If you flip a fair coin four times, what is the probability of flipping at least three heads in a row? Express your answer as a common fraction.

There are 3 ways to do this.

They are HHHT, THHH, and HHHH.

There are 16 total possibilities, so the probability is 3/16.

Also, we could do:

$ \left(\dfrac{1}{2}\right)^3 \plus{} \left( \dfrac {1}{2}\right)^4 \equal{} \boxed{\frac {3}{16}}$

since we add the probability of getting heads 3 times to the probability of getting heads 4 times.

EDIT: ducky, I meant the fraction, not 1. I forgot parentheses, just added them.
EDIT 2: Fixed parentheses.

[quote="isabella2296"]Also, we could do:

$ \frac {1}{2}^3 \plus{} \frac {1}{2}^4 \equal{} \boxed{\frac {3}{16}}$

since we add the probability of getting heads 3 times to the probability of getting heads 4 times.[/quote]

Is there really a point doing $ 1^3$ and $ 1^4$ since it's just one?


function trigonometry calculus derivative

Assume that a block lies on a sufficiently long incline. If one side of the incline is gradually lifted up from $ 0^\circ$ to $ 90^\circ$, then draw and explain the graph which represents the frictional force acting on the block as a function of the inclined angle. Here the coefficient of static friction is larger than the coefficient of kinetic friction.

Since friction is uN (where N is the normal force and u is the coefficient) or umg, and $ g\equal{}g\cos{\theta}$, then the graph is simply $ f\equal{}umg\cos{\theta}$

@Immanuel: comment when you solve the problem

@pascal: This is a problem from IJSO'08, multiple choice, posting picture of the problem. It looks like those lines are straight and curved...

When $ tan \theta < \mu_s \;\;,\;\;f \equal{} mg.sin\theta\;$.
After the discontinuity, i.e. $ tan \theta > \mu_s \;\;,\;\;f \equal{} \mu_k.mg. cos \theta \;\;.$
So, only B) could be the answer.
In the multiple choice there's no need to analyze the curves, but the sign of the first and second derivatives confirms the sketch.

Yeah, I checked, the answer is B. Thanks!


Find $\cot 70^\circ + 4\cos 70^\circ$ without calculators.


geometry
symmetry
analytic geometry
rectangle

Two curves have equations $ x^{4} \plus{} y^{4} \equal{} u$ and $ xy\equal{}v$ where $ u$ and $ v$ are positive constants. State the equations of the lines of symmetry of each curve.

The curves intersect at the distinct points A, B, C and D (taken anticlockwise from A). The coordinates of A are $ (\alpha, \beta)$, where $ \alpha > \beta >0$. Write down in terms of $ \alpha$ and $ \beta$, the coordinates of B, C and D.

Show that the quadrilateral ABCD is a rectangle and find its area in terms of $ u$ and $ v$ only. Verify that, for the case $ u\equal{}81$ and $ v\equal{}4$, the area is $ 14$.

For this I have $ x \equal{} 0, y \equal{} 0, y \equal{} x, y \equal{} \minus{} x$ and $ y \equal{} x, y \equal{} \minus{} x$.

Then $ (\beta, \alpha), ( \minus{} \alpha, \minus{} \beta), ( \minus{} \beta, \minus{} \alpha)$.

Opposite sides equal in length and product of gradients $ \equal{} \frac {\alpha \minus{} \beta}{\beta \minus{} \alpha} \frac {\alpha \plus{} \beta}{\beta \plus{} \alpha} \equal{} \minus{} 1$.

Area $ 2(\alpha^{2} \minus{} \beta^{2})$.

Then substitute:

$ (\alpha^{2} \minus{} \beta^{2})^{2} \equal{} \alpha^{4} \plus{} \beta^{4} \minus{}2(\alpha\beta)^{2} \equal{} u \minus{} 2v^{2}$

$ \alpha^{2} \minus{} \beta^{2} \equal{} \sqrt{u \minus{} 2v^{2}}$.

Therefore area $ \equal{} 2\sqrt{u\minus{}2v^{2}}$

For $ u\equal{}81, v\equal{}4$, this has the value $ 2\sqrt{81\minus{}16} \equal{} 2\sqrt{49} \equal{} 14$.


I was wondering, I know I can look at everyone's profile but nevertheless, how long everyone is a member of this site? I will soon be 2 years, on August 5 that is. Horray happy birthday for me being on this site.

I am here for 1 year and 8 months, I registred around new year 2004. Not like Arne since the very beginning of MathLinks, but still from quite early.
But to AoPS-norms that is not early I guess?

(but it is stated wrongly in my profile, it got overwritten by the later date when I registred on AoPS, so looking in people's profile doesn't always give the right result.)

For a month but I like this forum much!! :)

I am fond of maths and here I can find and give to other people interesting problems
and math material..... :lol:

I joined at the beginning of July, so I've been on this site for...

*thinks hard, then zones out*

a month! But I think it's the best forum I've been on so far!  :D

Hehe, my registration date is messed up too, because of the merge AoPS - ML :)

I've been on here for a year, and a month.

I googled this site  and joined it in the beginning of this Feb. And then started posting until now. Already half a year.

Bleumoose y do I barely recall u, I mean I dont remember u being here a long time ago. Even though it is good that a lot of users are joining, I think the percentage of advanced students declines because a lot of beginners join. I am still a beginner even though I dont even post on math topics anymore.

I think I beat all of you. I am approaching 2 years and 2 months. (Actually, that's pretty close to the earliest possible.)

I've been on 5.5 months  :P

Valentin Vornicu was the earliest person on 2/03/03.

I think I joined April 2004.  I'll have to check my profile, though.

EDIT: Yep, April 11, 2004.

you know, its right by every single one of your posts? right there, under your username?

asdfgh

right.. now I can see my join date, it is May 21st, 2004... but I have been here like 2-3 months before I registered, i was too lazy to do it

I've been browsing this site since the San Diego Math Circle started and actually joined the site six months ago.

I post about 6 times a day?  :huh:

for about a week now.. and I love this forum.. I didn't know how I lived without knowing about AoPS  :blush:

i joined last week
was looking for the official solutions of the IMO 2005, google led me to this site. so i try to post here as often as i can, but we don't have internet access in our school.
->i have to learn French instead of posting at AoPS
I like this forum very much

[quote="ndb"]i have to learn French[/quote]
What's your native language?

[quote="ndb"]my native language is German, Swiss German. there are many different dialects though. but Swiss people usually understang each other, while Germans have a lot of problems to understand us :lol:[/quote]
So French is your third language? It is impressive...

[quote="shobber"]So French is your third language? It is impressive...[/quote]

I can't tell if chinese came first or did english for me (I was born in canada, but my parents spoke very little english). All I know now is that I stink at Chinese.

I have three languages: English, Chinese, Spanish,
I mean four: Latex  :D

i have been a member for about 6 months now.  I have two languages, english and $\LaTeX$


geometry
geometric transformation
reflection
articles

There must be another way to remove heat from electronic equipment rather than passing cool air over the surface? What about light waves? Can they cool a surface? Can they absorb heat, reflect away heat? Any one have any lateral thoughts?

The only relevant thing that comes to my mind is the so called "laser cooling" (just type these words in Google and you'll see a variety of articles about it).


function
algebra
polynomial
logarithms
real analysis
real analysis unsolved

This problem is from the NMC (Nordic Mathematical Olympiad) 1988. Consider the functions $ f_n(x)\equal{}\sum_{k\equal{}0}^{2n} x^k$ for any $ x \in \mathbb R$.
Prove that the minimum values $ m_n$ of these functions converge to $ \frac{1}{2}$ as $ n \rightarrow \infty$.

Does anybody know to solve this?

$ f_n(x)$ can be written as:

$ f_n(x) \equal{} \frac {x^{2n \plus{} 1} \minus{} 1}{x \minus{} 1}$

Now, if we let $ n \rightarrow \infty$, then it becomes:

$ f_\infty(x) \equal{} \frac {1}{1 \minus{} x}$

iff $ |x| < 1$. And since $ f_\infty(x)$ is strictly increasing on the interval $ \left< \minus{}\infty,1 \right>$, it has it's local extrema (minimum) at $ x \equal{} \minus{} 1$ so:

$ f_\infty(\minus{}1) \equal{} \frac {1}{2}$

The polynomial $ f_n$ does not have its minimum exactly at $ \minus{}1,$ and it changes rapidly enough in that neighborhood that the result is sensitive to the exact location of the minimum.

Numerically, $ x\equal{}\minus{}1\plus{}\frac{1\plus{}\ln n}{n}$ seems like a fair approximation of the location of the minimum of $ f_n(x).$


[quote="toanhocmuonmau"]Let $a,b,c$ be positive real numbers such that $a+b+c=1$, determine the maximum value of
\[P=a\sqrt{8b^{2}+c^{2}}+b\sqrt{8c^{2}+a^{2}}+c\sqrt{8a^{2}+b^{2}}\]
:)[/quote]
The following reasoning is very ugly:
$\sum_{cyc}a\sqrt{8b^{2}+c^{2}}\leq1\Leftrightarrow\sum_{cyc}(a^{4}+4a^{3}b+4a^{3}c-3a^{2}b^{2}+12a^{2}bc)\geq2\cdot\sum_{cyc}ab\sqrt{(8b^{2}+c^{2})(8c^{2}+a^{2})}.$
But $\sum_{cyc}ab\sqrt{(8b^{2}+c^{2})(8c^{2}+a^{2})}\leq\sqrt{\sum_{cyc}ab(8b^{2}+c^{2})(8c^{2}+a^{2})(ab+ac+bc)}.$
Hence, remain to prove that
 $\sum_{cyc}(a^{4}+4a^{3}b+4a^{3}c-3a^{2}b^{2}+12a^{2}bc)\geq2\cdot\sqrt{\sum_{cyc}ab(8b^{2}+c^{2})(8c^{2}+a^{2})(ab+ac+bc)},$ which very ugly.
 Very interesting fact!

[quote="pvthuan"]I prefer the following hard one
\[(a+b+c)^{2}\geq a\sqrt{8b^{2}+c^{2}}+b\sqrt{8c^{2}+a^{2}}+c\sqrt{8a^{2}+a^{2}}. \]
[/quote]
and $\sum_{cyc}a\sqrt{8b^{2}+c^{2}}\geq \sum_{cyc}9ab\sqrt\frac{c}{7c+2b}$
:P

[quote="Zamfirmihai"]$(\sum_{cyc}a\sqrt{8b^{2}+c^{2}})^{2}=(\sum_{cyc}\sqrt{a}\sqrt{8ab^{2}+ac^{2}})^{2}$ Now, Cauchy[/quote]

After Cauchy we are left with

$\sum a^{3}+6abc+3\sum ab(a+b) \geq 8\sum ab^{2}+\sum ac^{2}$

and now Zamfirmihai what to do !!!!

Thank you very much.

Have you a nice and direct solution, toanhocmuonmau ? :wink:

I tried it but I failed :(

It is interesting problem. Does anyone have solution?

[quote="cancer"]Can the proposer show?  :wink:[/quote]
Oh yes, by Cauchy Schwarz, we have
\[\left(\sum a\sqrt{8b^{2}+c^{2}}\right)^{2}\le \sum a(51a+100b+2c)\cdot \sum \frac{a(8b^{2}+c^{2})}{51a+100b+2c}=51 (a+b+c)^{2}\cdot \sum \frac{a(8b^{2}+c^{2})}{51a+100b+2c}\]
It suffices to prove that
\[51 \sum \frac{a(8b^{2}+c^{2})}{51a+100b+2c}\le (a+b+c)^{2}\]
By expanding, we can easily check this inequality. :)

Your solutions are really excellent. Congratulations !!! :wink:

Well, how are the numbers $51,100,2$ chosen?  :D

[quote="Vasc"][quote="toanhocmuonmau"][quote="Zamfirmihai"]hmmm...this problem also appears in Valentin Vornicu`s book. but it is $\sum_{cyc}b\sqrt{8b^{2}+c^{2}}$ instead of $\sum_{cyc}a\sqrt{8b^{2}+c^{2}}$ maybe toanhocmuonmau had a mistake in writting the problem:P[/quote]
No, my problem is right![/quote]
It is right, of course; and very nice; but not easy one. There is a nice solution by SOS method.[/quote]

Can you, please, give a hint for your SOS prove?   :oops:

Thank you very much.

[quote="manlio"]
Can you, please, give a hint for your SOS prove?   :oops:
Thank you very much.[/quote]
This is a hint. It is enough to show that

$\sum (b-c)^{2}(3-\frac{4a}{b+2c}) \ge 0$.

Thank you very much, Vasc. I will think how to solve it using your hint.

Dear Vasc, I really cannot figure  :blush: how to deduce your SOS form

 $\sum (b-c)^{2}(3-\frac{4a}{b+2c}) \ge 0$

from the original inequality, so, please, I need another help.  :blush:

My trouble is that your SOS form is of fifth grade, while after Cauchy application we obtain a third grade form.

[quote="manlio"][quote="Zamfirmihai"]$(\sum_{cyc}a\sqrt{8b^{2}+c^{2}})^{2}=(\sum_{cyc}\sqrt{a}\sqrt{8ab^{2}+ac^{2}})^{2}$ Now, Cauchy[/quote]

After Cauchy we are left with

$\sum a^{3}+6abc+3\sum ab(a+b) \geq 8\sum ab^{2}+\sum ac^{2}$[/quote]

Thank you very much.

[quote="manlio"][quote="Zamfirmihai"]$(\sum_{cyc}a\sqrt{8b^{2}+c^{2}})^{2}=(\sum_{cyc}\sqrt{a}\sqrt{8ab^{2}+ac^{2}})^{2}$ Now, Cauchy[/quote]

After Cauchy we are left with

$\sum a^{3}+6abc+3\sum ab(a+b) \geq 8\sum ab^{2}+\sum ac^{2}$[/quote]

The last inequality is not true. See $a=0$ and $b=c$.

Dear Vasc, do you deduce your SOS form from

$ \sum \frac{a(8b^{2}+c^{2})}{51a+100b+2c}\le (a+b+c)^{2}$ ?

If not, can you please give me another hint?   :blush:

Thank you very much.

Thank you very much, Vasc and Arqady. You are very kind. :)

[quote="arqady"][quote="Vasc"]
 It is enough to show that
$ \sum (b-c)^{2}(3-\frac{4a}{b+2c}) \ge 0$.[/quote]
It's equivalent to $ \sum_{cyc}(b-c)^{2}\left(3+\frac{4(b+2c)}{a}\right)\geq0.$ :wink:[/quote]

Sorry, I cannot understand why these two expressions are equivalent.

Can you please explain it , Arqady?

Thank you very much.

[quote="manlio"][quote="arqady"][quote="Vasc"]
 It is enough to show that
$ \sum (b-c)^{2}(3-\frac{4a}{b+2c}) \ge 0$.[/quote]
It's equivalent to $ \sum_{cyc}(b-c)^{2}\left(3+\frac{4(b+2c)}{a}\right)\geq0.$ :wink:[/quote]
Can you please explain it , Arqady?
[/quote]
I have not found a nice proof for it.
$ \sum (b-c)^{2}(3-\frac{4a}{b+2c}) \ge 0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(b^{2}-2bc+c^{2})(3b+6c-4a)(c+2a)(a+2b)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3b^{3}+6c^{3}-9c^{2}b-4b^{2}a-4c^{2}a+8abc)(ac+2a^{2}+2bc+4ab)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3b^{3}ac+6c^{4}a-9c^{3}ab-4a^{2}b^{2}c-4c^{3}a^{2}+8a^{2}c^{2}b+$
$ +6b^{3}a^{2}+12c^{3}a^{2}-18a^{2}c^{2}b-8a^{3}b^{2}-8a^{3}c^{2}+16a^{3}bc+$
$ +6b^{4}c+12c^{4}b-18c^{3}b^{2}-8b^{3}ac-8c^{3}ab+16b^{2}c^{2}a+$
$ +12b^{4}a+24c^{3}ab-36b^{2}c^{2}a-16b^{3}a^{2}-16a^{2}c^{2}b+32a^{2}b^{2}c)\geq0\Leftrightarrow$
$ \Leftrightarrow2\cdot\sum_{cyc}(a^{4}b+2a^{4}c-3a^{3}c^{2})+3\cdot\sum_{cyc}(a^{3}bc-a^{2}b^{2}c)\geq0\Leftrightarrow$
$ \Leftrightarrow4\cdot\sum_{cyc}ab(a^{3}+2b^{3}-3ab^{2})+3\cdot\sum_{cyc}abc(a-b)^{2}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a-b)^{2}(4ab(a+2b)+3abc)\geq0\Leftrightarrow\sum_{cyc}(a-b)^{2}\left(3+\frac{4(a+2b)}{c}\right)\geq0.$

Thank you very much, Arqady.

I think that SOS method of Vasc more expansion method of Arqady is a very nice proof for this hard inequality.

Congratulations. :)

[quote="cancer"]Well, how are the numbers $ 51,100,2$ chosen?  :D[/quote]

Dear Can , please answer to this question . Because this time I can't found the reason . At one inequality of Vasc you have said me that you solve the system at the equality case to find the constants . But here we have only one equality case . Again the same ?


limit
linear algebra
matrix
ratio
function

Find the value of:

$ 3 \minus{} \frac{1}{3 \minus{} \frac{1}{3 \minus{} \frac{1}{3\minus{}...}}}$

And prove that this is indeed the unique value. I hope somebody comes up with a better proof than I did, mine's ugly.

[hide]$ x \equal{} 3 \minus{} \frac {1}{3 \minus{} \frac {1}{3 \minus{} \frac {1}{3 \minus{} ...}}}$
$ x \equal{} 3 \minus{} \frac {1}{x}$
$ x^2 \equal{} 3x \minus{} 1$
$ x^2 \minus{} 3x \plus{} 1 \equal{} 0$
$ x \equal{} \frac {3 \plus{} sqrt(5)}{2}$
$ x \equal{} \frac {3 \minus{} sqrt(5)}{2}$

[/hide]

oh, right :blush:
both aren't correct.
$ \frac{3\minus{}\sqrt{5}}{2}$?

[quote="firecricket91"]It should be (3+sqrt5)/2

becuase 1/x is from (0,1) so our answer has to be from (2,3).

Only that one is from (2,3).

The other answer is like 0.3ish.[/quote]

You're using circular reasoning. 1/x is only from (0,1) if x is $ \frac{3\plus{}\sqrt{5}}{2}$ and not $ \frac{3\minus{}\sqrt{5}}{2}$

The second root works fine if you check it like that.

$ \frac{1}{x} \in \left( 0 , 1 \right) \ \text{iff} \ x > 1$.
What if $ x < 1$?

Unless I've made a giant blunder:

Smallish hint:
[hide]
That line of reasoning doesn't work. You're going from the equation phase -- that is, $ x \plus{} \frac{1}{x} \equal{} 3$. At this point, the ambiguity has already been introduced and both roots are 100% valid. You'll have to look to the original expression to find which root is valid. This is where I wasn't able to achieve elegance :([/hide]

To make the expression well-defined:  let $ \{ a_n \}$ be a sequence such that

$ a_0 \equal{} 3$
$ a_{n \plus{} 1} \equal{} 3 \minus{} \frac {1}{a_n}$.

Then the desired value is $ \lim_{n \to \infty} a_n$.  The transformation $ a_n \mapsto a_{n\plus{}1}$ has two fixed points, but given an initial value of $ 3$, to which of the fixed points does the recursion actually tend?  :)

[hide="An easy finish"]Using t0r's notation, prove $ 1/3 < \frac{3 \minus{} \sqrt{5}}{2} \leq a_n \leq 3$ for all $ n$. So it can only tend to $ \frac{3 \minus{} \sqrt{5}}{2}$.[/hide]

Let t(x)=3-1/x=(3x-1)/(x)

Consider the corresponding matrix thing:

[3,-1]
[1,0]

If we take this to the nth power, we get a recurrence...We can write the entries out explicitly (use eigenvalues...)

these are equivalent to the fixed points of t(x), call them a,b. Then a>1>b. The terms will be in the from c_1 a^n+c_2 b^n. The ratio will clearly be dominated by a because it is greater than 1.

The x_n has a form something like (ratio of adjacent terms)+1/(nth term)^2* /(reciprocal of ratio of terms -k)

So for most k, we get that the the reciprocal ratio of terms (which goes to b) will evenutally get dominated by the nth term. However, when k=b, we get successive rational approximations to b, so it that fraction doesn't evenutally get smaller.

In summary of my not so rigirous, but still generally correct logic, the sequence converges to the larger fixed point provided that the initial term is not the lower fixed point or a value that makes the function undefined (any of the rational approximations of the lower fixed point).

Consider, for $ a \geq 2$, the graph of $ f(x) \equal{} a \minus{} 1/x$ and $ x$, with two solutions $ f_2 \geq f_1$.

$ x < f_1 \implies f(x) < x$, and the recursion $ f^n (x)$ does not converge.
$ x > f_2 \implies f_2 < f(x) < x$, and the recursion converges to $ f_2$.
$ f_1 < x < f_2 \implies x < f(x) < f_2$, and the recursion converges to $ f_2$.

Note that $ f_1 \equal{} \frac {a \minus{} \sqrt {a^2 \minus{} 4}}{2} \implies a > \frac {a}{2} > f_1$.

Hence, the recursion $ f^n (a)$ converges to $ f_2$.

[hide="Continued"]For $ |a| < 2$, the discriminant of $ x^2 \minus{} ax \plus{} 1$ is negative.

For $ a \leq \minus{} 2$, we have the opposite situation: the recursion converges to $ f_1$ unless $ x \geq f_2$.
Hence, the recursion converges to $ f_1$ unless $ a \geq f_2 \implies \frac a \geq \sqrt {a^2 \minus{} 4}$, which is  not possible. So, for $ a \leq \minus{} 2$, $ f^n (a)$ converges to $ f_1$.[/hide]

[quote="The Zuton Force"]Consider, for $ a \geq 2$, the graph of $ f(x) \equal{} a \minus{} 1/x$ and $ x$, with two solutions $ f_2 \geq f_1$.

$ x < f_1 \implies f(x) < x$, and the recursion $ f^n (x)$ does not converge.
[/quote]

This statement is not correct.

Ex f(x)=3-1/x, f_1=(3-5^.5)/2=.38...

take x=.3, then f(x)=-1/3 and f(f(x))=6...which converges by your other intervals

Ah, right.

So it does converge, then, as soon as the value drops below 0. Lovely. Thanks --

I remember that my friend gave me a problem like that before, except it was 4's instead of 3's. I still haven't gotten it. :D


Given triangle $ ABC$ and a point $ M$ lies inside. $ A'B'C'$ is the cevian triangle of $ ABC$ wrt $ M$. Denote $ G_a, G_b, G_c, G'_a, G'_b, G'_c$ the centroid of triangle $ AB'C', BA'C', CA'B', MB'C', MA'C', MA'B'$. Prove that $ G_aG'_a, G_bG'_b, G_cG'_c$ are concurrent.

Denote the point G0 is the centroid of triangle A'B'C'
It is easy to know that
GaGa' // AA'  GaG0 // AA'   =>  GaGa'G0 are collinear
analgously, GbGb'G0  GcGc'G0  are collinear
thus GaGa' GbGb' GcGc' are concurrent at G0

Other concurrencies:
+$ AG'_a, BG'_b, CG'_c$ are concurrent.
+$ A'G_a, B'G_b, C'G_c$ are concurrent.


abstract algebra
group theory
Galois Theory
superior algebra
superior algebra unsolved

Fix a prime number $ p$ and a $ p$-power $ q\equal{}p^d$, with $ d \geq 1$. Let $ F$ be a field which contain a primitive $ q^{th}$ root of unity and $ G\equal{}Gal(F_{sep}/F)$ its absolute Galois group, where $ F_{sep}$ is the separable closure of $ F$.

I want to prove that the kernel $ M$ of the epimorphism $ f: G \rightarrow G(p)$ with $ G(p)$ the maximal pro-$ p$ quotient of $ G$ is such that $ H^1(M) \equal{} H^1(M,\mathbb{Z}/p\mathbb{Z})\equal{}0$. In the paper I'am reading this fact is presented as a "standard result". If one know book reference for this result or can gives me hints, I thank you in advance.

If M is finite, then M is generated by its Sylow r-subgroups for r not equal to p (that is, M = O^p(G) = O^p(M)). Since Z/pZ is implied to be centralized by M, Z^1(M,Z/pZ) is just the group of homomorphisms from M to Z/pZ.  However, an r-element must be sent to an r-element, so every Sylow r-subgroup of M is sent into a Sylow r-subgroup of Z/pZ, that is, to 0.  Since M is generated by these elements, every element of M is sent to 0.  Hence Z^1(M,Z/pZ) = 0 = H^1(M,Z/pZ).

If M is profinite, then it is generated by its Sylow pro-r-subgroups for r not equal to p, and Z^1(M,Z/pZ) should consist of continuous group homomorphisms, and so should send Sylow pro-r-subgroups into Sylow pro-r-subgroups, so should send all of M to 0.


algebra
equation
IMO Shortlist
number theory

[quote="April"]Let $ n$ be a positive integer and let $ p$ be a prime number. Prove that if $ a$, $ b$, $ c$ are integers (not necessarily positive) satisfying the equations
\[ a^n \plus{} pb \equal{} b^n \plus{} pc \equal{} c^n \plus{} pa\]
then $ a \equal{} b \equal{} c$.[/quote]

Let $ a \equal{} max (a;b;c)$.
Case I: $ a > c > b$.
First $ if p\equal{}2$.
If $ a$ or $ b$ or $ c$ is equal to $ 0$ . It is trivial to see $ a\equal{}b\equal{}c\equal{}0$.
Now $ |a|;|b|;|c|\ge 1$.
We have $ a^n \minus{} b^n \equal{} (a \minus{} b)(a^{n \minus{} 1} \plus{} .. \plus{} b^{n \minus{} 1}) \equal{} p(c \minus{} b)$. Because $ a\minus{}b>c\minus{}b>0$ thus $ 2>a^{n\minus{}1}\plus{}..\plus{}b^{n\minus{}1}$. So that $ n\equal{}1$ or $ n\equal{}2$. Both cases imply $ a\equal{}b\equal{}c$.
If $ p>3$.
 It is obvious to see that if $ x \equal{} y$ ($ x;y$ are chosen in $ (a;b;c)$ ) then $ a \equal{} b \equal{} c$.
Thus $ a^n \minus{} b^n \equal{} (a \minus{} b)(a^{n \minus{} 1} \plus{} .. \plus{} b^{n \minus{} 1}) \equal{} p(c \minus{} b)$;$ c^n \minus{} b^n \equal{} p(a \minus{} c) \equal{} (c \minus{} b)(c^{n \minus{} 1} \plus{} .. \plus{} b^{n \minus{} 1})$;$ a^n \minus{} c^n \equal{} (a \minus{} c)(a^{n \minus{} 1} \plus{} .. \plus{} c^{n \minus{} 1}) \equal{} p(a \minus{} b)$.
It means $ (a^{n \minus{} 1} \plus{} .. \plus{} b^{n \minus{} 1})(c^{n \minus{} 1} \plus{} .. \plus{} b^{n \minus{} 1})(a^{n \minus{} 1} \plus{} .. \plus{} c^{n \minus{} 1}) \equal{} p^3 \equal{} ABC$ where $ C \equal{} (a^{n \minus{} 1} \plus{} .. \plus{} b^{n \minus{} 1})$ .etc.
If there exists one of $ A;B;C$ be equal to 1, Assume that It is C (The same agruments will work when it is either B or A).
Thus $ a^n \minus{} b^n \equal{} a \minus{} b$. If $ B$ or C is also equal to $ 1$.It is very easy to see that $ a \equal{} b \equal{} c$ contradiction.
So that    $ A \equal{} p$; $ B \equal{} p^2$ or $ B \equal{} p$ ;$ A \equal{} p^2$.
If $ A \equal{} p$; $ B \equal{} p^2$ henxe $ c^n \minus{} b^n \equal{} p(c \minus{} b) \equal{} p(a \minus{} c)$ and $ a^n \minus{} c^n \equal{} p^2(a \minus{} c) \equal{} p(a \minus{} b)$. The first equation implies $ 2c \equal{} a \plus{} b$, the second imlpies

$ p(a \minus{} c) \equal{} a \minus{} b$ which is equivalent to $ (p \minus{} 2)a \equal{} (p \minus{} 2)b$. Then $ a \equal{} b \equal{} c$ contradiction.
If $ B \equal{} p$ ;$ A \equal{} p^2$ thus $ a^n \minus{} c^n \equal{} P(a \minus{} c) \equal{} p(a \minus{} b)$ which implies $ a \equal{} b \equal{} c$.
In conclusion, None of $ A;B;C$ is equal to 1.
It means $ A \equal{} B \equal{} C \equal{} p$. Now it shows immediately $ a \equal{} b \equal{} c$.
Case II: $ a > b > c$. The proof is absolutely similar to the above!.
We have done!
Sorry for my poor English. Warm me if I make any silly mistakes (Grammar or Maths)

Let $ P(a)\equal{}a^n\minus{}pa$. Note: $ P(a)$ is a polynomial of $ a$, and $ p(a)$ is $ p\cdot a$.

Assume that not all of them are equal, and I'll leave out trivial cases (one is equal to zero, etc.).

We have that $ |a\minus{}b| \mid |P(a)\minus{}P(b)|$ for all $ a,b$.

Without loss of generality,  let $ a\ge b\ge c$ (the case $ a\ge c\ge b$ is similar).

$ P(a)\equal{}c^n\minus{}pb\equal{}c^n\minus{}pc\plus{}p(c\minus{}b)\equal{}P(c)\plus{}p(c\minus{}b)$.

$ a\minus{}c\mid p(b\minus{}c)$, similar we get $ a\minus{}b\mid p(a\minus{}c)$ and $ b\minus{}c\mid p(a\minus{}b)$.

Let $ a\minus{}b\equal{}x,b\minus{}c\equal{}y,a\minus{}c\equal{}x\plus{}y,gcd(x,y)\equal{}d,x\equal{}rd,y\equal{}sd$. $ r,s$ are positive integers.

$ x\plus{}y\mid py,x\mid p(x\plus{}y),y\mid px$.

$ r\plus{}s\mid ps,r\mid pr\plus{}ps,s\mid pr$.
As $ gcd(r,s)\equal{}1$, we get $ r\mid p,s\mid p$.
But, as $ p$ is a prime, we must have $ r\equal{}1$ or $ s\equal{}1$. Let $ r\equal{}1$.

$ 1\plus{}s\mid ps$ so $ 1\plus{}s\mid p$, also $ s\mid p$.

Again we conclude $ s\equal{}1$, and $ p\equal{}2$. We also conclude $ a\minus{}b\equal{}b\minus{}c$, or $ 2b\equal{}a\plus{}c$

$ a^n\plus{}2a\plus{}2c\equal{}b^n\plus{}2c\equal{}c^n\plus{}2a$.

$ (a\minus{}b)(a^{n\minus{}1}\plus{}\ldots\plus{}b^{n\minus{}1})\equal{}2(c\minus{}b)$, or $ a^{n\minus{}1}\plus{}\ldots\plus{}b^{n\minus{}1}\equal{}\minus{}2$. It's easy to see that we get a contradiction, and so $ a\equal{}b\equal{}c$.

I've solved a generalisation of this problem, but I think there might be some mistake, please check my solution:

Instead of prime number, let $ p$ be a positive odd number (well, we miss the case $ p\equal{}2$, so it's not really a generalisation :P ).

Note that $ \frac{a^n\minus{}b^n}{c\minus{}b}\equal{}\frac{b^n\minus{}c^n}{a\minus{}c}\equal{}\frac{a^n\minus{}c^n}{a\minus{}b}\equal{}p$. So $ \frac{(a^n\minus{}b^n)(b^n\minus{}c^n)(a^n\minus{}c^n)}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}\equal{}p^3$. If $ n$ is odd, then $ a^n\minus{}b^n,b^n\minus{}c^n,c^n\minus{}a^n$ have the same signs with $ a\minus{}b,b\minus{}c,c\minus{}a$ respectively, which implies that $ p$ is negative, a contradiction. Therefore $ n$ is even. Note that at least two of $ a,b,c$ have the same parity, wlog assume $ a\equiv b\pmod2$. We have $ \frac{a^n\minus{}b^n}{a\minus{}b}\equal{}a^{n\minus{}1}\plus{}a^{n\minus{}2}b\plus{}\ldots\plus{}ab^{n\minus{}2}\plus{}b^{n\minus{}1}$. If they are both even, clearly this expression is even. If they are both odd, the expression equals the sum of $ n$ odd integers, and hence is even. So $ \frac{a^n\minus{}b^n}{a\minus{}b}$ is even. But $ \frac{b^n\minus{}c^n}{b\minus{}c}$ and $ \frac{a^n\minus{}c^n}{c\minus{}a}$ are integers, so $ \frac{(a^n\minus{}b^n)(b^n\minus{}c^n)(a^n\minus{}c^n)}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}$ is an even integer, contrary to the fact that $ p$ is odd. QED

I think you are right.  :)
I have proved the same (for all $ p\equiv 1 \mod 2$) this way: Let $ P(x) \equal{} a^n \plus{} pb \minus{} x^n$ be a polynomial and note that $ P$ has integer coefficients. Now, $ P(a) \equal{} pb$, $ P(b) \equal{} pc$ and $ P(c) \equal{} pa$.

Assume wlog $ a \equiv b \mod 2$. Now $ P(a) \equiv P(b) \mod 2$, so $ pb \equiv pc \mod 2$, $ b \equiv c \mod 2$ since $ p$ is odd. So $ a \equiv b \equiv c \mod 2$. We can now prove by induction on $ m$ that $ a \equiv b \equiv c \mod 2^m$ for $ m \in \mathbb{N}$. Now choose $ m$ such that $ a,b,c \in ( \minus{} 2^m,2^m)$. wlog $ a \equal{} b$, so $ P(a) \equal{} P(b)$, so $ b \equal{} c$, thus $ a \equal{} b \equal{} c$ as desired.

So $ p$ isn't even required to be positive  :P

[quote="x164"] Let $ P(x) \equal{} a^n \plus{} pb \minus{} x^n$ be a polynomial and note that $ P$ has integer coefficients. Now, $ P(a) \equal{} pb$, $ P(b) \equal{} pc$ and $ P(c) \equal{} pa$. [/quote]I don't understand the definition of $ P(x)$. Why is $ P(b)\equal{}pc$? If $ P(x)\equal{}a^n\plus{}pb\minus{}x^n$, then it should be $ P(b)\equal{}a^n\plus{}pb\minus{}b^n$.

You are right, $ P(b) \equal{} a^n \plus{} pb \minus{} b^n$.
Now we use the fact that $ a^{n} \plus{} pb \equal{} b^{n} \plus{} pc$:

Since $ a^n \plus{} pb \equal{} b^n \plus{} pc$, $ P(b) \equal{} (a^n \plus{} pb) \minus{} b^n \equal{} (b^n \plus{} pc) \minus{} b^n \equal{} pc$. :)

[quote="manifestdestiny"]There may be something wrong here, but anyways...

$(a-b)(a^{n-1}+\ldots +b^{n-1})=p(b-c)$
$(b-c)(b^{n-1}+\ldots +c^{n-1})=p(c-a)$
$(c-a)(c^{n-1}+\ldots +a^{n-1}) =p(a-b)$.

First, it is clear that if any two of $a,b,c$ are equal, then all three are equal as we see that if, WLOG $a=b$, then $a^n=b^n$ so $pb=pc$ or $b=c=a$.

Multiplying the above three equations, we get that if $X=(a^{n-1}+\ldots +b^{n-1})$, $Y=(b-c)(b^{n-1}+\ldots +c^{n-1})=p(c-a)$, $Z=(c-a)(c^{n-1}+\ldots +a^{n-1}) =p(a-b)$, then $XYZ=p^3$. Clearly $X,Y,Z>0$.

If one of $X,Y,Z$ is one, then n=1, as for example, if $a^n-b^n=a-b$, this must imply n=1. Then $a+pb=b+pc=c+pa$. So we get $a-b=p(b-c)=p^2(c-a)=p^3(a-b)$ so either $p=1$ - impossible, or $a=b=c$.

Otherwise, $X=Y=Z=p$ and we get
$p(a-b)=p(b-c)=p(c-a)$, so we get $a-b=b-c$ or $2b=a+c$  and similarly $2c=a+b$ which gives $a=2c-b$, and substituting into the earlier equation, we get $3b=3c$, which gives us $b=c=a$, as desired.

Hence $a=b=c$.[/quote]

What about $5^2-(-4)^2=5-(-4)=9$, it isn't correct at all, when you say $X=1$ for example.

for the solution of Gunardi: if they heve the same signs, $p$ has to be positive, hence that proof isn't good.

solution with $P(x)$ isn't correct, you define $P(a)$ with help of $b$ and similar.

[quote="mousavi"]assume that $a>b>c$ (if two of are equal,it's easy too check that all of them are equal)

$\frac{a^n-b^n}{c-b}=\frac{c^n-b^n}{c-a}=\frac{c^n-a^n}{b-a}=p$

1)$p>2$:

$\left \| c-b \right \|_p=x\Rightarrow \left \| a^n-b^n \right \|_p=x+1$

$\left \| c^n-b^n \right \|_p=x+\left \| n \right \|_p\Rightarrow \left \| c-a \right \|_p=x+\left \| n \right \|_p -1$

$\left \| c^n-a^n \right \|_p=x+2\left \| n \right \|_p-1\Rightarrow \left \| a-b \right \|_p=x+2\left \| n \right \|_p-2$

$\Rightarrow x+1=\left \| a-b \right \|_p+\left \| n \right \|_P=x+3\left \| n \right \|_p-2\Rightarrow \left \| n \right \|_p=1$

$\Rightarrow \left \| a-b \right \|_p=\left \| b-c \right \|_p=\left \| c-a \right \|_p=x$

$a-b=p^x .k_1$ , $a-c=p^x.k_2$ , $b-c=p^x.k_3$

$k_1\mid k_3,k_3\mid k_2,k_2\mid k_1\Rightarrow k_1=k_2=k_3\Rightarrow a=b=c$ contradiction

2)p=2 is not different with the 1[/quote]

Are you sure, if $p \not| v_p(b-c)$ you aren't correct and where would you conclude that from?

[quote="Allnames"]
It means $ A \equal{} B \equal{} C \equal{} p$. Now it shows immediately $ a \equal{} b \equal{} c$.
[/quote]

I was doing this question, and similarly arrived at $A\equal{} B\equal{} C\equal{} p$, however, I could not reduce this down to $a\equal{} b\equal{} c$ immediately, could you elaborate?

I shall present my proof.(I have skipped trivial calculations)
Note that equality of any two of $a,b,c$ implies equality of all the three,so we assume that no two of them are equal.
Subtracting the terms taking two at a time we get

$p(c-b)=a^n-b^n$
$p(a-c)=b^n-c^n$
$p(b-a)=c^n-a^n$

Multiplying the three equalities yeilds
$\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{a-c}=p^3$

Note that $\frac{a^n-b^n}{a-b}$ never equals $1$ unless $n=1$ or $a,b \in (0,1,-1)$. If $n=1$ then $p=1$,a contradiction.Work out the details in the second case to show impossibility.

Since no term can equal $1$,the factors must be $(p,p,p)$ or $(-p,-p,p)$ in some order.In any case atleast one of them equals $p$,say $\frac{a^n-b^n}{a-b}$.Then $a^n-b^n=p(a-b)=a^n-c^n \implies b^n=c^n \implies p=0$ a contradiction.

The case when first two terms are $-p$ and the last one is $p$ needs to be addressed seperately.Then if $n$ is odd then  $a<b<c$.But this is not possible since then $0<p(c-b)=a^n-b^n<0$.If $n$ is even then $\frac{a^\frac{n}{2}-b^\frac{n}{2}}{a-b} \cdot (a^\frac{n}{2}+b^\frac{n}{2})=-p$ and similarly the LHS of the other equality is factorized.Work out the details to show that this case is also not possible.

In conclusion we must have $a=b=c$

[quote=April]Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a$, $b$, $c$ are integers (not necessarily positive) satisfying the equations \[ a^n + pb = b^n + pc = c^n + pa\] then $a = b = c$.

[i]Proposed by Angelo Di Pasquale, Australia[/i][/quote]

I have a surprisingly easy proof for the required result which might very well be wrong.It would help if you could tell me what you think of this.
So here goes:

First if $a=b=c$ then we are clearly done.
If not,then assume that two of these are equal(without loss of generality, say $a=b$)
Then replacing $b$ with $a$,the first equality becomes,
$a^n+pa=a^n+pc$ which just simplifies to $a=c$,implying the required result.

Now consider the case where all three are distinct.
WLOG let $a>b>c$
Then if $n$ is odd,we would have
$a^n>b^n$ and $pb>pc$,whose addition would contradict the first equality.(Note that this holds even when $a$ and $b$ are negative because still $a^n>b^n$ when $n$ is odd)

1.Therefore $n$ is even
Now assume that all three of these are positive.Then
$a^n>b^n$ and $pb>pc$ and addition would give us the previous contradiction.
So one of $a,b,c$ must be negative.
Since $a>b>c$,then $c<0$
Now consider three cases:
The first- $b>0$
Then $a>b$ gives $a>0$
Once more,$a^n>b^n$ and $pb>pc$ would be a contradiction.
The second- $b<0$
Then since by 1, $n$ is even,
$b^n<c^n$ and $pc<pa$ whose addition would contradict the second equality.
The third- $b=0$
Then the entire equality becomes
$a^n=pc=c^n+pa$
Divisibility conditions could be used to simplify this trivially(eg: since $p|a$ put $a=pk$ and simplify)
Doing so we arrive at another contradiction.

Therefore,our original assumption must have been wrong so $a=b=c$
$QED$

This is a huge generalization of the original problem so there must be something wrong with my proof.Looking forward to any comments :)

I think that this is not a generalization.[quote]This is a huge generalization of the original problem[/quote]
Becouse every thing fails when b=0

First we show that there are two of the variables that are equal. Note that $p=\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{c^n-a^n}{b-a}$ so $-p^3=\prod_{\text{sym}}(a^{n-1}+a^{n-2}b+\dots+b^{n-1})c^0.$ (The $c^0$ is there to avoid confusion about the number of terms in that product.)

Note that now $a^{n-1}+a^{n-2}b+\dots+b^{n-1}$ and the same expression with different variables, they must be $mp^x,np^y,qp^z$ with $x+y+z=3$ and $m,n,q=\pm 1$ with $mnq=-1.$

If all three are of power $1$ then two of the expressions are equal, say $a^{n-1}+a^{n-2}b+\dots+b^{n-1}=c^{n-1}+c^{n-2}b+\dots+b^{n-1}.$ If $c>a$ then $c^{n-1}+c^{n-2}b+\dots+b^{n-1}>a^{n-1}+a^{n-2}b+\dots+b^{n-1}$ and vice versa. Thus, $a=c$ as desired.

Otherwise there must exist one of the expressions with power $0$ so it's equal to one or negative one. In this case, the other expressions are also just $1$ since $n=1,$ but now the powers don't add to $3$ and that's a contradiction. Hence, two of the variables are equal. WLOG $a=c,$ and $a^n+pb=b^n+pa=a^n+pa.$ This implies $b^n=a^n$ so $a=b=c,$ and we're done.

If we have $a=b$ then we get $pb = pc$ which implies that $a=b=c$. Hence we can safely assume that $a,b$ are distinct. Thus, $a^n-b^n$ and cyclic multiplication gives us \[(a^n - b^n)(b^n-c^n)(c^n - a^n) = \prod_{\text{cyc}} \left(\frac{a^n-b^n}{a-b}\right) = -p^3\] Since we know that $\frac{a^n-b^n}{a-b} \in \mathbb{Z}$ , therefore we get
\begin{align*}
\frac{a^n-b^n}{a-b} &= \frac{p(c-b)}{a-b} \\
&= p_c \in \{\pm 1 , \pm p , \pm p^2 , \pm p^3 \}
\end{align*} define $p_a$ and $p_b$ similarly and $p_ap_bp_c = -p^3$
However the system
\begin{align*}
p(c-b) &= p_c(a-b) \\
p(b-a) &= p_b(c-a) \\
p(a-c) &= p_a(b-c)
\end{align*}
is true if and only if \[(b-c)(p_1p-p^2-p_1p_2) = 0\] Since we know that
\begin{align*}
p_1p-p^2-p_1p_2 &= 0 \\
\iff p_1(p-p_2) &= p^2 \\
\implies p - p_2 &\mid p^2 \\
\implies p &= 2
\end{align*}
Hence if $p \ge 3$ , it implies that $b=c$ and we are done.
[rule]
For the case $p = 2$ it must hold true that \begin{align*}
p_2 &= 4 \\
\implies p_1 &= -2
\end{align*}
However, the fact that \[p_1p-p^2-p_1p_2 = 0\] is also true. Contradiction. $\square$

Let:
$$(x,y,z)=(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+ab^{n-2}+b^{n-1},b^{n-1}+b^{n-2}c+b^{n-3}c^2+\ldots+bc^{n-2}+c^{n-1},c^{n-1}+c^{n-2}a+c^{n-3}a^2+\ldots+ca^{n-2}+a^{n-1}).$$
We rearrange the equations into:
\begin{align*}
x(a-b)&=p(c-b)\\
y(b-c)&=p(a-c)\\
z(c-a)&=p(b-a).
\end{align*}
If some two of $a,b,c$ are equal, WLOG $a=b$. Then from the first equation, $b=c$, so the conclusion holds.
Otherwise, assume FTSOC that $a\ne b\ne c\ne a$. Multiplying these three equations gives:
$$xyz=p^3.$$

$(\text a)$ If $p\in\{x,y,z\}$, then WLOG $x=p$. Then $p(a-b)=p(c-b)$, so $a=c$, contradiction.
$(\text b)$ If $\{-1,1\}\subseteq\{x,y,z\}$, then WLOG $|x|=1$.
[list]
[*] Suppose $x=1$ and that $n$ is odd, then $a^n-b^n=a-b$, or $a^n-a=b^n-b$. If $ab=0$, WLOG $b=0$, then $a=\pm1$ and we reach a contradiction upon concluding that $a^n=pc$. If $n$ is odd then $a,b$ must have the same sign. WLOG $a,b$ are both positive, then since $ab\ne0$ we have $a,b\ge1$. Now let $f(x)=x^n-x$. By expansion, $(x+1)^n>x^n+1$. Then $f(x+1)>f(x)$, so since $f(a)=f(b)$ we need $a=b$, contradiction.
[*] Suppose $x=1$ and that $n$ is even. If $a,b$ have the same sign then, as before, $a=b$. So $a$ and $b$ must have different signs, then WLOG $a<0$ and $b>0$, and $a+b\ge0$. Let $d=-a$, we have $d^n+d=(-d)^n+d=b^n-b$. Note that $b\ge d$. If $b=d$ then $b=0$ which is a contradiction, so $b\ge d+1$. Using the result we got above, $d^n+d\ge(d+1)^n-d-1$. But $(d+1)^n>d^n+2d+1$ by expansion, so we reach a contradiction.
[*] The same approach can be used for the case $x=-1$; it is impossible.
[/list]

Since one of the cases $(\text a)$ or $(\text b)$ must hold, we're done.

After trying to derive relations that allows applying LTE, I found this solution which do not use it at all.

Assume for contradiction that $a,b,c$ are distinct. We can easily see that if $a,b,c$ are not pairwise distinct, then they must be all equal. Then, we must have $a \ne b, b \ne c, c\ne a.$

If $n=1,$ we will have $$ a-b = -p(b-c), b-c = -p(c-a), c-a= -p(a-b)$$ and multiplying everityhing and using the pairwise distinc assumption yelds $p^3 = -1,$ contradiction since $p$ is prime. Then $n \ge 2$ and we can write,
$$ (a-b)T(a,b) = -p(b-c) ~(1A)$$ $$ (b-c)T(b,c) = -p(c-a)~(1B)$$ $$ (c-a) T(c,a) = -p(a-b)~(1C)$$
where $T(x,y) = x^{n-1} + x^{n-2}y + \dots +y^{n-1}.$

Multiplying everything again yelds $$T(a,b)T(b,c)T(c,a) = -p^3~(*)$$
[b][color=#f00]Claim:[/color] Neither of $T(a,b),T(b,c), T(c,a)$ can be equal to $p$ or $-p.$[/b]
[hide=proof]Suppose otherwise and, without loss of generality, assume $T(a,b) = \pm p.$

If $T(a,b) = p,$ then $(1A)$ will give $a=c$ which contradicts our pairwise distinct assumption.

On the other hand, if $T(a,b) = -p,$ then $(1A)$ will give $b = \frac{a+c}{2}$ and $(1B)$ will give $T(a,c) = {p/2}.$ But since $T(a,c)$ must be an integer it will imply $p=2,T(a,b) = -2$ and $T(a,c) =1$

Now, if $p=2,$ the original equation will imply $a^n \equiv b^n \equiv c^n \pmod 2,$ hence $a \equiv b \equiv c \pmod 2$ and $T(a,b) \equiv T(a,c) \equiv T(a,b) \equiv n\cdot a \pmod 2$ which is a contradiction given the values of $T(a,b)$ and $T(a,c).$

So done.
[/hide]

The Claim and the fact that $p$ is prime now implies that $(*)$ can only be satisfied if one of $|T(a,b)|, |T(a,c)|, |T(b,c)|$ be equal to $p^3$ and the remaining two be equal to $ 1.$ Then assume without of generality that $|T(a,b)| = p^3$ and $|T(b,c)| = |T(c,a)| =1.$ Thus, $(1A), (1B)$ and $(1C)$ will implie that
$$ v_p(a-b) + 3 = 1 + v_p(b-c) $$
$$ v_p(b-c) = 1 + v_p(c-a) $$
$$ v_p(c-a) = 1 + v_p(a-b) $$
That is $v_p(b-c) = 2 + v_p(a-b)$ and $v_p(c-a) = 1 + v_p(a-b);$ Then we will have
$$ a-b = p^{\alpha}k$$
$$ b-c = p^{\alpha + 2}s $$
$$ c-a = p^{\alpha+1}r $$
where $\alpha$ is an non-negative integer and $k,s,r$ are integers not divisible by $p.$ (Note that this can be done since, $a-b, b-c, c-a$ are assumed to be non-zero integers.) Then, adding up this equations gives
$$ 0 = p^{\alpha}(k + ps + p^2r) \Rightarrow k = -ps -p^2r,$$
contradiciton since $k$ is an integer not divisible by $p.$ So done. $\square$

Solved with [b]pyramid_fan90[/b].

Assume for the sake of contradiction that $a$, $b$, and $c$ are pairwise distinct. If $n$ is odd then suppose that $a>b$. It follows that $c>b$ since otherwise we would have $a^n+pb>b^n+pc$. We also obtain $c<a$ because otherwise $c^n+pa>a^n+pb$. However, this implies $b^n+pc<c^n+pa$, a contradiction. Then $a<b$. Similarly, $b<c$ and $c<a$, a contradiction.

Henceforth assume that $n$ is even. Notice that we may obtain
\[\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{c^n-a^n}{b-a}=p\]
which implies that
\[\prod\frac{a^n-b^n}{a-b}=\prod(a^{n-1}+a^{n-2}b+\dots+b^{n-1})=-p^3.\]
Notice that each of these sums is even when the parity of the two variables is the same, and each one is odd when the parity of the two variables differs. This implies that the product is even, or that $p=2$.

Now, notice that the parity of all the variables is the same. This implies that each of the sums was an even number. Now, the three sums can only be $(-2, -2, -2)$ or $(-2, 2, 2)$. FTSOC, suppose that the sums are $(-2, 2, 2)$ and WLOG set $\frac{a^n-b^n}{a-b}=2$. Then we also have that $\frac{a^n-b^n}{c-b}=2$ which implies that $a=c$, a contradiction.

Finally, suppose the sums are $(-2, -2, -2)$. Then clearly $b-a=c-b$, $c-b=a-c$, and $a-c=b-a$, which implies $a=b=c$. We are done. $\blacksquare$

For the sake of contradiction we will have that $a,b,c$ are not equal to each other.

[color=#f00][b]CLAIM:[/b][/color] We have that if $n$ is odd then there is not $a,b,c$ that satisfy the problems equation.

[color=#00f][b]PROOF:[/b][/color] : WLOG $a>b$ and $c>b$ and we have $c<a$ but after plugging our values in we get that $b^n+pc<c^n+pa$ which is a contradiction. Similarly we have that the other cyclic variants are also a contradiction. QED.

[color=#0f0][b]$N$ is even case [/b][/color] For this case after some rearrangements we get that $\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{c^n-a^n}{b-a}=p$. Since we know that $a-b|a^n-b^n$ we have that $\prod(a^{n-1}+a^{n-2}b+\dots+b^{n-1})=-p^3.$  We will notice that $(a^{n-1}+a^{n-2}b+\dots+b^{n-1})$ has a divisor of $2$ which implies that $p=2$. Since the product is $-p^3$ we have that the sums could be $(-2,-2,-2)$ or $(-2,2,2)$ we would like to have $(-2,2,2)$ as the sums but we see that there is a immediate contradiction that $a=c$ so we have that $a=b=c$ works .  QED $\blacksquare$

[color=#f0f][b]REMARKS[/b][/color]: This problem was really nice but it took me some  time for me to notice that $p=2,$

If two of the variables are equal, say $a=b,$ then we get $pb=pc$ or $a=b=c$. Assume for the sake of contradiction that $a\neq b\neq c$. We find that $p = \frac{a^n-b^n}{c-b} = \frac{b^n-c^n}{a-c} = \frac{c^n-a^n}{b-a}$. It is easy to check that if $n$ were odd then one of the terms is negative, contradiction since $p$ is positive. This means $n$ is even. We must have two of the variables congruent mod 2 by pigeonhole, so WLOG, $a\equiv b \pmod 2.$ But since $\frac{a^n-c^n}{a-b}$ is an integer, $a\equiv c \pmod 2$ as well. Taking the product of the three values, we see that $$p^3 = - \prod_{cyc} \frac{a^n-b^n}{a-b}.$$ If $2\mid a,b$, let $\nu_2(a) = k, \nu_2(b)=m$. If $k\neq m$, then $\nu_2(a-b) = \min(m,k)$ and $\nu_2(a^n-b^n) = \min(mn, kn)$ and since $n$ is even, $\min(mn,kn) > \min(m,k)$ implying $2\mid p$ or $p=2.$ If $m=k$, let $a = 2^m \cdot x, b=2^m \cdot y.$ Then we see that $\nu_2(a-b) = m + \nu_2(x-y)$ and $\nu_2(a^n-b^n) = mn + \nu_2(x^n-y^n) = mn + \nu_2(x-y) + \nu_2(x+y) + \nu_2(n) - 1 > m+\nu_2(x-y)$, so $2\mid p$ or $p=2.$ If $p=2$, then if 3 non 1 integers have a product of $-8$, they must be $(-2,-2,-2)$ or $(-2,2,2)$. However it is easy to check that one of the cyclic values $\frac{a^n-b^n}{c-b}$ must be equal to the corresponding cyclic value $\frac{a^n-b^n}{a-b},$ or $a=b$, contradiction. This means $a=b=c.$

Subtracting pairs of equations from each other,
\begin{align*}
a^n-b^n &= p(c-b) \\
b^n-c^n &= p(a-c) \\
c^n-a^n &= p(b-a).
\end{align*}
This implies $$-p^3 = \frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{c-a}.$$ All three terms must equal $-p$, as they are all integers and otherwise two of them would contradict the third. But now, cross-equating, $$-p(a-b) = a^n-b^n = p(c-b) \implies a+c=2b.$$ Thus, each of $a, b, c$ is the average of the other two, so $a=b=c$.

Took me $5$ hours to type the LaTeX along with fixing all the fakesolves I had made  :stretcher: .

Don't someone dare say this was easy. Might be easy for some omegaorsmax guy like you :omighty: but I being a nub felt that this problem was very very instructive as it connects all the different ways of algebraic manipulations and joins all of them together.

We case bash this :-D. Firstly, if any two of the ones become equal, it's easy to see that all of them become equal. FTSOC assume that none of them are equal. Before we actually begin, notice that the equations are just cyclic and not symmetric. So we cannot take $a>b>c$. What we can actually do is take $\max\{a,b,c\}=a$. Now manipulating the first and the second equalities, we get,\[\dfrac{a^n-b^n}{c-b}=\dfrac{b^n-c^n}{a-c}=\dfrac{c^n-a^n}{b-a}=p.\tag{$\spadesuit$}\] Cross multiplying and simplifying further we get, $a^{n+1}+b^{n+1}+c^{n+1}=ab^n+bc^n+ca^n$. That is,\[a\cdot a^n + b\cdot b^n+c\cdot c^n=c\cdot a^n+a\cdot b^n+b\cdot c^n.\tag{$\clubsuit$}\] Now if $n$ is odd, note that for any ordering of $a$, $b$ and $c$ (like say for $a>b>c$), we are going to have a similar ordering for $a^n$, $b^n$ and $c^n$ (like say for $a^n>b^n>c^n$). This however forces a contradiction to the rearrangement inequality applied to $(\clubsuit)$. Thus we must have that $n$ is even.

Now $a$ must be positive, otherwise we arrive at a contradiction for $\dfrac{c^n-a^n}{b-a}=p$ as the LHS becomes -ve. Now we divide the rest into two cases :P :
[list]
[*] $a>b>c$.
[/list]
From $(\spadesuit)$ we have $\dfrac{a^n-b^n}{c-b}=p$ which then forces $|a|<|b|$. This then implies that $c<b<-a$ as $a$ is +ve which further gives that $|c|>|a|$ which however is a contradiction to $\dfrac{c^n-a^n}{b-a}=p$ as the LHS again becomes -ve.

So we must have:
[list]
[*] $a>c>b$.
[/list]
Now moving onto the final part of the solution, manipulating $(\spadesuit)$ we get the following identities,
\begin{align}
    \dfrac{(a-b)(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})}{(c-b)}&=p\\
    \dfrac{(b-c)(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})}{(a-c)}&=p\\
    \dfrac{(c-a)(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})}{(b-a)}&=p
.\end{align}

Now from $a>c>b$ we are going to have $\dfrac{b-a}{c-a}>1$ from which we can deduce $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=\dfrac{p(b-a)}{(c-a)}>p$. Now multiply $(1)\times(2)\times(3)$ to get,\[(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=-p^3\] which along with our inequality that we arrived at above forces $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})\in\{p^2,p^3\}$.

Now again case bash OOOOOOF!!!!! If $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=p^2\implies\dfrac{b-a}{c-a}=p$. Then we multiply $(1)\times(2)$ to get, \[(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})=-p.\] Now note that $(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})$ can't be $=p$ otherwise this forces $a=c$. Thus we must have $(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})=1$ and $(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})=-p$. Thus our final equations change to,
\begin{align}
    \dfrac{(a-b)}{(c-b)}&=p\\
    \dfrac{(b-c)}{(a-c)}&=-1\\
    \dfrac{(c-a)}{(b-a)}&=\dfrac{1}{p}
.\end{align}

Now from $(3)$ and $(1)$ we get $\dfrac{1}{p}=\dfrac{c-a}{b-a}=\dfrac{b-c}{b-a}$ which upon adding gives $\dfrac{1}{p}=\dfrac{1}{2}$ that is $p=2$ and $a+b=2c$. So now our problem statment becomes \[(2c-b)^n+2b=b^n+2c=c^n+2(2c-b).\] From this we get $2(c-b)=(2c-b)^n-b^n$ which gives $|2c-b|> |b|$. Now note that the RHS is an increasing function (can be easily proved by differentiating and checking off the rest of the cases manually) and the equality clearly holds at $n=1$ which thus forces $n=1$ but we had that $n$ is even which forces a contradiction.

Now for the other case, if $(c^{n-1}+c^{n-2}\cdot a+\cdots+c\cdot a^{n-2}+a^{n-1})=p^3\implies\dfrac{b-a}{c-a}=p^2$. Again multiply $(1)\times(2)$ to get $(a^{n-1}+a^{n-2}\cdot b+\cdots+a\cdot b^{n-2}+b^{n-1})=1$ and $(b^{n-1}+b^{n-2}\cdot c+\cdots+b\cdot c^{n-2}+c^{n-1})=-1$ and then we have our final equations as follows,
\begin{align}
    \dfrac{(a-b)}{(c-b)}&=p\\
    \dfrac{(b-c)}{(a-c)}&=-p\\
    \dfrac{(c-a)}{(b-a)}&=\dfrac{1}{p^2}
.\end{align}

So from $(1)$ and $(2)$, we get $p-\dfrac{1}{p}=\dfrac{a-b}{c-b}+\dfrac{a-c}{b-c}=1$ which upon solving gives a clear contradiction.

lol what is this proof

Define \[f(a,b)=a^{n-1}+a^{n-2}b+\dots+b^{n-1}.\]

If two of these are equal then obviously all three are equal. So assume FTSOC that $a$, $b$, and $c$ are pairwise distinct. Then
\[p=\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{c^n-a^n}{b-a}\]
so
\[-p^3=f(a,b)\cdot f(b,c)\cdot f(c,a).\]

[b]Lemma 1.[/b] If $|f(a,b)|=1$, then $p\mid a-b$ and $p\nmid n$.
[b]Proof.[/b] \[|a^n-b^n|=|a-b|\rightarrow p=\frac{|a-b|}{|c-b|}\rightarrow p\mid a-b.\] So by LTE, $p\nmid n$.

[b]Claim.[/b] The absolute values of the factors being $1$, $1$, and $p^3$ is impossible.
[b]Proof.[/b] If $|f(a,b)|=1$, then we can use lemma 1 to get $p\mid a-b$ and $p\nmid n$. Similarly, WLOG $|f(b,c)|=1$, so $p\mid b-c$. So $p\mid c-a$.
We have $|f(c,a)|=p^3$, so
\[\nu_p\left(|c^n-a^n|\right)>\nu_p\left(|c-a|\right).\]
But $c-a$ is a multiple of $p$, contradiction by LTE.

Therefore, at least one of the absolute values must be $p$.

[b]Claim.[/b] None of the factors can be $p$.
[b]Proof.[/b] Assume WLOG that this is $f(a,b)$. Then
\[p=\frac{a^n-b^n}{a-b}=\frac{a^n-b^n}{c-b}\rightarrow a=c,\]
contradiction.

[b]Lemma 2.[/b] If $f(a,b)=-p$, then $a+c=2b$.
[b]Proof.[/b] Assume WLOG that this is $f(a,b)$. Then
\[p=\frac{a^n-b^n}{b-a}=\frac{a^n-b^n}{c-b}\rightarrow a+c=2b.\]

[b]Claim.[/b] The factors can't be all $-p$.
[b]Proof.[/b] Otherwise, by lemma 2, we have $a+c=2b$ and cyclic variations so $a=b=c$, contradiction.

[b]Claim.[/b] The factors can't be $1$, $-p$, and $p^2$ nor $-1$, $-p$, and $-p^2$.
[b]Proof.[/b] Let $|f(a,b)|=1$. By lemma 1, $p\mid a-b$ and $p\nmid n$. So with an argument similar to the first claim, $p\nmid b-c$ and $p\nmid c-a$. But we know that $a+b=2c$ or $b+c=2a$, both of which are enough to show that $a$, $b$, and $c$ are all equivalent mod $p$, contradiction.

We are done!

Storage from a while ago

We have that $a^n-b^n=p(c-b)$ and cyclically. If n=1 it's evident that a=b=c, if WLOG $a=b$ it's evident that a=b=c. Assume henceforth $n>1$ and $a,b,c$ distinct. Then $$\frac{a^n-b^n}{a-b}\frac{b^n-c^n}{b-c}\frac{c^n-a^n}{c-a}=-p^3.$$ Hence (noting $|a^n-b^n|>|a-b|$ $a^n-b^n=\pm(a-b)p$ (otherwise if it has $p^2$ or smth then some other term will have same magnitude). If it's equal to the positive version, then $a-b=c-b$, contradiction, hence all of them must be negative multiplied by p, implying $a+c=2b$ and cyclically, so $2b-2c=c-b$ or $c=b$ again, contradiction! $\blacksquare$

Note that $n=1$ iff $a=b=c$. Then, assume that $n \neq 1$. We have $$p=\frac{a^n-b^n}{c-b}=\frac{b^n-c^n}{a-c}=\frac{a^n-c^n}{a-b} \Rightarrow -p^3=\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{a^n-c^n}{a-c}.$$ Note that each of these terms must be integers.

$\textbf{Claim:}$ Each of the terms contain a factor of $p$.

$\emph{Proof.}$ Assume for the sake of contradiction that instead at least one of the terms is $\pm 1$. But since $n \neq 1$, we always have $|x^n-y^n| > x-y$ for integers $x$ and $y$, contradiction.

From here, it immediately follows that each term has exactly one factor of $p$. There are now $2$ cases to check: either $1$ term is equal to $-p$ or all $3$ terms are equal to $-p$. Considering the first case, assume $$a^n-b^n=-p(a-b),$$ $$b^n-c^n=p(b-c),$$ $$a^n-c^n=p(a-c).$$ But from above, we also see that $b^n-c^n=p(a-c)$, so $a=b$, violating the condition that $n \neq 1$. Now, consider the second case in which all terms are equal to $-p$. We have $$a^n-b^n=-p(a-b),$$ $$b^n-c^n=-p(b-c),$$ $$a^n-c^n=-p(a-c).$$ Again from above, we have $b^n-c^n=p(a-c)$, so $a+b=2c$. Similarly, $a+c=2b$ and $b+c=2a$. Solving, we get $a=b=c$ again. We have covered all cases, so we are done.

This is more alg than NT (as usual bruh)

By rearranging we have that
\begin{align*}
    a^n-b^n &=p(c-b)\\
    b^n-c^n &= p(a-c)\\
    a^n - c^n &= p(a-b)\\
\end{align*}
Multiplying these together we get that,
\[\frac{a^n-b^n}{a-b}\frac{b^n-c^n}{b-c}\frac{a^n-c^n}{a-c}=-p^3\]
First note that $n$ is clearly even, as if $n$ was odd, then the denominator and numerator of each fraction would have the same sign, making the product positive. Thus, $n$ is in fact even.\\
Note that one of these fractions can be $1$ or $-1$ if and only if $a,b,c\in{0,1,-1}$. We can check the cases when $a=0,-1,1$ by substituting back into the equations, to see that these cases in fact do not work. Thus, each fraction is either $p$ or $-p$.\\
Now, there are two cases $(-p,-p,-p)$ or $(p,p,-p)$.\\
$Case \ 1:$ In this case we first consider
\begin{align*}
    \frac{b^n-c^n}{b-c} &= -p\\
    b^n-c^n = -p(&b-c) = b^n - a^n
\end{align*}
Thus, in-fact $a^n=b^n \implies a=b$ and in turn $a=b=c$.\\
$Case \ 2:$ We know that always, atleast one of $\frac{a^n-b^n}{a-b}$ and $\frac{a^n-c^n}{a-c}$ is equal to $p$. Assume WLOG, that $\frac{a^n-b^n}{a-b} = p$. Then, we have that
\begin{align*}
    \frac{a^n-b^n}{a-b} &= p\\
    a^n-b^n = p(&a-b) = a^n - c^n
\end{align*}
Thus, we must have that $b^n = c^n \implies b=c$, and thus $a=b=c$. Now, substituting $a=b=c$, we see that all three equations are satisfied without contradictions. Therefore, $a,b,c\in \mathbb{Z}$ satisfy the given equations if and only is $a=b=c$.

First, for completeness, we note $a=b=c$ can hold. Assume by contradiction that $a$, $b$, and $c$ are not all the same.

Suppose two of the three are equal; WLOG let $a=b$. Equating the first two expressions, we get \[pb = pc \implies b=c,\]

contradiction. Next, suppose all three are distinct. Then
\[(a^n-b^n)(b^n-c^n)(c^n-a^n) = p(c-b) \cdot p(a-c) \cdot p(b-a)\]
\[\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{a^n-c^n}{a-c} = -p^3,\]
implying that all three factors must be $\pm p$, and it's easy to verify all cases lead to $a=b=c$. $\blacksquare$

 We will rewrite the equations as follows

\begin{align*}
a^n-b^n &= p(c-b) \\
b^n-c^n &= p(a-c) \\
c^n-a^n &= p(b-a)
\end{align*}

Multiplying and simplifying, we get

\[\frac{a^n-b^n}{a-b} \cdot \frac{b^n-c^n}{b-c} \cdot \frac{c^n-a^n}{c-a}=-p^3\]

Then, $a^n-b^n=\pm p(a-b)$ since $a-b \mid a^n-b^n$ and they cannot be equal. (This is the same for cyclic variations)

If $a^n-b^n=p(a-b)$, we have

\[a^n-b^n=p(a-b)=p(c-b)\]

so $a=c$, implying through some simple calculations that $a=b=c$.

If $a^n-b^n=-p(a-b)$, we have

\[a^n-b^n=-p(a-b)=p(c-b)\]

Then, $2b=a+c$ and same with cyclic variations, meaning $a=b=c$.

This covers all cases, so our proof is complete. $\square$


Asymptote
trigonometry
limit
function
calculus
calculus computations

How do you finsd the [b]Asymptotes[/b] of a given curve, say,
$ (\sqrt {1 \plus{} x^2}) \cdot \sin {\frac {1}{x}} ?$

hello, substituting $ \frac{1}{x}\equal{}t$ we have $ \sqrt{1\plus{}t^2}\cdot\frac{\sin(t)}{t}$ for $ t>0$ or $ \minus{}\sqrt{t^2\plus{}1}\cdot \frac{\sin(t)}{t}$ for $ t<0$. If $ x$ tends to infinity then $ t$ tends to zero.  So we get
$ \lim_{t \to 0\plus{}0}\sqrt{t^2\plus{}1}\cdot \frac{\sin(t)}{t}\equal{}1$. In the other case we have the result $ \minus{}1$.
So we get $ 1$ as the asymptote for $ x$ tends to $ \plus{}\infty$ or $ \minus{}1$ for $ x$ tends to $ \minus{}\infty$.
Sonnhard.

1) If $ \lim_{x \to \infty} f(x) \equal{} L$, then the line $ y \equal{} L$ is an horizontal asymptote for f(x).

On the present case,

$ \lim_{x \to \pm \infty} \sqrt {1 \plus{} x^2} \sin \frac1x \equal{} \lim_{x \to \pm \infty} \sqrt {1 \plus{} \frac {1}{x^2}} \frac {\sin \frac1x}{\frac1x} \equal{} 1$

and so $ y \equal{} 1$ is an horizontal asymptote.

2) If $ \lim_{x \to \infty} \frac{f(x)}{x} \equal{} a$ and $ \lim_{x \to \infty} [f(x) \minus{} ax] \equal{} b$, then the line $ y \equal{} ax \plus{} b$ is an oblique asymptote of f(x).

On the present case, its easy to check that this will also lead to y = 1.

3) If $ \lim_{x \to c} f(x) \equal{} \infty$, then the vertical line $ x \equal{} c$ is an asymptote of f(x).

On the present case there are no vertical asymptotes.

hello, for $ x<0$ we have $ \lim_{x \to \minus{}\infty}\sqrt{x^2\plus{}1}\cdot\sin\left(\frac{1}{x}\right)\equal{}\minus{}1$ so the searched asymptote is $ \minus{}1$ in this case.
Sonnhard.

Hello
Even Dr. and Carcul has explain all the things .Here is one very easy example.
suppose $ f(x) \equal{} x \plus{} \frac {1}{x}$
Now
$ \lim_{x\to 0^{ \plus{} }}f(x) \equal{} \infty$ and $ lim_{x\to 0^{ \minus{} }} \equal{} \minus{} \infty$
so $ x \equal{} 0$ is a vertical asymptote.
$ lim_{x\to\infty}\frac {f(x)}{x} \equal{} 1;lim_{x\to \minus{} \infty}\frac {f(x)}{x} \equal{} 1;$
$ lim_{x\to \plus{} \minus{} \infty}(f(x) \minus{} x) \equal{} 0 ;$
$ y \equal{} x$ is an inclined asymptote.
Thank u.

[quote="Carcul"]And in general, g(x) is an asymptote of the function f(x) if $ \lim_{x \to \infty} [f(x) \minus{} g(x)] \equal{} 0$.[/quote]
Ok thanks, that is precisely what i wanted. thank you.


what's the best specific area of chemistry that's best for research for high school students?

[quote="amirhtlusa"]Organic
and Electro-chemsitry which has always been my favorite areas...[/quote]

I hate them the most  :P

I like stochiometry and the gas laws.

you mean organic and electro chemistry are the most approachable topic for high school research. from research i'm meaning siemens.
thanks.

I dont know. Chemsitry is ok I guess its not as fun as physics. if you ever look for a topic in physics to study about do atomic or nuclear hgih energy. Atomic physics is awesome and so is nuclear physics. For chemsitry I would say organic. But physics chemsitry is also ok


induction
function
algebra proposed
algebra

The Fibonacci $ (F_n): F_0\equal{} F_1\equal{} 1, F_{n\plus{}2}\equal{} F_{n\plus{}1}\plus{} F_n$ for $ n\geq 0$. Determine all pairs $ (k,m)$ of integer, with $ m> k\geq 0$, for which the sequence $ (x_n)$ defined by $ x_0\equal{}\frac{F_k}{F_m},x_{n\plus{}1}\equal{}1$ for $ x_n\equal{}1,x_{n\plus{}1}\equal{}\frac{2x_n\minus{}1}{1\minus{}x_n}$ for $ x_n\neq 1$ contains the number $ 1$.

If we have $x_{n+1}=1$, we must have $x_{n}=\frac{2}{3}$
If we have $x_{n}=\frac{2}{3}$, we must have $x_{n}=\frac{5}{8}$
It is easy to prove by induction that if $x_{n+1}=\frac{F_{2k)}{F_{2k+1}}}$, we must have $x_{n+1}=\frac{F_{2k+2)}{F_{2k+3}}}$
As a result if $x_0=\frac{F_{2k}}{F_{2k+1}}$ for some integer $k$, we will get $x_k=1$. Note that for all $k$, $\frac{F_{2k}}{F_{2k+1}}$ is larger than the golden ratio.
Consider that $\frac{F_{2k+1}}{F_{2k+2}}$ is a increasing function for k, its limit is to the golden ratio. Therefore it is impossible for some $i,j$ such that $\frac{F_{2i}}{F_{2i+1}}=\frac{F_{2j+1}}{F_{2j+2}}$
As a result, if and only if $(k,m)$ can be written as the form of $(2n,2n+1)$ for any integer $n\ge 0$


function
algebra unsolved
algebra
functional equation

Find all functions $f: \mathbb{Q}^{+}\rightarrow\mathbb{Q}^{+}$ satifying all following conditions:
1) $f(x).f\left(\frac{1}{x}\right)=1$  $\forall x\in\mathbb{Q}^{+}$
2) $f(2x)=f(f(x))$  $\forall x\in\mathbb{Q}^{+}$
where $\mathbb{Q}^{+}$ stands for set of positive rational numbers

hien, the problem is wrong. The correct is:

Find all functions $f: \mathbb{Q}^{+}\rightarrow\mathbb{Q}^{+}$ satifying all following conditions:
1) $f(x).f\left(\frac{1}{x}\right)=1$  $\forall x\in\mathbb{Q}^{+}$
2) $f(2x)=2f(f(x))$  $\forall x\in\mathbb{Q}^{+}$
where $\mathbb{Q}^{+}$ stands for set of positive rational

[quote="hien"]Thank for help me correcting the problem! Who can suggest any ideas?[/quote]
I'm trying... I hope I can help you soon... :)

I think that is only solution for that problem but up to now i can't prove it

i don't think so
in fact,let S={x|x=p/q, p,q are odd number,q>0,(p,q)=1}
then for any T,a subset of S
define g(x) as follow:
if x doesn't belongs to T,  g(x)=y,y is any one of elements of T
if x belongs to T,             g(x)=x,
let's define F(x) as follow:
for any x=m$2^{k}$,where k is an integer and m belongs to S
F(x)=$2^{k}$g(m)

F(x) defined as such are all suitable
and i think that should be the answer to this question

There are infinetely many solutions. We can find only multiplicative solutions
\[f(\prod_{i}p_{i}^{k_{i}})=\prod_{i}p_{i}^{g_{i}(k_{i})}. \]
For these solutions we had functions $g_{p}: Z\to Z$ satisfyed $g_{p}(-n)=-g(n)$. Second condition for our functions is $g_{2}(n+1)=1+g_{2}(g_{2}(n)), \ g_{p}(n)=g_{p}(g_{p}(n))$.
It mean, that $g_{p}$ is projector. All functions satisfyed it can be build in these way. Let $A_{p}$ is subset N, let $t_{p}: N/A_{p}\to A_{p}$ any function, we can define $g_{p}(n)=n, n \in A_{p}, g_{p}(n)=t_{p}(n), n\in N/A_{p}$ and $g(0)=0,g(-n)=-g(n) \ \forall n\in N$.
There are infinetely many functions satisfyed your conditions. For example $g_{2}(n)=n, \ g_{p}(n)=a_{p}sig(n),a_{p}\in Z \ \forall p\in P$.

There is typo in the first equation, it should contain addition instead of multiplication.
[hide=For a solution]follow the white rabbit [aops][url=http://artofproblemsolving.com/community/c6h127357p11003332]>[/url][/aops][/hide]


Let  $ a,\ b,\ c\in\mathbb{R}^{*}$ such that $ ab\plus{}bc\plus{}ca\equal{}0$. Prove that the follow inequality holds: \[ \frac{4}{3}\left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)\ge a^{2}\left(b\minus{}1\right)\left(c\minus{}1\right)\plus{}b^{2}\left(c\minus{}1\right)\left(a\minus{}1\right)\plus{}c^{2}\left(a\minus{}1\right)\left(b\minus{}1\right).\]

[i]Claudiu Mindrila[/i]

[quote="kunny"]What do you mean $ \mathbb{R^{*}}$?[/quote]

$ R^*$ represents the set of the real numbers not equals with zhero .

Thanks.

$ ab\plus{}bc\plus{}ca\equal{}0\Longleftrightarrow \minus{}a(b\plus{}c)\equal{}bc$, we have $ a^2(b\minus{}1)(c\minus{}1)\equal{}a^2bc\minus{}a\cdot a(b\plus{}c)\plus{}a^2\equal{}a^2bc\plus{}abc\plus{}a^2$,

$ \therefore 4*(L.H.S.\minus{}R.H.S.)\equal{}a^2\plus{}b^2\plus{}c^2\minus{}3abc(a\plus{}b\plus{}c)\minus{}9abc$

$ \equal{}(a\plus{}b\plus{}c)^2\minus{}3abc(a\plus{}b\plus{}c)\minus{}9abc$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

How can i show that S_{4} has a quotient group ismorphic to S_{3}
but S_{5} has noquotien group isomorphic to S_{4}

Ok S_4 / V is isomorphic to S_3

now the other...

mm i need to learn how to paste the code

and speak english as well hehe

Ok!!
so the only normal subgroups in S_5 are 1 and A_5
so S_5 / A_5 has order 60
and S_4 has order 24

but i still dont see why S_4 / V is cyclic so is not Z_6 and it ahs to be S_3

[quote="Keenan"]so S_5 / A_5 has order 60[/quote]

Er, it has order 2, actually.

[quote="Keenan"]but i still dont see why S_4 / V is cyclic so is not Z_6 and it ahs to be S_3[/quote]

I'm assuming that you mean "why S_4 / V is [i]not[/i] cyclic".  Proofread your posts, please, it takes time to do mind-reading.

Anyway, for the answer to your question, just observe that $ S_4$ has no elements of order 6.


Hi.. This may sound stupid but... How many point pairs (p1,q1), (p2,q2), ..., (pn,qn) do I need in order to calculate a 3x3 linear transform so that this is true:

For all i=1,...,n : T·p(i) = q(i)

I think it should be an easy question, but I'm a little confused..
A little help about "how to" calculate this 3x3 transform would be REALLY appreciated.

Thanks.

You need three points, and you need those three $p_i$ to be linearly independent. If we know the action of $T$ on a spanning set, we know it everywhere and $T$ is unique. For existence, see below.

Suppose $p_1,p_2,p_3,q_1,q_2,q_3$ are column vectors in $\mathbb{R}^3$, and $p_1,p_2,p_3$ are linearly independent.
Let $P$ be the matrix with columns $p_1,p_2,p_3$; $P$ is invertible. Let $Q$ be the matrix with columns $q_1,q_2,q_3$. We have
$P\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}=p_1$, $P\cdot\begin{bmatrix}0\\1\\0\end{bmatrix}=p_2$, $P\cdot\begin{bmatrix}0\\0\\1\end{bmatrix}=p_3$,
$Q\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}=q_1$, $Q\cdot\begin{bmatrix}0\\1\\0\end{bmatrix}=q_2$, and $Q\cdot\begin{bmatrix}0\\0\\1\end{bmatrix}=q_3$.

Now define $T=Q\cdot P^{-1}$. $Tp_1=Q\cdot(P^{-1}p_1)=Q\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}=q_1$, and similarly $Tp_2=q_2,Tp_3=q_3$.

If the $p_i$ aren't linearly independent, existence may fail. If the $p_i$ don't span the space, uniqueness fails.
This construction can be extended to $\mathbb{R}^m$ for any $m$, in obvious ways.

Hi! thanks...
Now I realized that I want to have a transformation like this:

For all p,q that match, then  q = T·p + d
So... I would have:

[q1 q2 q3] = T·[p1 p2 p3] + [d d d] ??

And then... I would need 6 pairs in order to make 2 equations?

Q1 = T·P1 + D
Q2 = T·P2 + D

=> [Q1-Q2] = T·[P1 - P2]
=> T = [Q1-Q2]·[P1 - P2]^(-1)

And then calculate D ??

My post is all about questions because I'm not really sure if i'm doing ok...
Could you "light" me?

Do I need 6 l.i. points to do that?
Thanks![/code]

We only need four points, not all on one plane.

If we have four noncoplanar points $p_1,p_2,p_3,p_4$, we can find constants $c_1,c_2,c_3,c_4$ such that $c_1p_1+c_2p_2+c_3p_4=0$ and $c_1+c_2+c_3+c_4=1$.

Let $F(x)=Tx+d$.
$c_1F(x_1)+c_2F(x_2)+c_3F(x_3)+c_4F(x_4)=F(c_1x_1+c_2x_2+c_3x_3+c_4x_4)$
$=F(0)=d$.
Subtract off $d$ and find $T$ as before.


Prove that $ \sin \hat{A} \plus{}\sin \hat{B} \plus{} \sin \hat{C} \equal{} \dfrac{p}{2r}$, where $ \hat{A}$, $ \hat{B}$ and $ \hat{C}$ are angles of a triangle $ ABC$, $ p$ is the semiperimeter of triangle $ ABC$ and $ r$ is the radius of incircle.

I think the problem isn't right. It should be $ \sin A\plus{}\sin B\plus{}\sin C\le \frac{p}{2r}$ or $ \sin A\plus{}\sin B\plus{}\sin C\equal{} \frac{p}{R}$.

[hide="Solution"]$ \frac{a}{\sin A}\equal{}\frac{b}{\sin B}\equal{}\frac{c}{\sin C}\equal{}2R\implies \sin A\plus{}\sin B\plus{}\sin C\equal{}\frac{a\plus{}b\plus{}c}{2R}\equal{}\frac pR\le \frac{p}{2r}$. Where the last inequality follows from the well known $ R\ge 2r$.[/hide]

Perfect nayel, that's why i posted it here.


integration
calculus
logarithms
trigonometry
calculus computations

$\int \textrm{sech}(\theta) d\theta$

I looked at this in the table of integrals, and couldn't find a method to solve it.  Does anyone have a suggestion for a substitution I could make?  Please don't send the complete solution if you reply, just a hint.

Thanks a lot,

Alex.

$\textrm{sech} \theta = \frac{2}{e^\theta + e^{- \theta}}$

There are two reasonable methods I see. First, we look at the analogy with $\int \sec t\, dt$. Working with that analogy, we use $\mathrm{sech}' t=-\mathrm{sech } t\tanh t$ and $\tanh' t=\mathrm{sech}^2 t$ to get $(\mathrm{sech } t+i\tanh t)'=\mathrm{sech } t(i\mathrm{sech } t-\tanh t)$, so $\int\mathrm{sech } t\, dt=-i\ln(\mathrm{sech } t+i\tanh t)$. That complex expression doesn't look like what we really want, so we try something else.

Second, we try a hyperbolic-algebraic-trig substitution chain.
$u=\sinh t$ transforms $\int\mathrm{sech } t\,dt$ to $\int\frac1{u^2+1}\,du$
This gives $\int\mathrm{sech } t\, dt=\tan^{-1}(\sinh t)$
Alternately, $u=\cosh t$ transforms the integral to $\int\frac1{u\sqrt{u^2-1}}\,du$.
This gives $\int\mathrm{sech } t\,dt=\sec^{-1}(\cosh t)$.

Is that enough different forms for you?

Yes that is great.  I was looking for the $u=\textrm{sinh}(t)$ substitution I guess because the answer is what my table of integrals has.

Thanks again,

Alex.


How many different numbers can be constructed using the digits 0, 1, 2, 2? All of the digits must be used each time and no number can begin with 0.

[hide] If there were no restrictions, then we would have $ 4!$ numbers. But there are two $ 2$'s so we must divide this result by $ 2!$. Doing so yields $ \frac{4!}{2!}\equal{}12$. However, $ 3$ of these will have $ 0$ as the leading digit ($ 0122$, $ 0212$, $ 0221$) so our answer is $ 12\minus{}3\equal{}\boxed{9}$.[/hide]


everytime I insert a picture, it leaves a huge vertical space behind it (like half the page is blank).  How can I reduce the space (I don't want the pictures to take up three pages of my work)????

What command are you using to insert the picture? I know how to fix the problem if you're using the includegraphics command, but there are a ton of ways to put pictures in $\LaTeX$ documents.

yes, I use the include graphics one, but i figure out the problem (my view port thing is too small so it include like a dozen space...)


Hello,

I have seen a few puzzle posts in the Round Table and Fun & Games forum lately (reality is that not too many, but some) and I was wondering if the administrators would consider to create a Puzzle forum. By puzzles I mean things similar to the books of Martin Garner (I think the most popular autor on the "area"). Some of those problems (criptosums or "the Einstein puzzle" kind of problems) would probably partially fit on some olympiad or mathcounts forum, but not all of them, and it would be nice to have a forum grouping all of them (and posting them without feeling that we are posting things slighly out of the topic :P ).

Any way, it is just a suggestion.


inequalities
calculus
algebra
polynomial
geometry
function
number theory

Hi. I'm looking for a good algebra PDF with solutions. Can you help me?

Wich level you want ? I know a lot of ! Just enter the level ! :roll:

Thank you!  I'm high school level, close to MMO but still not that good.

[quote="A. S. K."][quote="PenguinIntegral"]What specifically do you need help with?[/quote]

Well, high school algebra contains inqualities, equations, identities and so...[/quote]
But what specific topics are you unframiliar with? If you just have no clue about Inequalities, Equations, functional itdentities etc, then go buy AoPS II. (But it lacks for anything classically hard in the inequality department (ie calculus), so go buy the cauchy master class book if you want inequalities. It's a good intro to creative inequality stuff though).

okay, thanks, but at the moment i'm looking for internet stuff (pdf). does anybody know one?

Again, what exactly are you looking for? Please enumerate the subjects, if I find something on polynomials it is no help if need help with functional equations.

For a basic Number Theory reference http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf
might be a good start. You can post the practice problems on the boards if you need help with them.

Ok I need help with:

1. equations
2. polynomials
3. inqualities

That could be a start. Any pdf?

For Inequalities, review the thread in the AMC section; it has plenty of pdfs for you to munch on :)

For Equations, just solve a lot of problems. If you know your algebra basics and how to factor, there is nothing else to teach besides creativity.

For Polymomials, there is really no good .pdf circulating specifically for problem solving techniques. You might be able to find an Illegal pdf of a book by Barbeau that is very good, but also has some things that you would never use on a problem solving competition. (eg how to solve quintic equations by hand, and some calculus stuff). I wouldn't worry *too* much for now if you know at least root-coefficient relations, the rest is only used on a few kinds of problems.

If you are willing to throw down some money, AoPS 2 has a good review of the basics in all of these areas plus some very good example problems (which will be much better for your learning than any lecture or book, the problems teach creativity and oddball techniques).

But I must stress that once you know your algebra basics, you just need to do problems. There are no secrets besides a good understanding of basic math a loads of creative thinkings. So just practice.

btw, do you know of any good resources for functions? dealing with generating, manipulating, arbitrary, etc. functions?

well for polynomials i wrote [url=http://www.artofproblemsolving.com/Resources/Papers/PolynomialsAK.pdf]this[/url], but you will probably find it too easy


I took BC Calculus at my school using Stewart's [i] Single Variable Calculus: Concepts and Contexts [/i] and was not that impressed. I'm looking for a more rigorous and in depth treatment. I have heard Apostol's book called the second best one. Which might be considered the best?

I liked Spivak's [i]Calculus[/i] a lot.  It is rigorous and has lots of challenging problems.

Thanks for the reply, Ravi B. I googled Spivak's book and it looks great.

I would also be greatly ingratiated to anyone to could offer a comparison between Spivak and Apostol. In particular, I'm looking for interesting, non-routine problems in calculus, not the "same idea, different numbers" sort of exercises that encumber most textbooks. Would these books offer problems which might be considered at a pre-olympiad/olympiad level, if such contests encompassed calculus?


I would appreciate it if you guys can solve these three problems and if you have time write a solution.
Thanks in Advance  :lol:

[hide="picture 3"]
Notice you can rearrange the picture so that you have half the square. The area is simply $ \fbox{(B) 2}$
[/hide]

[hide="picture 2"]
Notice here that each one except for $ (C)$ is one less than a factor of 1000. By guess and check, we find that $ (C)$ indeed does not work, while all the others work if they are the second number. The answer is $ \fbox{(C) 299}$[/hide]

Something is messed up with the last one...

[hide="Some reasoning behind Picture 2"]
Let the 2nd number be $ a$ Therefore the numbers will be

$ 1,a,a \plus{} 1,2(a \plus{} 1),4(a \plus{} 1),8(a \plus{} 1),\ldots,2^k(a \plus{} 1)\ldots$
[/hide]

[hide="Problem 1"]
The area of one semi circle is $ \frac{\pi}{2}$

Therefore the area of the two semicircles minus the extra bit is $ 2 \frac{\pi}{2} \minus{} 2(\frac{\pi}{2}\minus{}1) \equal{} \boxed{2}$
[/hide]

These sound like homework problems.
But anyways...
[hide="Problem 1"]You can rearrange the unshaded parts of the semicircles to cover exactly one half of the square. Since the area of the square is $ 2^2\equal{}4$, the area of the shaded region is half that, or $ 2$.[/hide]

[hide="Problem 3, Picture 1 -- Is there an easier way to do this?"]If we draw a line from $ F$ to $ C$, we see two congruent triangles $ \Delta BFC$ and $ \Delta EFC$. The area of $ AFDC$ is also split into two congruent halves. Because $ \Delta FCA$ and $ \Delta FCE$ share the same altitude to their base $ CE$, the ratio of their areas is the ratio of their bases. Therefore:
\begin{align*} & \frac {[FCA]}{[FCE]} = \frac {2[FCA]}{2[FCE]} = \frac {[AFDC]}{2S - [AFDC]} = \frac {CA}{CE} = \frac {1}{4} \\
& \implies \frac {[AFDC]}{2S - [AFDC]} = \frac {1}{4} \\
& \implies 4[AFDC] = 2S - [AFDC] \\
& \implies 5[AFDC] = 2S \\
& \implies [AFDC] = \boxed{\frac {2}{5}S} \end{align*}
[/hide]


[hide]
i'm assuming (a,6) means the greatest common factor, i don't know symbols that well :)
You can change this problem around to prove that if $(a,6)=1$
Then $a^2\equiv 1(\bmod4)*1(\bmod3)*1(\bmod2)$
you also know that a must be prime greater than 6 to have $(a,6)=1$
So then you must prove that
$a^2\equiv1(\bmod4)$
This is always true because any number squared is either 0 or 1 mod 4, and because $a$ must be prime, it will be odd, meaning if squared it will be 1 mod 4
$a^2\equiv1(\bmod2)$
This is always true because any odd number squared will have an odd one's digit, making it 1 mod 2
finnally
$a^2\equiv1(\bmod3)$
This is always true because any odd number must either be $0\bmod3$ or $2\bmod3$ and 0 mod 3 doesn't work because then $(a,6)\not=1$
so it must be 2 mod 3, then if that number is squared you can just square the remainder.. which becomes
$4\bmod3\equiv1\bmod3$
And that's it
[/hide]

[hide="Solution"]
Consider the numbers $a-1, a,$ and $a+1.$  Since $(a,6)=1$ we know that $a$ divisible by neither 2 nor 3.  Since $a$ is not divisible by 2, both $a-1$ and $a+1$ must be.  Moreover, one of $a-1$ and $a+1$ must be divisible by 4.  Among three consecutive integers, one is a multiple of 3.  Since $a$ is not a multiple of three, one of $a-1$ and $a+1$.  Thus, we must have that $(a-1)(a+1)$ is a multiple of $2\cdot4\cdot3=24.$  Which gives $(a-1)(a+1)\equiv 0\mod 24\Rightarrow a^2\equiv 1\mod 24.$[/hide]

AMPIS, $a$ doesn't have to be prime (it is still odd though).

Oh yeah, sorry.  That's a pretty bad mistake i made... :(


algebra
polynomial
ratio
limit
complex analysis
complex analysis unsolved

If P(n) is not [u]=[/u] 0 is a polynomial in n, show that P(n+1) / P(n) --> 1 as n --> infinity.

Conclude that the power series with coefficients an = P(n) has radius of convergence R=1.

Thank you in advance

p.s. [u]=[/u] is the equals with the three bars.  I am not sure really what that means.

The three bars is just saying $ P$ is not the zero polynomial. It sort of means P is not "identically" zero.

Now if $ P(x)\equal{}a_k x^k\plus{}...$ Divide the top and bottom of the fraction by $ n^k$. Then the ratio that you have is

$ \frac{ a_k (1\plus{}1/n)^k\plus{}...}{a_k\plus{}...}$ where the $ ...$ go to zero. It is pretty obvious that this goes to one so if you don't see this, check for yourself...

Then the second part using the ratio test for
$ a_0\plus{}a_1 x\plus{}a_2 x^2\plus{}...$
, we get that the ratio of terms as n goes to infinity is

$ \lim_{n\rightarrow\infty}\frac{|a_{n\plus{}1} x^{n\plus{}1}|}{|a_n x^n}\equal{}\lim_{n\rightarrow\infty}\frac{|a_{n\plus{}1}x|}{|a_n|}\equal{}|x|$

So if $ |x|>1$, this series diverges by the ratio test. So $ |x|\le 1$. Furthermore, If $ |x|<1$, the series converges by the ratio test. So the radius of convergence is $ 1$. (Note $ x\equal{}\pm 1$ are not determined by the ratio test, but they are not needed to find the radius of convergence).


If positive numbers $ x_1,x_2,...,x_n$ satisfies $ x_1\plus{}x_2\plus{}...\plus{}x_n\equal{}1$. Prove the inequality?
$ (\frac{1}{x_1^2}\minus{}1)(\frac{1}{x_2^2}\minus{}1)...(\frac{1}{x_n^2}\minus{}1) \ge (n^2\minus{}1)^n$

${ \prod ({\frac{1}{x_1 }-1}}) \geq (n-1)^n$

Similarly

${ \prod ({\frac{1}{x_1  } +1}}) \geq (n+1)^n$

Multiplying we obtain the desired