sqrt{2}}{4} \qquad \textbf{(B)} \; \frac{3\sqrt{3}\plus{}2}{8} \qquad \textbf{(C)} \; \frac{2 \sqrt{2} \minus{} 1}{2} \qquad \textbf{(D)} \; \frac{2\plus{}\sqrt{3}}{4} \qquad \textbf{(E)} \; \frac{4 \sqrt{3} \minus{} 3}{4}$

[hide]Right, so we need to find some sort of maximum horizontal distance apart the two points could possibly be.  To do this, we draw two circles which are tangent, and draw the radii to the point of tangency.  This creates a 30-60-90 triangle, and we see that the horizontal distance between the centers is thus $ \sqrt {3}$.  This is the maximum horizontal distance.
[img]12728[/img] (sorry for bad image)
Now we use geometric probability to find out the probability that the two points are horizontally less that $ \sqrt {3}$ apart.  To do this we make a graph, with the x-axis representing the x-coordinate of the first circle, and the y-axis representing the x-coordinate of the second circle.  Then we simply find the area of the shaded region shown divided by the total region, which is $ \frac {4 \minus{} (2 \minus{} \sqrt {3})^2}{4} \equal{} \frac {4\sqrt {3} \minus{} 3}{4}$, which is $ E$.
[img]12727[/img][/hide]

Let $ A'=(x,0)$ be the center of circle $ A$ and let $ B'=(y,0)$ be the center of circle $ B$. Lets find the probability that $ A$ and $ B$ do not intersect. It is equivalent that the distance between the centers is greater than $ 2$. In otherwords, ${ y\ge \sqrt{3}+x}$. That is $ y\in(x+\sqrt{3},2]$ and $ x\in[0,2-\sqrt{3})$. Consider some differential $ dx$, the probability of that $ x$ occuring is $ \frac{dx}{2}$. The probability for $ y$ is \[ \int dy=\frac{2-\sqrt{3}-x}{2}\]  Hence we integrate the product $ p(x)\cdot p(y)$. So the probability is \[ \int_{0}^{2-\sqrt{3}} \frac{2-\sqrt{3}-x}{4}\, dx=(2-\sqrt{3})^2\cdot \left(\frac{1}{4}-\frac{1}{8}\right)=\frac{7-4\sqrt{3}}{8}\] We have to multipy by $ 2$ for the case for $ x>y$. Then the answer is the complement of that probability or \[ 1-\frac{7-4\sqrt{3}}{4}=\frac{4\sqrt{3}-3}{4}\] so the answer is $ \boxed{E}$.

Ummm I don't know calculus, so any other alternate solutions?

Is this what the diagram should look like? o.O

I saw the answer was B. Anyone knows why?

\bump is there any non calc solution?  I only got as far as showing the half of the time it is guarenteed, but I couldn't find a good way to compute the other half...

[quote=archimedes1][hide]Right, so we need to find some sort of maximum horizontal distance apart the two points could possibly be.  To do this, we draw two circles which are tangent, and draw the radii to the point of tangency.  This creates a 30-60-90 triangle, and we see that the horizontal distance between the centers is thus $ \sqrt {3}$.  This is the maximum horizontal distance.
Now we use geometric probability to find out the probability that the two points are horizontally less that $ \sqrt {3}$ apart.  To do this we make a graph, with the x-axis representing the x-coordinate of the first circle, and the y-axis representing the x-coordinate of the second circle.  Then we simply find the area of the shaded region shown divided by the total region, which is $ \frac {4 \minus{} (2 \minus{} \sqrt {3})^2}{4} \equal{} \frac {4\sqrt {3} \minus{} 3}{4}$, which is $ E$.[/hide][/quote]

I believe this doesn't involve calculus

Oops missed that one... I guess I just saw the solution which shows up on the collection


Let $ X,Y$ be on circle $ (O)$ such that $ BX\equal{}XY\equal{}YC, AX\cap BC\equal{}\{M\}, AY\cap BC\equal{}\{N\}$. We will show that $ BM\equal{}MN\equal{}CN$
It's easy to see that $ BM\equal{}CN (1)$ and $ BX\equal{}\frac{1}{2}AB$
$ BC$ is the bisector of angle $ ABX$ thus $ \frac{AM}{MX}\equal{}\frac{AB}{BX}\equal{}2$
By Thales's theorem: $ \frac{MN}{XY}\equal{}\frac{AM}{AX}\equal{}\frac{2}{3} \Rightarrow MN\equal{}1/3BC (2)$
From (1) and (2) we complete the proof.

[quote="Nebraska boy"]on the side $ BC$ of the equilateral triangle $ ABC$ we construct outside $ ABC$ a semicircle $ T$ . On the segment $ BC$ two points $ M,N$ divide $ BC$ in three equal parts. Prove that the  rays $ AN,AM$ cut the semicircle $ T$ in three congruent arcs[/quote]
my solution for discution
 :play_ball:
[url=http://www.servimg.com/image_preview.php?i=247&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/image110.jpg[/img][/url]


calculus
integration
calculus computations

$\int x^{3}e^{4x}dx=\frac{1}{4}\left [ e^{4x}x^{3}-3 \int x^{2}e^{4x}dx\right ]=\frac{1}{4}\left[ e^{4x}x^{3}-\frac{1}{4}\left [ 3x^{2}e^{4x}+6\int x e^{4x}dx \right ]\right ]= \frac{1}{4}\left[ e^{4x}x^{3}-\frac{1}{4}\left [ 3x^{2}e^{4x}+\frac{6}{4}\left[ xe^{4x}-e^{4x}\right ] \right ]\right ]+C$
There may be some calculations mistakes, but this is just straightforward integration by parts.


Given a circle $ (O)$ and a line $ d$ which has not common points with $ (O)$. Let $ M$ be a point on $ d$. From $ M$, we draw tangents $ MA$ and $ MB$ of $ (O)$. Let $ H$ be the projection of $ O$ to $ d$. Let $ P,Q$ be the projections of $ H$ to $ MA,MB$.

Show that: $ PQ$ passes through a fixed point while $ M$ runs on $ d$
(Sorry for my bad English)

if PQ cuts OH at S then S is the fix point and we have
OS= ((OH)/(2)) + ((R^2)/(2*OH))
R is the radius of (O).

Since $ \angle MHO$ is right, then $ H$ lies on the circumcircle of $ \triangle MAB$ $\Longrightarrow$ projection $ R$ of $ H$ onto $ AB$ is collinear with $ P,Q$ on the Simson line with pole $ H$ WRT $ \triangle MAB.$ $ AB$ goes through the fixed pole $ D$ of $ d$ WRT $ (O),$ thus $ R$ moves on the circle with diameter $ DH.$ Let $ V \equiv OH \cap PQ.$ From the cyclic quadrilaterals $ AMHO$ and $ AQHR,$ we get $ \angle MOH \equal{} \angle MAH \equal{} \angle QRH,$ which implies that $ \angle VDR \equal{} \angle VRD$ $ \Longrightarrow$ $ \triangle DVR$ is isosceles with apex $V$ $ \Longrightarrow$ $V$ is midpoint  of $ DH.$


Do you really like that? I don't know, but it's too ugly to be difficult...

I wasn't wrong , it is trivial:
   It becomes (sum b_k*p_k)*(sum (a_k)^2)/(sum a_k*p_k)+(sum a_k*p_k)*(sum (b_k)^2)/(sum b_k*p_k)>=2 sum a_k*b_k. Now apply AM-GM for the LHS and then use Cauchy.

when it comes to ineqs I think u can never be wrong!!

thank you harazi!


probability
linear algebra
matrix
vector
geometric series

p(2)=1/3

p(x)=1/3*(1-p(x-1)) which means the probablity of winning on the xth move is a third of the probability of landing on an edge point on the x-1st move (obviously if x-1st is center, he loses).

Too lazy to find the explicit formula, but it does exist, and is very easy to find.

rofler, nice job.  To find the explicit formula, lets look for a pattern:

$ p(2) \equal{} 1/3$
$ p(3) \equal{} 1/3 \minus{} 1/3^2$
$ p(4) \equal{} 1/3 \minus{} 1/3^2 \plus{} 1/3^3$

so it appears that we have partial sums of a geometric series: $ p(x \minus{} 2) \equal{} 1/3\frac {1 \minus{} ( \minus{} 1/3)^{x \minus{} 1}}{4/3} \equal{} \frac {1}{4}(1 \minus{} ( \minus{} 1/3)^{x \minus{} 1})$

So the recursion quickly approaches $ \frac {1}{4}$ (after about $ 21$, my calculator can't tell the difference, so after 1000000 moves the difference between the actual probability and 1/4 is extremely negligible).

[quote="azn6021023"]Pentagon $ ABCDE$ with center at $ O$ is drawn such that each edge is drawn and all line segments connecting the center to the vertices are drawn.  Now, (insert name here) is playing a game with the pentagon involving moves, or moving along the line segments drawn.  He/she will win if the final move lands on the center.  Assuming that the player starts in the center and that all paths chosen are equally likely, what is the probability that the player will win after 1,000,000 moves.[/quote]

[hide="Using stochastic matrices."]
-----
Number the vertices $ A$ through $ E$ as $ 1$ to $ 5$ and label $ O$ as $ 6$. Then consider the matrix $ A_{6x6} \equal{} [a_{ij}]$, where $ a_ij$ is the probability of moving to the $ i^{th}$ vertex from the $ j^{th}$ vertex on the next move. From the structure of the pentagon, $ A$ is:
\[ \left( \begin{array}{cccccc} 0 & \frac {1}{3} & 0 & 0 & \frac {1}{3} & \frac {1}{5} \\
\frac {1}{3} & 0 & \frac {1}{3} & 0 & 0 & \frac {1}{5} \\
0 & \frac {1}{3} & 0 & \frac {1}{3} & 0 & \frac {1}{5} \\
0 & 0 & \frac {1}{3} & 0 & \frac {1}{3} & \frac {1}{5} \\
\frac {1}{3} & 0 & 0 & \frac {1}{3} & 0 & \frac {1}{5} \\
\frac {1}{3} & \frac {1}{3} & \frac {1}{3} & \frac {1}{3} & \frac {1}{3} & 0 \end{array} \right)
\]
Now let $ \vec{p_k} \in \mathbb{R}^6$ be a vector whose $ i^{th}$ entry is the probability of being at the $ i^{th}$ vertex after $ k$ moves. Then $ p_{k} \equal{} A^k\vec{p_0}$, where $ \vec{p_0}$ is:
\[ \left( \begin{array}{c} 0 \\
0 \\
0 \\
0 \\
0 \\
1 \end{array} \right)
\]
The answer to the question, therefore, is the $ 6^{th}$ entry of $ A^{1000000}\vec{p_0}$. Unfortunately, I don't have my calculator with me, so I'll leave the satisfaction of getting the same answer as the posts above to someone else.  :wink:

However, since $ 1000000$ is so large, we can assume that everything is at equilibrium; that is, the probabilities are constant and multiplying by $ A$ keeps $ p$ constant: $ Ap \equal{} p$, so we seek an eigenvector of $ A$ associated with eigenvalue $ 1$ whose entries sum to $ 1$. Doing this by hand gives me $ p$ as:
\[ \left( \begin{array}{c} 0.15 \\
0.15 \\
0.15 \\
0.15 \\
0.15 \\
0.25 \end{array} \right)
\]
Therefore the probability of being in the center approaches $ \frac {1}{4}$.

-----
[/hide]

Very nice solutions all.  Of course, there is an easier one.

[hide]Let $ P(x)$ represent the probability of landing in the center.  This is $ 1/3$ the probability of landing on a vertex, since a move to the center must come from the vertex.  Using complements, $ P(x)\equal{}(1\minus{}P(x))/3$. Solving for $ P(x)$ yields $ 1/4$.[/hide]

wasn't this the solution that facis posted??

No.  My solution used the complement property of probability.  facis used an infinite geometric series.

@azn6021023: Both of your solutions use complementary probabilities, but your solution is not correct. You assume that the probability of being in the center is constant after all moves, which is not true; you do not use the quantity 1000000 at all. facis used the series to show that the probability [i]approaches[/i] $ \frac{1}{4}$.

On a contest though, I think it would be safe to assume that the probability approaches a constant, especially since $ 1000000$ is arbitrary.

considering how quickly the probability converges on $ 1/4$, it is safe to think of the problem as if the probability is constant after a large number of moves (at least $ 20$).  However, though the probability is almost constant, it is not exactly constant.

azn6021023, what you really found is the number the probability has to converge on if it converges.  You do not address your assumptions that (1) the probability converges, and that (2) $ 1000000$ moves is sufficiently large that the limit will approximate the desired probability.  If you were asked for a proof, what you have is not really enough; as for just finding the answer, it is pretty well done.

Actually, none of these solutions quite works.  They all show the probability approaches 1/4, but the geometric series should really go out to 1000000 terms, not be infinite.  The stochastic matrix solution and azn602...'s solution both assumed that it reached equilibrium.

I agree that in a contest it's reasonable to assume that 1,000,000 is going to be large enough that you can use any of these methods (and put down 1/4 as a final answer), given that you don't have serious computing power.

I plugged matrices into Mathematica and found that the actual answer is about $ 4.171974074367847*10^{\minus{}477122}$ less than 1/4.

btw,

[hide="much simpler (and enlightening) stochastic solution"]
You can think about this problem as just being two verticies - one for the center and one for the outside.  On each turn, if you start at the center you must move to the outside.  If you start at the outside then there's a 1/3 chance you move to the center, or a 2/3 chance you stay where you are.

This leads to the matrix:
0   1/3
1   2/3

which is easier to work with.

If the problem is done with any regular polygon, the answers are all the same for any number of turns because any polygon can be turned into the two-vertex case.

[/hide]

Follow-up question:  What if it were an octagon instead of a pentagon?[/hide]

jeez, just plug 1,000,002 into Facis's explicit formula for x to get p(1,000,000)

Here's a simple solution
Solving the equation
$ x_{n \plus{} 1} \equal{} \frac {1 \minus{} x_n}{3}$
gives
$ x_{n \plus{} 1} \equal{} \frac {1}{4} \plus{} a\left( \minus{} \frac {1}{3}\right)^n$
as a solution
Plugging in $ x_n \equal{} 0$ gives
$ a \equal{} \frac {3}{4}$
plugging n=1000000 into the equation gives
$ x_{1000000} \equal{} \frac {1}{4}\left(1 \plus{} \left(\frac {1}{3}\right)^{999999}\right)$

or approximately $ \frac {1}{4} \plus{} 10^{ \minus{} 477121.38}$

CalculusLover's answer is almost the same except s/he got the sign wrong

actually, I just realized that my explicit formula isn't quite correct.  It should be $ P(x) \equal{} 1/4(1\minus{}(\minus{}1/3)^{x\minus{}1})$.  The $ x\minus{}2$ was to shift the series so that $ P(2)\equal{}1/3$ would be the first term, but we still end up with $ P(x)$ and not $ P(x\minus{}2)$.  So the actual probability is $ 1/4\plus{}1/4(1/3)^{999999}$ (which is approximately $ 1/4$) as 0714446459923 found.

I'm not quite sure I understand the stochastic stuff that well since I really don't know much linear algebra, but I think you guys have the same idea, but with matrices to powers instead of the recursive functions.

[quote="blahblahblah"][quote="calculuslover800"]Actually, none of these solutions quite works.  They all show the probability approaches 1/4, but the geometric series should really go out to 1000000 terms, not be infinite.  The stochastic matrix solution and azn602...'s solution both assumed that it reached equilibrium.[/quote]

that the matrix limit exists follows (e.g.) from the perron-frobenius theorem.[/quote]

Yes, the limit is guaranteed to exist, but you don't know that $ A^{1000000} \equal{} A^{1000001}$.  That is, you don't know that it's reached equilibrium.

@071444....: I'll check my work, but I think you have a sign error in your work when -1/3 turns into 1/3 once you plug in n = 1,000,000.  But a nice solution.  Could you explain how you got from the equation in your first line to the general form of the formula for $ x_{n\plus{}1}$ though?  (or provide a link). (This might just be something very basic about recursive formulas I've never learned...)

@Facis: yes, you're right.  It's exactly the same calculations.

[quote="calculuslover800"][quote="blahblahblah"][quote="calculuslover800"]Actually, none of these solutions quite works.  They all show the probability approaches 1/4, but the geometric series should really go out to 1000000 terms, not be infinite.  The stochastic matrix solution and azn602...'s solution both assumed that it reached equilibrium.[/quote]

that the matrix limit exists follows (e.g.) from the perron-frobenius theorem.[/quote]

Yes, the limit is guaranteed to exist, but you don't know that $ A^{1000000} \equal{} A^{1000001}$.  That is, you don't know that it's reached equilibrium.[/quote]

i didn't assert anything of the sort.

[quote="calculuslover800"]
@071444....: I'll check my work, but I think you have a sign error in your work when -1/3 turns into 1/3 once you plug in n = 1,000,000.  But a nice solution.  Could you explain how you got from the equation in your first line to the general form of the formula for $ x_{n \plus{} 1}$ though?  (or provide a link). (This might just be something very basic about recursive formulas I've never learned...)
[/quote]
The reason it's positive is because it is raised to an even power (1,000,000). I just took out the negative sign when I simplified

$ x_{1000000} \equal{} \frac {1}{4} \plus{} \frac{3}{4}\left( \minus{} \frac {1}{3}\right)^{1000000}$
$ \equal{} \frac {1}{4} \plus{} \frac{3}{4}( 3^{\minus{}1000000})$
$ \equal{} \frac {1}{4} \plus{} \frac{1}{4}( 3^{\minus{}999999})$
$ \equal{} \frac {1}{4}\left(1 \plus{} \left(\frac {1}{3}\right)^{999999}\right)$

I am not aware of any ways to solve general recursive equations either. I just tried substituting $ x \equal{} u \plus{} \frac {1}{4}$ into the equation, which made the solution apparent.

I see what you're saying now.  Yes, you're right.  I didn't look at the signs closely enough.

The way to get that explicit formula is probably the characteristic equation method.  Notice $ x_{n\plus{}1} \equal{} 1/3\minus{}1/3 x_n \implies x_{n\plus{}1}\plus{}1/3x_n \equal{} 1/3$ becomes $ a\equal{}\minus{}1/3$ (we ignore the constant in the recursive formula, and substitute $ a^y$ for every $ x_{n\plus{}y}$).  The next step, which you are asking about, is where the guess came from.  I can't really say for sure, but because of the constant in the recursive formula, I think we are supposed to guess the form $ x_{n\plus{}1} \equal{} c\plus{}d(\minus{}1/3)^n$.  Then plug in $ x_2\equal{}1/3$ and $ x_3\equal{}2/9$ and solve for the constants.  I would then check it to make sure it fits the recursive formula.


What do you mean by 'a small line passing through the center'? And how can the answer be $B_{0}R/2$ when the units are wrong?

Of course the units are correct.You probably misunderstood the unit of $B_{0}$ as tesla, the equation $B ={B_{0}}t$ says that its unit is $tesla/s$ .By the way , I meant that there would actually be no field lines at the centre.

Look, $B_{0}$ is in $\text T/\text s$. That's ok. But $B_{0}R$ is in $\text{Tm/s}$, which is the unit of the strength of an electric field, not a volt as it should be.

Additionally, I don't see how one can find the EMF induced in a point...

OK...OK.....I got it now.Question wrong ....CHAPTER CLOSED

Just a second, lets think on this ..WHAT WILL BE THE INDUCED ELECTRIC FIELD AT THE CENTRE?

[quote="ashwinkrishnan"]Consider a cylindrical region containing a uniform time varying magnetic field as $\vec{B}={B_{0}}t$with its direction allong the direction of the axis of the cylinder.A disc of radius $R$ and uniform charge distribution is placed in this field so that its axis coincides with the axis of the cylindrical region.Determine the induced EMF at the centre of the ring.

I found this in a book where the answer was given to be $\frac{{B_{0}}R}{2}$.But I feel that there should be no induced at the centre since the field lines if any, will be perpendicular to the small line passing thru the centre.[/quote]
hey, how can you define $\vec{B}={B_{0}}t$? the unit of equation on the left and right side aren't same. You should define as $\vec{B}=k{B_{0}}t$. Where $k=$(some constant)$/s$.

We've already resolved this, read the rest of the topic.


Euler
geometry
incenter
analytic geometry
geometry proposed

let $ ABC$ be a triangle, let $ D,E,F$ be the feet of the internal angle bisectors from vertices $ A,B,C$, respectively. Denote by $ I$ the incenter of triangle $ ABC$, $ H_{1}$ the orthocenter of triangle $ DEF$. Prove that line $ IH_{1}$ is parallel to the Euler line of triangle $ ABC$.

[quote="BaBaK Ghalebi"]but $ IH_1$ passes through $ O$...
so the problem is true whenever $ IO$ passes through $ H$ which isn't always true... :huh:[/quote]

No, $ IH_1$ doesn't pass through $ O$! You are wrong!

[img]http://i227.photobucket.com/albums/dd72/TPFC/unsolve.png[/img]

Barycentric coordinates kills it :wink: But i don't like such ugly solutions,so i'll try to find another proof...
Thank you.
P.S:Congratulations Fang-jh,all of your geometry problems are great.

[quote="BaBaK Ghalebi"]but I'm sure that the Euler line of triangle DEF passes through $ O$ :maybe:[/quote]

Read the problem again. $ DEF$ is not the intouch triangle. ;)

In [url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html]Clark Kimberling[/url]'s terms, the problem states that $ X\left(500\right)$ lies on the line $ X\left(1\right)X\left(30\right)$. But that does not bring us nearer to the synthetic proof...

  darij

[quote="darij grinberg"][quote="BaBaK Ghalebi"]but I'm sure that the Euler line of triangle DEF passes through $ O$ :maybe:[/quote]

Read the problem again. $ DEF$ is not the intouch triangle.
[/quote]
 :wallbash:
sorry :oops:

Similarly, consider the excenter of triangle $ \bigtriangleup ABC.$ it holds:
Let $ ABC$ be a triangle, let $ I_{a}$ be the excenter on the side $ BC, I_{a}A$ intersects $ BC$ at $ D, I_{a}B$ intersects $ CA$ at $ E, I_{a}C$ intersects $ AB$ at $ F.$ Let $ H_{a}$ be the orthocenter of triangle $ DEF.$ Prove that the Euler line of triangle $ ABC$ is parallel to $ I_{a}H_{a}$.

[quote="Zhang Fangyu"]Can anyone solve this problem?[/quote]
You mean problemS. :) Because both of this problem are still unsolved now.I'm interested in their proof too.

I have found a proof of this problem.
I will put it on my web site which I will open next.
Jean-Louis

When do you open your website?
thank you!

Erken, can you outline how you used barcyentrics to bash this one out?
I don't know much about barycentrics, not sure how to approach it.

In few days, I will open my website.
You will find a synthetic solution of the problem and more.
Jean-Louis

When you open your website, please you inform us
thank you!

My website is open to day :
http://perso.orange.fr/jl.ayme
or Google: Geometry Geometrie Geometria Jean-Louis Ayme
You will find the proof in volume 2 : le point de Gray et l'alignement Gra-I-X500
Sincerely
Jean-Louis

Could you translate your solutions into English? then post it in this forum?
thank you a lot!
 Wish you Merry Christmas! Happy the coming new year!

[quote=Erken]Barycentric coordinates kills it :wink: But i don't like such ugly solutions,so i'll try to find another proof...
.[/quote]
Does anybody have a strategy for proof with bary-bash?


Find all functions $ f: \mathbb R\rightarrow\R$ such that $ f(x^2-y^2)=xf(x)-yf(y)$

try $ y\equal{}0$
The result is $ f(x)\equal{}kx$

Could you elaborate?  $ f(x^2) \equal{} x f(x)$ does not immediately imply $ f(x) \equal{} kx$, at least to me.

It's because $ f\left(x^2\right) \equal{} xf(x)$ implies
\[ f\left(x^2 \minus{} y^2\right) \equal{} f\left(x^2\right) \minus{} f\left(y^2\right)\,.
\]
With some careful manipulation (and with $ f\left( \minus{} y^2\right) \equal{} \minus{} yf(y)$), you should be able to show that $ f( \minus{} x) \equal{} \minus{} f(x)$ for all $ x$ and that $ f$ satisfies the Cauchy equation:
\[ f(x \plus{} y) \equal{} f(x) \plus{} f(y)\,.
\]
Therefore,
\[ f\left(x^2\right) \plus{} 2f(xy) \plus{} f\left(y^2\right) \equal{} f\left((x \plus{} y)^2\right) \equal{} (x \plus{} y)\left(f(x) \plus{} f(y)\right)\,.
\]
Consequently,
\[ 2f(xy) \equal{} xf(y) \plus{} yf(x)\,.
\]
Substituting $ y \equal{} 1$, we obtain
\[ 2f(x) \equal{} f(x) \plus{} xf(1)\,,
\]
which shows that $ f(x) \equal{} xf(1)$ for all $ x$.

Sorry,here is my full solution
Thus $ f(x)\equal{}kx$ for all $ x\in R,k\in N$
$ f((x\plus{}1)^2\plus{}(x\minus{}1)^2)\equal{}f(2x^2\plus{}2)\equal{}2f(x^2)\plus{}2f(1)$
Other way,$ f((x\plus{}1)^2\plus{}(x\minus{}1)^2)\equal{}f((x\plus{}1)^2)\plus{}f((x\minus{}1))^\equal{}(x\plus{}1)(f(x)\plus{}1)\plus{}(x\minus{}1)(f(x)\minus{}1)$
it is quite is to find $ f(x)\equal{}kx$

here is good answer for your problem! :wink:
http://www.mathlinks.ro/Forum/viewtopic.php?t=150112


linear algebra
matrix
linear algebra unsolved

Let $ A \in M_{n}(\mathbb{R})$ a invertible  matrix with properties :
1) $ \det(A) \in \mathbb{Q}$
2) all $ n \minus{} 1$-minors $ \in \mathbb{Q}$.
Prove that   $ A \in M_{n}(\mathbb{Q})$.

$ \sqrt{2} \sqrt{2} \sqrt{2}$
 $ \sqrt{2} \sqrt{2} \sqrt{2}$   I think is a counterexample because $ detA\equal{}0$ and also all the minors of order $ 2$ are  $ 0$.
 $ \sqrt{2} \sqrt{2} \sqrt{2}$

Yes, you need the additional condition $ \det A\neq 0$.

  darij

I am a bit confused, only $ n \minus{} 1$ minors?
If you refer to all the minors, then we can see that $ A^{ \minus{} 1}$ is in $ M_n(\mathbb{Q})$ by Cramer's rule. Then of course $ A \in M_n(\mathbb{Q})$.

[quote="Soarer"]I am a bit confused, only $ n \minus{} 1$ minors?
.[/quote]
I believe he wanted to say "all $ (n\minus{}1)  \times (n\minus{}1)$ minors $ \in \mathbb{Q}$" and as you said, in this case $ A^{\minus{}1}\in\mathcal{M}_n {\mathbb{Q}}$ and  $ A^{\minus{}1}\in\mathcal{M}_n {\mathbb{Q}} \Rightarrow A\in\mathcal{M}_n(\mathbb{Q})$

[quote="Svejk"][quote="Soarer"]I am a bit confused, only $ n \minus{} 1$ minors?
.[/quote]
I believe he wanted to say "all $ (n \minus{} 1) \times (n \minus{} 1)$ minors $ \in \mathbb{Q}$" and as you said, in this case $ A^{ \minus{} 1}\in\mathcal{M}_n {\mathbb{Q}}$ and  $ A^{ \minus{} 1}\in\mathcal{M}_n {\mathbb{Q}} \Rightarrow A\in\mathcal{M}_n(\mathbb{Q})$[/quote]
beautiful solution...very simple :(
$ adj(A) \equal{} A^{ \minus{} 1}det(A)$ .But $ adj(A) \in {M}_n(Q)$ so $ A^{ \minus{} 1} \in {M}_n(Q)$

[quote="posabogdan"]beautiful solution...very simple :(
$ adj(A) \equal{} A^{ \minus{} 1}det(A)$ .But $ adj(A) \in {M}_n(Q)$ so $ A^{ \minus{} 1} \in {M}_n(Q)$[/quote]

Why the frowny face?  Anyway, this is identical to Soarer's proof.


Euler
function
number theory proposed
number theory

Find all the positive integers $ n$ such that $ n \equal{} \phi_{n} \plus{} 402$ where $ \phi$ is the Euler function. It is known that $ (p_1...p_k)$ are all different prime divisors of $ n$, thus:$ \phi_{n} \equal{} n.\left(1 \minus{} \frac {1}{p_1} \right).....\left(1 \minus{} \frac {1}{p_k} \right)$  ;                                  $ \phi_{1} \equal{} 1$


limit
LaTeX
abstract algebra
real analysis
real analysis unsolved

[quote="litton"]First: Show that the convergence of $\{s_n\}$ implies convergence of $\{|s_n|\}$. Does it suffice to say that $\epsilon > |s_n - s| > ||s_n|-s|$? Maybe one can use Majornans' theorem to motive it further?

Second: How do I calculate $\lim_{n \to \infty}(\sqrt{n^2+n}-n)$? It seems to me that i should be zero but I don't really know how to show it.

One can see that $\sqrt{n(n+1)}>n$ but this difference will decrase and become smaller than $\epsilon$ when $n > N \in \mathbb{N}$. Hence it should converge to zero, any thoughts? Thanks.

(I have an exam in a couple of days and I really need to get some of these things straight)

By the way, how do I install a LaTeX module into phpBB? :)[/quote]

For the first question, you're right, $||s_n|-|s|| \leq |s_n - s|$ and the RHS tends to 0 and so does the LHS.

For the second, you say $\sqrt{n^2+n}-n = n (\sqrt{1+1/n}-1)$ and $\sqrt{1+1/n}-1 = 1/2n + O(1/n^2)$. Then the limit is 1/2.

Thanks. I first wrote $\sqrt{n^2-n}-n$ and then updated the post so it looks a bit weird :).

[quote="litton"]First: Show that the convergence of $\{s_n\}$ implies convergence of $\{|s_n|\}$. Does it suffice to say that $\epsilon > |s_n - s| > ||s_n|-s|$? Maybe one can use Majornans' theorem to motive it further?
[/quote]
What is Majornan's theorem?

Hmm, I propably got the name of it wrong then. Anyhow its one of these theorems that sate that if $|a_n| < c_n$ and $\sum c_n$ converge then $\sum a_n$ converges.

I'm having some problem seeing how to apply it to this situation though, now when I think about it

I don't really know that name  :blush: In China, we simply call it 'Comparison Test'.

By the way, what do you need series here?

Yeah, good point. I just engage in name-dropping so I can sound as cool as real mathematicians ;)

[quote="litton"]Second: How do I calculate $\lim_{n \to \infty}(\sqrt{n^2+n}-n)$? It seems to me that i should be zero but I don't really know how to show it.
[/quote]
Using the fact that $a-b = \frac{a^2-b^2}{a+b}$.
$\lim (\sqrt{n^2+n}-n) = \lim \frac{(n^2+n) - n^2}{\sqrt{n^2+n}+n} = \lim \frac{n}{\sqrt{n^2+n}+n}  = \lim \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac 12$

Push it a little harder. Let $k>1.$ What is

$\lim_{n\to\infty}\left(n^k+n^{k-1}\right)^{\frac1k}-n$ ?


calculus
integration
floor function
algebra
polynomial
number theory unsolved
number theory

[b]Problem[/b]: Prove that there exists an infinitely many positive integer $n$ such that $[\sqrt{34}n]$ is a perfect square.

Cheer ~ :D

Using a fact that:Pell-equation :$x^2-34y^2=\pm 1$ has infinite integral solutions

I don't think so, because the equation $x^2-34y^2=-1$ has no solution .

[i]Hong Quy[/i], how do you prove that $x^2-34y^2=-1$ has no integer solutions?

Anyway, to [i]shimizu-nin[/i], even if the equation $x^2-34y^2=-1$ has no integer solutions, more work has to be done.
The equation $x^2-34y^2=1$ has primitive solution $(x,y) = (35, 6)$ and so has infinitely many integer solutions, which implies that $\lfloor \sqrt{34} y \rfloor = x-1$. How do you show then that the sequence $\{ x_n \}$ generated by $x_n + \sqrt{34} y_n = {(35 + 6 \sqrt{34})}^n$ contains infinitely many integers of the form $a^2-1$? This does not seem trivial to me.

The nescessary and suffices condition for $x^2-Dy^2=-1$ has integer solution is:
  $a^2+Db^2=xo ;2ab=yo$
has integer solution,whee (xo;yo) is minimal integer solution of x^2-Dy^2=1.
And if x^2-Dy^2=-1 has integer solution then call all solution (xn;yn)
->$[\sqrt{D}x_ny_n]=(x_n)^2$
Pluricomplex's problem is harder so much,i believe

If there is at least one solution to the Pell's equation $x^2 - 34y^2 = -1$, there will definitely be infinitely many solutions. My question is, what is the primitive solution to this Pell's equation? I have checked for all values of $x <1000$ and found none to work. I'm too lazy to continue checking. And if no solutions exist, how do we prove that?

But $x^2-34y^2=-2$ has infinetly many solutions, so take them and get $\sqrt{34} xy=x \sqrt{x^2+2}$. Now looking careful ($x^2+1 > x \sqrt{x^2+2} > x^2$) at it we get that $\left\lfloor \sqrt{34} xy \right\rfloor =x^2$.

[quote]But $x^2-34y^2=-2$ has infinetly many solutions[/quote].

could you show me any solutions of the equation, ?

Ups...
seems that the program miscalculated. Sorry. :blush:

I'm not too familiar wih density, so I'm not too sure whether my solution is correct.  :oops: Here's my solution anyway:

First, we note that the Pell's equation $x^2-34y^2=1$ has infinitely many solutions for $x,y$ with primitive solution $(x,y) = (35,6)$. Manipulating this equation, we get $\sqrt{34}y = \sqrt{x^2-1}$. Note that $\sqrt{x^2-1} \rightarrow x$ as $x \rightarrow \infty$, so for arbitrarily large values of $x$, we have $\lfloor (x-4)\sqrt{x^2-1} \rfloor = (x-4)x - 1 = x^2-4x-1$. Thus, $\lfloor \sqrt{34} (x-4)y \rfloor = \lfloor (x-4)\sqrt{x^2-1} \rfloor = x^2-4x-1$ for arbitrarily large $x$.

By Weyl's Theorem, $\{ \sqrt{34} (x-4)y \}$ is dense in the interval $(0,1)$, so $\exists$ infinitely many $x,y$ such that $0 < \{ \sqrt{34} (x-4)y \} < 0.16 \Rightarrow 5.83 < \sqrt{34} + \{ \sqrt{34} (x-4)y \} < 5.99$.

Thus, $\exists$ infinitely many $x,y$ such that $\lfloor \sqrt{34} ((x-4)y+1) \rfloor = x^2-4x+4 = {(x-2)}^2$ and we are done.

Thanks for nice try  :)  but I do not see why
[quote="dblues"]so $\exists$ infinitely many $x,y$ such that $0 < \{ \sqrt{34} (x-4)y \} < 0.16 \Rightarrow 5.83 < \sqrt{34} + \{ \sqrt{34} (x-4)y \} < 5.99$.
Thus, $\exists$ infinitely many $x,y$ such that $\lfloor \sqrt{34} ((x-4)y+1) \rfloor = x^2-4x+4 = {(x-2)}^2$ and we are done.[/quote]
:blush:  ???

Fact i am afraid that $\sqrt{34}(x-4)y$ is not density in  $(0;1)$ when $(x,y)$ is taken from the equation $x^2-34y^2=1$  :?:

Well, like I said, I'm not too sure of my solution.  :?
How do you prove that an expression $\{ f(x) \}$ is dense in $(0,1)$?

Here is my solution.

firstly,we will show that if n satisfy $[\sqrt{34}n]=m^2$ then we can find a positive integer k such that k>n and $[\sqrt{34}k]$ is a perfect square.

Denote :   $\epsilon=\sqrt{34}n-m^2>0.$

because $\{\{\frac{t^2+2mt}{\sqrt{34}}\}|t\in N\}$ is dense in $(0;1)$.

then there exists t such that  $0< \{\frac{t^2+2mt}{\sqrt{34}}\}<\frac{\epsilon}{\sqrt{34}}$ and denote

$k=[\frac{t^2+2mt}{\sqrt{34}}]$.  We have: $\frac{t^2+2mt}{\sqrt{34}}=k+\{\frac{t^2+2mt}{\sqrt{34}}\}<k+\frac{\epsilon}{\sqrt{34}}<\frac{t^2+2mt+1}{\sqrt{34}}$.
$\Rightarrow t^2+2mt< \sqrt{34}k+\epsilon< t^2+2mt+1$ (1)
besides,  $m^2=\sqrt{34}n-\epsilon$.(2).
Since (1) and (2) we have : $(m+t)^2<\sqrt{34}(n+t)<(m+t)^2+1$.

Nice solution [i]Hong Quy[/i]!! :D
Just a side question, we know that $\{\{\frac{t^2+2mt}{\sqrt{34}}\}|t\in N\}$ is definitely dense in the interval $(0,1)$ by Weyl's Theorem (or is it by Weyl's Criterion??), but how do you prove Weyl's Theorem (or Weyl's Criterion) in the general sense (as in $\{f(x) \}$ is equidistributed in the interval $(0,1)$ for $x = 1, 2, \ldots$ iff $f(x)$ is a polynomial with at least one irrational coefficient? Sorry, I'm very new in this and I want to find out more.

Btw [i]Hong Quy[/i], you have a small typo error in the last line of your solution. It should be $(m+t)^2<\sqrt{34}(n+k)<(m+t)^2+1$. Hence, you should use another variable other than $k$ in your 2nd line, as you have shown that $\lfloor \sqrt{34}(n+k) \rfloor$ is a perfect square, and not $\lfloor \sqrt{34}k \rfloor$. It doesn't affect the solution though.  :P


Find the reminder on the division of $ 43!$ by $ 47$, in other words:

$ 43! \equiv  ?  (\mod 47)$

I think it's nice (i know it doesn't sound very humble, but it isn't against the law to say your unharmful beliefs, is it?! :D  ), and i will post my solution in about 2 / 3 weeks.


calculus
integration
geometry
LaTeX
function
calculus computations

Well I recently learned Definite Integration but I have been struggling with some basic concepts.
I state below what I believe is true about the properties of a definite integral.Please explain to me where I am wrong .
Thank you for the help.

We know,
\[ \begin{array}{l}
 \int {f(x)} dx \equal{} \varphi (x) \\
 \int\limits_a^b {f(x)dx}  \equal{} \varphi (b) \minus{} \varphi (a) \\
 \end{array}\]

Well then , are the following relations true..?
\[ \begin{array}{l}
 \int\limits_a^b {f(x)dx}  \\
  \equal{} \int\limits_0^{b \minus{} a} {f(x \plus{} a)} dx \\
  \equal{} \int\limits_{2a}^{b \plus{} a} {f(x \minus{} a)dx}  \\
  \equal{} \int\limits_0^{b \minus{} a} {f(x)d(x \minus{} a)}  \\
  \equal{} \int\limits_{2a}^{b \plus{} a} {f(x)d(x\plus{} a)}  \\
 \end{array}\] :blush:  :D

[hide="Hint"]The integral is nothing but area under the curve in the graph of $ f(x)$ versus $ x$ within the given limits.So imagine the graph and you will get your answer. [/hide]

I hope, except the second equality, others need not be true.

It looks like you have covered the Riemann integral $ \int_a^b f(x)\,dx$ and not the more general Riemann-Stieltjes integral $ \int_a^b f(x)\,d g(x)$ ... because it seems you wrote $ d(x\minus{}a)$ and $ d(x\plus{}a)$, using them incorrectly.  So stick to the ones you know, and only use $ dx$.

well can you elaborate on the more general integral with $ dg(x)$

The First two equalities are true . ie\[ \begin{array}{l}\int\limits_{a}^{b}{f(x)dx}\\ \equal{}\int\limits_{0}^{b\minus{}a}{f(x\plus{}a)}dx\\ \equal{}\int\limits_{2a}^{b\plus{}a}{f(x\minus{}a)dx}\\   \end{array}\]

To prove the firstone put $ x\equal{} y\plus{}a$ and to prove the second one put $ x\equal{} y\minus{}a$ . As definite integration is independent of variables we get the desired result .

About the other two equalties i dont think they are true .
If i am correct consider $ \int_{1}^{2} x dx  \equal{} \frac{3}{2}$ but this is not equal to $ \int_{0}^{1} x d(x\minus{}1) \equal{} \frac{1}{2}$

Please correct me if i am wrong

$ \text{\LaTeX}$ note: The [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Math#Aligning_Equations_.28align.29]align environment[/url] works better than an array for this purpose, as it uses full-size lines and display mode naturally.
[hide="Example"]\begin{align*}\int_a^b f(x)\,dx & =\int_0^{b-a}f(x+a)\,dx\\ & =\int_{2a}^{b+a}f(x-a)\,dx\end{align*}[/hide]

The problem with those last integrals is that $ d(x-a)$ is just equal to $ dx$. The full substitution rule applies; when making a substitution, we change the limits, the integrand, and the d_. Changing one of the limits and the function without the other is guaranteed to cause trouble.

Well I state what I understood, please tell me if I am right,

\[ \int\limits_a^b {f(x)dx}\]
 probably means the integral of $ f(x)$ as $ x$ varies from $ a$ to $ b$

Then,\[ \int\limits_a^b {f(g(x))d(g(x} ))\]
probably means the integral of $ f(g(x))$ as $ g(x)$ varies from $ a$ to $ b$ , which must be equal to the integral  of $ f(g(x))$ as $ x$ varies from $ g^{\minus{}1}(a)$ to $ g^{\minus{}1}(b)$

that is,,
\[ \int\limits_a^b {f(g(x))d(g(x} )) \equal{} \int\limits_{g^{ \minus{} 1} (a)}^{g^{ \minus{} 1} (b)} {f(g(x))dx}\]

But what about\[ \int\limits_a^b {f(x)} dg(x)\]
Well if i go back to the basics, and let \[ \frac{{d\varphi (x)}}{{dx}} \equal{} f(x)\] then
\[ \frac{{d\varphi (g(x))}}
{{d(x)}} \equal{} \frac{{d\varphi (g(x))}}
{{d(g(x))}}\frac{{d(g(x))}}
{{dx}} \equal{} f(g(x))g'(x)\]
Therefore,
\[ \int\limits_a^b {f(x)} dg(x) \equal{} \int\limits_a^b {f(x)} g'(x)dx\]
Am I right??

Yes, for differentiable functions $ g$; that's practically the definition. The interesting part of the Stieltjes integral is how it handles functions that aren't classically differentiable everywhere, including such things as jump discontinuities.

This still has nothing to do with the equations in your original post.

Well can you explain why...
\[ \int\limits_{a}^{b}{f(g(x))d(g(x})) \equal{} \int\limits_{g^{ \minus{} 1}(a)}^{g^{ \minus{} 1}(b)}{f(g(x))dx}\]
is true,

And one more thing ..how can  we relate\[ \int\limits_{}^{} {f(g(x))d(x} )\]
 and
\[ \int\limits_{}^{} {f(x)d(x} )\]

[quote="anorithdialga"]Well can you explain why...
$ \int\limits_{a}^{b}{f(g(x))d(g(x})) \equal{} \int\limits_{g^{ \minus{} 1}(a)}^{g^{ \minus{} 1}(b)}{f(g(x))dx}$
is true,[/quote]
It's not.

[Edit- quote added. The rest of the post wasn't there when I first wrote my response.]

Consider the case $ f(x) \equal{} e^x$, $ g(x) \equal{} x^2$. The antiderivative of $ e^{x^2}$ is not an elementary function.
There is nothing there.

please consider cases where $ f(g(x))$ has an elementary antiderivative, then what is the answer to my question :blush:
[quote="anorithdialga"]How can we express $ \int\limits_a^b {f(x)d(x} )$ which is equal to $ \int\limits_a^b {f(g(x))d(g(x)} )$ in terms of $ \int\limits_{}^{}{f(g(x))dx}$

Is it there no general method?[/quote]

Another question,
i believe$ \int\limits_a^b {f(x)} dg(x) \equal{} \int\limits_a^b {f(x)} g'(x)dx$ , but what effect will it have on the limits when we expand with integration by parts....

Please somebody clearly explain to me, I have wasted so much time pondering over these simple matters. :oops:

[b]There is no general rule!!!!![/b]
Please stop asking before I have to bang my head into a wall.

On the second question, what makes you think integration by parts does anything at all to the limits? You're confusing it with substitution.

So is this the expansion with integration by parts??
 :oops:
$ \int\limits_a^b {f(x)} g'(x)dx \equal{} f(x)\int\limits_a^b {g'(x)} dx \minus{} \int\limits_a^b {\left( {f'(x)\int\limits_a^b {g'(x)} dx} \right)} dx$

but then the right hand remains a variable while the left hand denotes a fixed number.... :(

What is the correct expansion then :?:

The correct expression is this:

$ \int_a^bf(x)g'(x)\,dx\equal{}f(b)g(b)\minus{}f(a)g(a)\minus{}\int_a^bg(x)f'(x)\,dx.$

So the correct expansion is this:
\[ \int\limits_a^b {f(x)} dg(x) \equal{} \int\limits_a^b {f(x)} g'(x)dx \equal{} \left[ {f(x)\int\limits_{}^{} {g'(x)} dx} \right]_a^b  \minus{} \int\limits_a^b {\left( {f'(x)\int\limits_{}^{} {g'(x)} dx} \right)} dx\]
Well again I have this problem
What does $ \int\limits_a^b {f(x)} dg(x)$ mean?
does it not mean that we are integrating $ f(x)$ wrt $ g(x)$, as $ g(x)$ varies from $ a$ to $ b$
But when we are expanding the limit is applied on $ x$ i.e., as $ x$ varies from $ a$ to $ b$.
How is that justified? :huh:

That expression has a specific technical meaning which (a) is beyond your current understanding of calculus, and (b) simplifies in nice cases to something already answered in this thread. Try to understand some of what has already been said rather than asking the same questions repeatedly in hopes that the answers will change.

:o
Well, I can see that I have been annoying you guys with my stupid questions.Well maybe this is not the right place for a stupid fellow like me. :wallbash_red:

I would be very grateful if at least anyone could cite a good book clearly explaining all this :maybe:


geometry
perimeter
Support
articles
geometric transformation
rotation

Same here, like three weeks ago I predicted that the Spurs over Pistons in 6.
Here are the advantages for both teams:

Pistons:
-first of all, confidence, after the win against Miami they have a lot of confidence they can win in the Finals. You didn't see that last year since many of these guys believed the Lakers would win.
-Rasheed's defense is going to be good against Duncan since Duncan hasn't really played against a superb defender this postseason.
-Detroit's defense matches up well with SA because remember, the Sonics were the best defensive team the Spurs played and gave them the most trouble in the POs.

Spurs:
-Tim Duncan's injury is getting better by the second.
-homecourt advantage (45-5 this entire year including playoffs at home).
-The Pistons are probably going to be fatigued since they just played against Miami
-They won't give the Pistons many offensive rebounds and second chance points.

Like I predicted a while back, SA over DET in 6 even though I'm rooting for Detroit, go Rip Hamilton!

SPURS SPURS SPURS SPURS SPURS SPURS

If anyone's read the latest SLAM, dont you gotta like that bunch of funloving guys? plus they can play ball... great coach, great management, great system, great players... no mefirst people, players who tkae their roles unquestioningly and just play for the team... great defense, good offense too (something many dont give them credit for), and great fans (98% home tickets sold... and people say the spurs are a boring team... wrong... why would so many fans come out to  watch... ok maybe cause theyalways win, but also cause they are exciting as well... at times)
hmmm...most of these things seem to hold true for pistons too.... ahhhhhhhhhh!!!

edit: at the beginning of the playoffs i put pistons winning finals in 7 (yea what a backstabbing fan i am :( ) but now id say spurs in 6 or 7

bah. i don't like either team. i guess i'll root for the spurs since i don't want to see the pistons win two in a row.

I know Spurs are gonna win, but isn't it surprising that I'm rooting for Pistons after they destroyed Philly? :o

[quote="Super Naterul"][quote="sheepwarrior"][quote="Super Naterul"]I am a sonics fan so I guess I am rooting for the spurs since they beat the sonics. But if the sonics had had lewis and radmonovic or whatever they could have beaten the spurs.[/quote]

huh????? if they beat your team wouldnt you be mad?? o well... cant wait til i get bacvk to america.. il be able to see all the games!![/quote]

If the spurs win, the sonics lost to the best team. If spurs lose, then sonics lost to 2nd or worse.[/quote]

Same thing for me, that's why I want Detroit to win  :blush:

[quote="Super Naterul"][quote="sheepwarrior"][quote="Super Naterul"]I am a sonics fan so I guess I am rooting for the spurs since they beat the sonics. But if the sonics had had lewis and radmonovic or whatever they could have beaten the spurs.[/quote]

huh????? if they beat your team wouldnt you be mad?? o well... cant wait til i get bacvk to america.. il be able to see all the games!![/quote]

If the spurs win, the sonics lost to the best team. If spurs lose, then sonics lost to 2nd or worse.[/quote]

yeah! the sonics rock! (they've gotten a lot better than last year) Who do you think would win, sonics or the suns? Sonics lost 4-2 to the spurs and suns lost 4-1 to the spurs.

I'm a HUGE pacers fan

I HATE the Pistons

GO SPURS!!!!!

If they don't win I'm gonna be SOOOOOOO mad

[quote="elvenchamp777"]Im so dissapointed in the Heat sigh... WADE CMON! go Spurs and get revenge for the Heat![/quote]

Er...just the opposite for me. I'm pissed that San Antonio beat Phoenix!

Go Pistons!

I am dissapointed in the Heat too. I was more dissapointed in Tiger Woods when he missed the cut a few weeks ago though. Sorry, the second statement is off the thread.

Here is what I think:

[size=150]Result: Spurs in 6[/size]

[b]Point Guard[/b]

Chancey Billups over Tony Parker....  I would say Chancey Billups because of 1. his experiences which edges Parker's 2. How much of an important player he is to the Pistons if they want to have some sort of penetration throught the Spurs' defense, and 3. He is a great defender (although u could argue that Parker is better)

[b]Shooting Guard[/b]

Manu Ginobili over Rip Hamilton.... Few thoughts: Manu scores more than Rip [i]coming off the bench[/i] Manu is a great shooter and a good penetrator into the post. He also is a good passer. The only thing that I would hold Rip over Manu is his ability to get off the defender by the means of the screen and shooting.

[b]Small Forward[/b]

Draw: Prince and Bowen. They are both great defender, and they might have to guard one of the guards, which negates the matchup between them. They dont shoot a lot a nd score a lot, although that has not been the case recently.

[b]power forward[/b]

Tim Duncan over Rasheed Wallace.... Tim Duncan, without a doubt, is the best all-around player in the NBA. I think Rasheed would even acknowledge that fact. Although Rasheed will seriously give Duncan a heck of a task defending, Rasheed will not score as much as Tim Duncan with Bowen on his face. Plus, Duncan in the post will 1. attract the players which opens up shooters and 2. eliminates a lot of chances for Rebounds for Pistons, on both ends.

[b]center[/b]

Ben Wallace over Massenburg/Mohammed/Rasho.... Ben Wallace can overplay all of these three at the same time... that is how great he is compared to all these players from Spurs

[b]bench[/b]

Spurs over Pistons.... Have you heard of a guy named Brent Barry and Robert Horry? Robert Horry will take pressure off Duncan by shooting and occasional defense and rebound. Brent Barry is a terrific shooter that maybe Heats needed to beat Pistons. Pistons dont have much of a bench player, Lindsay Hunter might be a great player. Certainly if Darko was as half good as Ben Wallace, I would put Pistons over Spurs.

[b]Coach[/b]

Draw. They are both great. They have both won championship, with the team they are in right now. They know each other's weaknesses and strength, as well as their own. There is not much bad thing to talk about either of them.

San Antonio [b]DESERVED[/b] to win :P HAH

[quote="MithsApprentice"]That and the fact that Billups couldn't get shots up.[/quote]
bowen was a wonderful defender, including the last shot block.

I think if blame needs to be put on, it has to be on refs making such biased calls against Detroit, Rasheed, Ben, Billups, McDyss all were in foul trouble. However, Rip Hamilton also jacked up careless shots and shots that he should have made. It's okay though, they'll return to the Finals next year ;)

[quote="noneoftheabove"]i highly doubt that detroit will make finals next year after losing larry brown[/quote]

You don't know that for sure, Larry might still coach next year...

Oh no...Pistons lost...but they have played 7 wonderful games!

[quote="stupidkid"][quote="math92"]No Detroit deserved to win :P[/quote]
How can you say which team deserved to win?  It's not like one team worked harder than the other to get to the finals, and besides, Detroit won last year so I don't see whats wrong with the Spurs winning.[/quote]
I was just saying that because Tunio said that the Spurs desereved to win and i like the Pistons so i was just joking around.

[quote="tetrahedr0n"]In terms of which team worked harder to get here, the answer is obviously the Pistons, who had to play two last series without home advantage, and it went to 7 games in both. Also, the Pistons have a 7 man rotation, compared to the Spurs 9-10 man rotation. So in terms of who worked mroe, the Pistons deserve it more. Of course, one could argue that the Spurs didn't need to work as hard beacause they were better.

As for Larry Brown, no, he won't be back. Keep in mind that that doesn't mean very much. The Pistons played nearly 20 games without him this year, so a good fifth of the season. He caused more harm than good during the Miami series with his various Cleveland rumors, etc... If the Pistons replace him with someone like Flip Saunders, I could see them making another championship run. I'd be concerned if I was a Spurs fan - Horry is a free agent and there are rumors that he wants to join Shaq. (Given, they are just rumors.) And without Horry, the Spurs would not have won this championship.[/quote]

Larry Brown was really tactless to talk about going to Cleveland during Detroit's playoff run. As long as Detroit keeps its players, they'll probably be fine.

There is no doubt Robert Horry was instrumental in the series win. A Robert Horry addition in the Heat lineup would give them another weapon vs. detroit. the penetration --> 3's killed the pistons and there's no doubt that horry can make 3 pointers.

Well I'd have to disagree that Brown's talk of going to Cleveland actually hurt the Pistons. I mean sure that got quite a bit of media attention but as far as how well they played, I didn't see too much of a shift.

Also if Horry were to join Miami, I'd be less worried for Detroit than San Antonio. San Antonio doesn't really have another 3-shooter (sure Tony Parker isn't bad) that matched Horry, in timing and the influence that his shots had on team morale - both ways. Detoit has its own long shooters (Billups, Hamilton, R. Wallace...) but they just couldn't get their shots in game 7. They definitely need to be more consistent.

I'd personally like to see Horry on Houston's lineup. I don't exactly like him since he did sink Detroit this year (remember game 5?) but it would be interesting to see what he could do with TMac and Yao.

During the first half I thought that there was no way that Detroit could lose, but they did however. Both sides had their lucky and unlucky plays, and it was a good game. Hamilton did put up several horrible shots where he could have just passed to open players with a much better position.


So which one do you use?
I've actually never used Linux before, but whatever.

Yeah it should  :wink: .

I've used all of them but I generally use XP.

I'm happy to say, as of now, my only windows is virtualized, on top of Linux. I have one OS X installation still, but I only use it about once a month. Then the remaining computers run Linux - one Ubuntu, many Gentoo.

Yay for gentoo! (I run two gentoo systems (well, they dual boot XP, but I never use it except when linux breaks and it's required for homework (typing Japanese is hard to do without X)))

Mac FTW ... windows apps run under DARWINE... which is sometimes a little buggy... :D

I like macs only for their simplicity.  Nothing so distracting like Windows.
But I have 10.2&3 so I don't know what 10.4&5 are like.
Happy $ \pi$ day!

Oh, I have 10.4... I would really love to upgrade to 10.5 though.

This topic reminds me of this:

[youtube]h4iyksLeo7w[/youtube]

It depends what mood I am in really. I generally go for XP or Vista though. :roll:

don't you have a mac?
Or was that someone else?

The problem with these comparisons is that it leaves out the consideration of what the computer is used for. For an average consumer, windows is great. It can game, do work, or almost anything you want. For a more technophobic consumer, macs are great. They have worked extensively on their user interface to make it easy to use, something that windows has not done and linux has left out completely (with the exception of a few specialized distributions). Mac and Windows are fighting for the desktop PC. Linux is not.

Mac OS X 10.5 and Windows XP. My laptop is running the former, the family desktop the latter.

Lmao at the vid, btw. It made my day.

Ubuntu and Fedora distros of Linux are really good. I don't know about the others. I like Ubie the ebst :)

Haha, I have a thing against Vista... too bad my laptop came with it, and I don't think some of the hardware is supported by XP...

I don't hate Macs as much as I used to, though.

Oh, I also have a sick Windows 98 computer!  :lol:

[quote="The Zuton Force"]Oh, I also have a sick Windows 98 computer!  :lol:[/quote]

We used to have Windows 98...before the monitor died and we had to get a new 22" widescreen computer  :lol:

[quote="7h3.D3m0n.117"]We used to have Windows 98...before the monitor died and we had to get a new 22" widescreen computer  :lol:[/quote]

I'm telling you, I did [b]not[/b] break that moniter!

We use all three, but the games already installed on Linux are my favorite  :) They have lots of penguin games.

[quote="davidyko"]Dude, we had a Windows 95 until 2003. Beat that.  :wink:[/quote]

I think my grandpa is still using Windows 95.

I'm typing this on a Mac laptop (10.4). The room I'm in has four unplugged old computers: a G4 tower (that I never upgraded to OS X), an iMac that's a little older than that, a Powerbook (I'm not sure how old exactly), and an ancient black-and-white Mac Plus. The latter still works- it's incredibly robust hardware.

im on my MacBook right now lol. i think macs are much better than windows or linux lol


MATHCOUNTS
probability
AMC
AMC 8
AMC 10
AMC 12
AIME
\/closed

So currently, this year, I've taken
Mathcounts Basics
Introduction to Counting and Probability
Introduction to Number Theory
and Introduction to Geometry.
Currently, at school, I got 28 and 29 after taking two sprint rounds, compared to last year's 8 and 15.
Unfortunately, I got a lower score on the AMC8 than what I would've expected (21, but misbubbled, so 20). And usually, I can score up to 130s on AMC10 and 105s on AMC12 (practice). I took one practice AIME and got a 4.
Any recommendations?

[quote="cheeseyicecream"]If you haven't yet learned trig, then you may want to consider the class Precalculus. It will help with AMC 12 and AIME (and USAMO).[/quote]

I'd look at Algebra 2 or 3 first; use the diagnostic tests to determine which is right for you.

I do not know much trigonometry, so I was wondering how much the class Precalculus would help me on the AMC contests.

I am taking Adv. Alg/Trig at school right now but we have not gotten to the trig segment yet.

Would you recommend Precalculus starting in March?

Thanks.


Brad bicycles from home at an average speed of 9 miles per hour until he gets a flat tire.  With no
way to fix the tire, Brad walks his bike back home by the same route, averaging 3 miles per hour.  If the entire round trip of biking and walking took a total of  6 hours, what was Brad's average speed in miles per hour for the entire round trip? Express your answer as a decimal to the nearest tenth.

Let $ t$ be the number of hours Brad spent biking.
Then, he spent $ 3t$ hours walking (because his biking speed is three times his walking speed).
$ 3t\plus{}t\equal{}4t\equal{}6$
$ t\equal{}\frac{3}{2}\equal{}1.5$
Thus, he traveled $ 1.5\cdot 9\plus{}1.5\cdot 3 \cdot 3\equal{}13.5\plus{}13.5\equal{}27$ miles.
So his average speed is $ \frac{27}{6}\equal{}\frac{9}{2}\equal{}4.5$ miles per hour.

Harmonic mean of 3 and 9 is 4.5

(2/(1/3+1/9))


I was doing Alcumus, and when I clicked Submit, it showed up like the print screen here. I tried going back to the homepage and trying again, but it kept on doing that. I'm pretty sure something's happening with the server, but I don't know what. Any ideas?

Note: the print screen was resized so it looks kind of weird.

They're probably updating.  It's not working for me either.

This has been fixed. In general, when you see something like this we are working on the system and already know about the issue. If it's not fixed within a half hour or so, please report it.


If you take logs of both sides, you get

$ (\log 2)^2\minus{}(\log 3)^2\equal{}(\log 3\minus{}\log 2)*\log x$

You should be able to finish from here.

yes i just did it.

except i used $ a^{\log{a}}\equal{} 10^{\log^2a}$

but i guess that's the same thing as you said :D


My school might start  doing the AMC tests next year and I'll be in 7th grade hopefully taking the AMC 8.  I want to prepare somehow.  I did the sample problems from the AMC 10 and 12 (had to guess on two in the AMC 12 sample, but the AMC 10 sample questions were easy) on their website, and I got a perfect on those, but you never know what new tricks they are going to put on the tests.

[quote="tokenadult"]Hey, forum moderator, what say we put pointers to more resources in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8715]Links to AMC Resources[/url] sticky thread that you started a while ago, and then answer all these other threads with pointers to that thread? There seem to be about a bazillion (conservatively estimated) questions about how to prepare for different levels of the AMC exams in the last month.

I'll try to post more information in the thread linked above, when I have time to dig it up and write it down.[/quote]

I notice that you say 'moderator' as opposed to 'moderators'.  Are you hinting at something ([i]like, that the other two moderators aren't doing their job?[/i])

Anyway, I'll see if I can put something together.

[quote="mathfanatic"]
I notice that you say 'moderator' as opposed to 'moderators'.  Are you hinting at something ([i]like, that the other two moderators aren't doing their job?[/i])
[/quote]

haha, yeah...in lots of forums there's multiple mods listed and one actually doing stuff :-) But it's all good...mods are there to help ppl anyways...and it's good t ohave backups when ur absent for long periods of time or something...

Alright, I've been a diligent moderator over the past hour, so I made a page describing [url=http://goodlymaths.0catch.com/contests/amc/amc_how_to_prepare.htm]how to prepare for the AMC contests[/url].  Any suggestions for what to add to this page should be posted here.

- mathfanatic

P.S. Does anyone have any objections to me posting this page on my math resource website, as opposed to in a (long) thread on the AoPS forum?  I think it's more convenient, because it doesn't move (so you don't have to hunt for it), and so on.


[quote="Remike"]May I send the solution of one problem? And if I solve another problem then I will send the solution of it. Or should I send everything at the very end?[/quote]The regulations specifically say you can send solutions as early as you like. For each problem only the latest (time wize) solution will be considered.

In the results tables a cell with "-" in it means that the contestant didn't send anything to that problem. So yes, you can send each problem at a time. But if you for example submit problem 1, and then in another email problems 1 and 2, problem 1 solution will be considered only from the latest email (you can correct your solution in time - if you figure that it's not correct).


SOLUTION-

$ \lim_{x\to 0}\frac {(x \plus{} 1)\sqrt {2x \plus{} 1} \minus{} (x^{2} \plus{} 2009)\sqrt [3]{3x \plus{} 1} \plus{} 2008}{x}$

Using L Hospital rule,we have,

$ \lim_{x\to 0}\frac {x\plus{}1}{\sqrt{2x\plus{}1}} \plus{} \sqrt{2x\plus{}1} \minus{} \frac{x^2\plus{}2009}{3x\plus{}1}^\frac{2}{3} \minus{} 2x(3x\plus{}1)^\frac{1}{3}$

$ \equal{} 1\plus{}1\minus{}2009$

$ \equal{}\minus{}2007$


Even though I highly doubt it is true, I was just watching Russian news and some Ilyin guy says he found a simple proof of FLT. Anyone here hear about that?

I haven't heard about it yet. I'll do some research about it though.

I read something like this in another forum. I cannot remember the name of the guy but it was long... :lol:

Imagine what would Wiles do if it is true.

Considering the hundreds and hundreds of false proofs that have been formulated over the hundreds of years that this problem was open, my reaction is sceptical to say the least.

mine is sceptical too but it is also written in wikipedia. Btw, Wikipedia is made by any people that want to make an entry right?

Yeah, anyone can edit Wikipedia entries.  If changes it to something inappropriate it is usually changed very quickly though.

[quote="bchui"]Anyway can we find the "New Proof" ? Does the pdf file available on the web in detail? :roll:[/quote]Of course not. Then we would see it's false. :)

If this is about  Omsk reserarcher, it is, IMO all bunk. From news sources, it appears that the the paper has not been published  in any journal, and the summary in  Novaya Gazeta, to put it in a nice way, is full of inaccracies, so much so that  one can can stop reading after a first few items like "eternal engines" ...:)

Of course, a few "mathematican" cited by Novaya Gaseta who "found no faults" with the proof, are not even known .. according to one news source .. about mathmaticans vouching about the proof:

[quote]However, Leonin Gorynin was not listed among the members of the Russian Academy of Science. Professor Sergei Chukanov, a doctor of technical sciences from Omsk who works at a department that deals in automated data processing, is clearly no specialist in number theory.
[/quote]

Well that on it's own may not be a proof for invalidness... I mean on our university, there even are no number theorists! If I would have to ask someone, it would be an algebraician or something, by lack of 'real' number theorists.

However I'm very curious to see that mysterious document :roll:

Did the techniques he used existed in Fermat's time btw? As I never really believed the Fermat story...

Yes, its the Omsk guy I was talkin about. I dont know if there is online document I never saw it, I saw it on Wikipedia but there is no document there and I saw it on Russian tv.

here is the ref to his proof (in russian). It's [b]obviously[/b] wrong. You can ask to find mistakes even a school-guy. But this can be due to to the journalist's mistakes though.

http://2005.novayagazeta.ru/nomer/2005/61n/n61n-s00.shtml

BTW -Professor Andrew John Wiles of Princeton University was awarded the Shaw Prize in mathematics for his proof of Fermat's Last Theorem ..last Fridaay in Hong Kong.   :!:

Also this is what mascownews.com says:
[quote]A summery was published by Novaya Gazeta, but a number of inaccuracies makes the text too susceptible to criticism. It’s also not exactly clear what has been proven at this point. The version (obviously false) published by Novaya Gazeta fully reproduces the remarks published on a personal Website dated from 2003 and signed by a different name. Then again, it’s possible that there a lot more such proofs.[/quote] :huh:

But this is just funny: In  [url=http://new.mn.ru/english/issue.php?2005-33-22]News you can afford to Miss :)[/url][quote]n engineering professor from Siberia's Omsk, Alexander Ilyin, proved the ever-elusive Fermat's Theorem live before TV cameras. Last week, NTV aired 15 minutes of the professor proving the theorem. The proof has yet to stand up to the tests of the Russian Academy of Science, and if it holds up, it will be published in one of the academy's scientific journals  Pierre de Fermat came up with the theorem in 1660. Since then, mathematicians have been struggling to prove it. Andrew Wiles finally proved the theorem in 1994, basing his proof on the Shimura-Taniyama-Weil Conjecture. But mathematicians from around the world are still trying their hand.

Those Russians are always coming up with something. Couple of years back, folks from Armenia astonished the world with their 100 percent effective AIDS cure. Oh, wait, Armenians aren't Russian.... Still, Lenin invented the light bulb, the steam engine, and I'll bet he even proved Fermat's Last Theorem back in grade school. What, like it's hard? I never could figure out why Yutaka Taniyama had to get into elliptic curves just to prove a theorem that would help prove Fermat's theorem. He committed suicide, by the way. So the theorem states that the equation xn+yn=zn has no non-zero integer solutions for x, y, and z where n > 2. What's there to prove? Let's have x=1, y=2, and z=3. For n=3, you get 1+8 =27. See? That wasn't so hard. I'd explain further, there's just not enough space on page 11 .[/quote] :lol:

BTW an old classic is to try out (may be around April 1st) with someone who  trusts his/her calculator a lot:

to calculate:

$(1782^{12} + 1841^{12})^{(1/12)} $ to get $1922$ which proves:
$1782^{12}+1841^{12} = 1922^{12}$
:lol:

Hahaha. Is it really close?

Many proofs have been attempted but assuming unique prime factorization. This is not true in the realm of complex numbers. But I don't see how that ties in with Fermat's Last.

[quote="gyan"]$(1782^{12} + 1841^{12})^{(1/12)} $ to get $1922$ which proves:
$1782^{12}+1841^{12} = 1922^{12}$
:lol:[/quote]

Here we used the obvious fact that the sum of an even integer and an odd integer is an even integer ;)

[quote="Arne"][

Here we used the obvious fact that the sum of an even integer and an odd integer is an even integer ;)[/quote]
True, of course, :)   if  one "trusts" the calculator and does not get  "side-tracked" by  these trivial facts :)

 :rotfl:

Stan  - The part I found funny is that there are enough gullible people in the word who are ready to belive any story without being least bit skeptic.  The item I posted, was not poking fun at   Russian  Newpaper but rather, t at those people are ready to believe  anything without being skeptic.  .. and  News Papers publishing report without even checking the basic facts like  if the guy claims "xyz of  abc academy has found this credible" at least check if xyz is really  a member of "abc" academy..

If u mean me since I first posted here then I can say that I never believed the story but still have heard it and wanted to post about it. Btw, Stan, do u speak on Russian as I see u have a Russian flag? Lets go talk in the Russian forums, no Russians visit here anymore, except me, especially that myth is gone now.

Hello, No, I certainly did no mean you...It is some  newspapers, when they report a story, they ought to do some basic checking of facts.

[quote="stan"]I don't like posts like the one from Gyan, where he quotes  "News you can afford to Miss". The english version of The Moscow News is http://english.mn.ru/english/ and not the http://new.mn.ru/english/. So I think the latter is a site that discredits the well known Russian newspaper. The author as follows from the article is absolutely uneducated and inadequate person and I believe knows nothing neither about maths as a whole, nor about Fermat's theorem in particular. And, Gyan, I wonder what is funny in this stupid "article"? If you find it funny I bet, guy, you lack the sense of humour.[/quote]

I think you're the one without sense of humour.... Especially when you read this part:

[i]So the theorem states that the equation xn+yn=zn has no non-zero integer solutions for x, y, and z where n > 2. What's there to prove? Let's have x=1, y=2, and z=3. For n=3, you get 1+8 =27. See? That wasn't so hard. I'd explain further, there's just not enough space on page 11 .[/i]

See? It's actually similar to what Fermat has written, that he didn't have enough space so.. That's why it's so funny.. :D

That's a good example of a "proof by example": the author gives only one case and "without loss of generality" states that the other 2,154,709 cases are analogous.

By the way, the LaTeX code is

[code]x^n+y^n+z^n[/code]

(with dollar signs around it) for

$x^n+y^n+z^n$


Four cashews and three walnuts weigh 10 grams; two walnuts and six pecans weigh 13 grams; and one cashew, one walnut and two pecans weigh 6 grams.  Assuming each cashew has the same weight, each walnut has the same weight and each pecan has the same weight, how much does one walnut weigh in grams?

We write some equations:
$ (1)$ $ 4c \plus{} 3w \equal{} 10$
$ (2)$ $ 6p \plus{} 2w \equal{} 13$
$ (3)$ $ 4c \plus{} w \plus{} 2p \equal{} 6$

From $ (2)$ and $ (3)$ we get:
$ (4)$ $ 2p \minus{} 2c \equal{} 1$

From $ (1)$ and $ (3)$ we get:
$ 6p \minus{} c \equal{} 8$
$ (5)$ $ 12p \minus{} 2c \equal{} 16$

From $ (4)$ and $ (5)$ we get:
$ p \equal{} \frac32$

so $ 9 \plus{} 2w  \equal{} 13$
$ 2w \equal{} 4$
$ w \equal{} \boxed{2}$


Let triagle $ ABC$. Prove that
$ \sum_{cyc}\frac{(b\plus{}c\minus{}a)^6}{c[3(a\minus{}b)^4\plus{}3c^4\plus{}10(a\minus{}b)^2c^2]}\geq\frac{a\plus{}b\plus{}c}{3}$
Have fun! It is great!

Not nice! Sorry because I haven't just finded a solution for it. But I think it can solve by setting $ a\equal{}x\plus{}y,b\equal{}y\plus{}z, c\equal{}z\plus{}x$ :D

Please! Can you post your proof. My solution is very nice! :roll:

nobody done? [b] arqady, vasc, Jichen,.. [/b]where are you? :(

Setting $ a\equal{}y\plus{}z,b\plus{}x\plus{}z,c\equal{}x\plus{}y$. We need to prove that
$ \sum_{cyc} \frac {x^6}{(x\plus{}y)(x^4\plus{}y^4\plus{}10x^2y^2)}\ge \frac {x\plus{}y\plus{}z}{24}$
Because $ 3(x^2\plus{}y^2)^2\minus{}(x^4\plus{}y^4\plus{}10x^2y^2)\equal{}2(x^2\minus{}y^2)^2\ge 0$,
so $ 3(x^2\plus{}y^2)^2\ge x^4\plus{}y^4\plus{}10x^2y^2$
Thus, the inequality will be true if
$ \frac {x^6}{(x\plus{}y)(x^2\plus{}y^2)^2}\ge \frac {x\plus{}y\plus{}z}{24}$
Apply AM_GM inequality we have
$ \frac {x^6}{(x\plus{}y)(x^2\plus{}y^2)^2}\plus{}\frac {x\plus{}y}{16}\ge \frac {x^3}{2(x^2\plus{}y^2)}$
Thus,
$ \sum_{cyc} \frac {x^6}{(x\plus{}y)(x^2\plus{}y^2)^2}\ge \frac 12 \sum_{cyc} \frac {x^3}{x^2\plus{}y^2}\minus{}\frac {x\plus{}y\plus{}z}{8}$
But by AM_GM inequality
$ \frac {x^3}{x^2\plus{}y^2}\equal{}x\minus{}\frac {xy^2}{x^2\plus{}y^2}\ge x\minus{}\frac {xy^2}{2xy}\equal{}x\minus{}\frac {y}{2}$
So,
$ \sum_{cyc} \frac {x^3}{x^2\plus{}y^2}\ge \frac {x\plus{}y\plus{}z}{2}$
$ \sum_{cyc} \frac {x^6}{(x\plus{}y)(x^2\plus{}y^2)^2}\ge \frac {x\plus{}y\plus{}z}{24}$
The inequality has proved. :wink:
P/s: The inequality of you is nice, and my solution is nice,too :D

I used T2's lemma after that i used AM_GM a lot  and the problem solved :wink:


One can show that for every quadratic equation (x-p)(x-q)=0 there exist constants a, b, and c such that the equation (x-a)(b-x)=c is equivalent to the original equation, and the faulty reasoning "either x-a = c or b-x = c" yields the correct answers"x=p or x=q".

Determine constants a, b, and c, with c not equal to 0, so that the equation (x-19)(x-97)=0 can be "solved" in this manner.

Spoiler:

[hide]The constants are:

a = 96

b = 20

c = -77

Since we are told that "either x-a = c or b-x = c" yields "x=p or x=q", we know that p = a+c, q = b-c

Substituting 19 and 97 or vice versa into these equations:

19 = a+c

97 = b-c

or

97= a+c

19=b-c

Solving for a or b, we have:

a = 19 - c

b = c + 97

or

a = 97 - c

b = c + 19

Now, we also have the equation (x-a)(b-x) = c.  Subsituting for a and b:

(x - 19 + c)(c + 97 - x) = c

or

(x - 97 + c)(c + 19 - x) = c

Expanding:

-x:^2: + 116x + c:^2: + 78c - 1843 = c

or

-x:^2: + 116x + c:^2: + 78c - 1843 = c (same thing)

Simplifying, we have

x:^2: - 116x + (1843 - c:^2: - 77c) = 0

Since we know that THIS equation is equivalent to (x-19)(x-97) = 0, we can equate the two:

x:^2: - 116x + (1843 - c:^2: - 77c) = (x-19)(x-97)

x:^2: - 116x + (1843 - c:^2: - 77c) = x:^2: - 116x + 1843

- c:^2: - 77c = 0

c:^2: + 77c = 0

c(c+77) = 0

c = 0 or -77.

But we were told that c wasn't 0.  Therefore, c = -77

Plugging c into these equations:

a = 19 - c

b = c + 97

gives

a = 96

b = 20

Hence our answer is a = 96, b = 20, c = -77.[/hide]

my my mathfanatic. use your head for once!

[hide]jk! correct!   [/hide] <---

more later my mouse is running out of batteries!

[quote="s0mp"]more later my mouse is running out of batteries![/quote]
 :shock: bettery-operated mouse!?

its an optical.
theres not much to say about this problem, its just sifting through equations...

I also have optical but it's not battery-operated...

optical cordless mice take batteries.  They are also heavy and I don't like them.

eh...i could care less about the mouse...i juss happen to have the best mousepad in the world...

my optical gets the power from the computer, but my other one's rechargable, cool? :)

This discussion has a lot to do with math, doesn't it?


modular arithmetic
inequalities
induction
algebra
polynomial
number theory proposed
number theory

If $ x$ and $ y$ are two distinct positive integers, and $ k$ is a positive integer, then $ k\mid\varphi\left(x^{k} \plus{} y^{k}\right)$.

I have seen this fact being proven using Zsigmondy's theorem; actually there is a not quite hard elementary proof (unless I am mistaken).

What is also nice is that, if $ x > y$, then $ k\mid\varphi\left(x^{k} \minus{} y^{k}\right)$, and, even stronger, $ k\mid\varphi\left(\frac {x^{k} \minus{} y^{k}}{x \minus{} y}\right)$.

Good luck!

  Darij

Variations on the same theme :):

First of all, if $u\mid v$, then $\varphi(u)\mid \varphi(v)$, so we may as well assume that $(x,y)=1$ in both problems (we apply the property above to $u=a^{k}+b^{k},v=x^{k}+y^{k}$ in the first problem, and to $u=\frac{a^{k}-b^{k}}{a-b},v=\frac{x^{k}-y^{k}}{x-y}$ in the second problem, where $a=\frac x{(x,y)},b=\frac y{(x,y)}$).

We assume that $x>y$. $x^{k}\equiv-y^{k}\pmod{x^{k}+y^{k}}$, so $(xy^{-1})^{2k}\equiv 1\pmod{x^{k}+y^{k}}$, where $y^{-1}$ makes sense because $(y,x^{k}+y^{k})=1$. Suppose now that $x^{k}+y^{k}\mid x^{n}-y^{n}$. Then $x^{k}+y^{k}\mid x^{(2k,n)}-y^{(2k,n)}$, which means that $(2k,n)$ must be greater than $k$. This implies that it's $2k$, so $2k$ is the order of $xy^{-1}$ in the group of units modulo $x^{k}+y^{k}$. Since this group has order $\varphi(x^{k}+y^{k})$, we get $k\mid 2k\mid \varphi(x^{k}+y^{k})$, as desired.

For the second problem (stronger version) the same idea works. Simple inequalities show that if $\frac{x^{k}-y^{k}}{x-y}\mid x^{n}-y^{n}$, then $n\ge k$, and since $(xy^{-1})^{k}\equiv 1\pmod{\frac{x^{k}-y^{k}}{x-y}}$, we find that $k$ is the order of $xy^{-1}$ modulo $\frac{x^{k}-y^{k}}{x-y}$, so $k\mid \varphi\left(\frac{x^{k}-y^{k}}{x-y}\right)$.

[quote="{x}"]@darij: Can you point us to the elementary proof ?[/quote]

See Grobber's post for an elementary proof... because mine was not more elementary, but [i]far[/i] more complicated. I began by reducing to the case when x and y are coprime, then strengthening the assertion to "the order of $\frac{x}{y}$ modulo $x^{k}+y^{k}$ is divisible by k", then reducing this assertion - by multiplicativity -, to the case of k being a prime power, and then proving it for $k=p^{a}$ by induction over a. Probably you have got an idea now why I am not intending to post the solution in details here... In fact, the problem (the $k\mid\varphi\left(x^{k}+y^{k}\right)$ one) was told to me by ZetaX with the comment that it was probably quite hard (he had a proof using Zsigmondy and one using cyclotomic polynomials), so I handled it like a hard problem and got a long solution...

  darij

Recently i proposed a similar ques as the second one which is a subcase.

To prove $ k\mid\varphi\left(x^{k}\minus{}y^{k}\right)$ needs only to prove it holds for every $ k\equal{}p^{n}$ where p is prime.

the most vital part is to prove for the case n=1, then by induction we get the whole proof.

first we can easily prove $ (x\minus{}y,\frac{x^{p}\minus{}y^{p}}{x\minus{}y})\equal{}p^{m}$ m>=0

suppose $ q\neq p$ is prime and $ q\mid \frac{x^{p}\minus{}y^{p}}{x\minus{}y}$, set r to be q's primitive root and $ x\equiv r^{x'} mod q$ $ y\equiv r^{y'} mod q$ $ p\equiv p' mod (q\minus{}1)$ ,because q can not be a factor of x-y, so $ x' \neq y'$. Since $ r^{p'x'}\minus{}r^{p'y'}\equiv 0 modq$ then $ q\minus{}1\mid p'(x'\minus{}y')$ since |x'-y'|<q-1 so $ (p',q\minus{}1)\equal{}p''\neq 1$. If q<p, we get p=k(q-1)+p'=up'' a contradict to p's primality. Thus p'=p and $ q\equiv 1mod p$ , thus $ p\mid \varphi(q)$, but $ q\mid \frac{x^{p}\minus{}y^{p}}{x\minus{}y}$which means  $ p\mid\varphi(\frac{x^{p}\minus{}y^{p}}{x\minus{}y})$
If q is the only factor of $ \frac{\left(x^{p}\minus{}y^{p}\right)}{x\minus{}y}$ it's obvious the order must be greater then 2.


We do have a few things in the works.  There's the independent study, which we expect to offer in the fall.  We are also developing some courseware, though it will be a while before it's available, and we likely won't catch up to your level (i.e. it would have been perfect for you to look at 6 months ago).  We expect to finish our first offerings of courseware by the end of the year.

The independent course is a combination of taking a couple of our online classes plus a series of assignments in various textbooks.  It will emphasize problem-solving, i.e. be outside the standard curriculum.


http://mathlinks.ro/Forum/viewtopic.php?t=47701

[quote="maxal"]$1^n + 2^n + 4^n$ is obviously prime for $n=1$.

Suppose that $n>1$.

Note that $1^n + 2^n + 4^n = \frac{8^n-1}{2^n - 1}$.

Let $n=3^k m$ where $m$ is not a multiple of $3$. Then $8^{3^k}-1=2^{3^{k+1}}-1$ divide $8^n-1=2^{m3^{k+1}}-1$ but not $2^n-1=2^{m3^k}-1$.

Hence, there exists prime $p\leq2^{3^{k+1}}-1$ dividing $\frac{8^n-1}{2^n - 1}$.
If $m>1$ then \[p \leq 2^{3^{k+1}}-1 < 2^{2m3^k} - 1 < \frac{8^n-1}{2^n} < \frac{8^n-1}{2^n-1}\] implying that $p$ is a non-trivial divisor of $\frac{8^n-1}{2^n - 1}$. Therefore, $m=1$ and $n=3^k$.[/quote]
I can't understand this proof,especially these last steps:[quote="maxal"]

Hence, there exists prime $p\leq2^{3^{k+1}}-1$ dividing $\frac{8^n-1}{2^n - 1}$.
If $m>1$ then \[p \leq 2^{3^{k+1}}-1 < 2^{2m3^k} - 1 < \frac{8^n-1}{2^n} < \frac{8^n-1}{2^n-1}\] implying that $p$ is a non-trivial divisor of $\frac{8^n-1}{2^n - 1}$. Therefore, $m=1$ and $n=3^k$.[/quote] smb help me!

Ok, let me give an overview of what [i]maxal[/i] meant...

Firstly, the problem asks us to prove that if $1^n + 2^n + 4^n$ is prime, the only possible way is for $n$ to be a power of $3$, thus letting $n=3^km$, it's equivalent to proving that if $1^n + 2^n + 4^n$ is prime, then $m$ must be $1$. [i]maxal[/i]'s approach is proof by contradiction, assuming that $m>1$ and then show that this implies that $\exists$ a non-trivial divisor of $1^n + 2^n + 4^n$, which then means that $1^n + 2^n + 4^n$ is not a prime, thus deriving a contradiction. What he had done is to show that we can find a $p$ that satisfies $1 < p < 1^n + 2^n + 4^n$, which is the non-trivial divisor we want for our contradiction.

I hope this makes [i]maxal[/i]'s solution clearer to you. :P