function
irrational number
real analysis
real analysis unsolved

Start with one number that you know for sure is irrational. I'll use $\sqrt{2}$ but that choice is entirely arbitrary.

We want to find an irrational number between $a$ and $b,$ where $a<b.$ Choose an integer $n$ such that $n>\frac{\sqrt{2}}{b-a}.$ (Note that we have directly invoked the Archimedean property of the reals.)

Now consider numbers of the form $\frac{k\sqrt{2}}n$ for integers $n.$ We want

$a<\frac{k\sqrt{2}}n<b$ or $\frac{an}{\sqrt{2}}<k< \frac{bn}{\sqrt{2}}$

Choose the smallest integer $k$ such that $\frac{an}{\sqrt{2}}<k.$

Then $k\le\frac{an}{\sqrt{2}}+1<\frac{an}{\sqrt{2}}+\frac{(b-a)n}{\sqrt{2}}=\frac{bn}{\sqrt{2}}.$

Hence the number $\frac{k\sqrt{2}}n$ lies in the interval $(a,b),$ and since it is a rational multiple of a known irrational number, it is irrational.

--

Adjustment: this proof doesn't work if $k$ turns out to be zero, but if that happens, just double $n$ and move on to the next $k.$

Thanks. I guess the main idea here is that we are using nested sequences of rational numbers to show that $x$ is irrational?

You don't need nested sequences. Here is my version of the proof:

Step 1. The rational numbers form a field, which means that the sums, products and quotients of rational numbers are also rational. Therefore, the sum of a rational number $a$ and an irrational number $c$ must be irrational (if $a+c$ were rational, then $(a+c)-a$ would be rational too). By a similar reasoning, $ac$ is irrational unless $a=0$.

Step 2. Consider a linear function $f(x)=kx+m$ with rational coefficients $k$ and $m$, where $k>0$. By the above argument, $f(x)$ is rational if and only if $x$ is rational.

Step 3. Given two rational numbers $a$ and $b$, where $a<b$, introduce a linear function $f(x)=(b-a)x+a$. We have $f(0)=a$, $f(1)=b$, and the coefficients of $f$ are rational. Therefore, $f(1/\sqrt{2})$ is an irrational number between $a$ and $b$.

[color=darkblue]For $\{a,b\}\subset Q\ ,\ a<b$ consider the sequence of the irrational numbers $a_{n}=\frac{a+b}{2}+\frac{\sqrt 2}{n}\searrow \frac{a+b}{2}\in (a,b)\ .$
Thus, $\left(\exists\right) n_{1}\in N^{*}\ \left(\mathrm{can\ choose\ }n_{1}=\left[\frac{2\sqrt 2}{b-a}\right]\right)$ so that $\left(\forall\right) n>n_{1}$ , $a_{n}\in (a,b)\ .$

[b]Remark 1.[/b] It is more difficulty to show that between any two irrational numbers exists at least a rational (irrational) number. For it, use the following simple lemma :
[b]Lemma.[/b] Let $\rho_{1}\ ,\ \rho_{2}$ be two irrational numbers so that  $1+\rho_{1}<\rho_{2}\ .$ Then exists $z\in Z$ so that $\rho_{1}<z<\rho_{2}\ .$
If $\rho_{2}-\rho_{1}< 1$, then exists $n\in N^{*}$ so that $\frac{1}{n}<\rho_{2}-\rho_{1}< 1$, i.e. $n\rho_{2}-n\rho_{1}>1\ .$
From the above lemma exists $m\in Z$ so that $n\rho_{1}<m<n\rho_{2}$, i.e. exists $r=\frac{m}{n}\in Q$ so that $\rho_{1}<r<\rho_{2}\ .$

[b]Remark 2.[/b] For the initial problem you can use (or show) the property (the density of the set of the rational numbers in the set of the real numbers) '' for any irrational number $\rho$ exists a sequence of the rational numbers $a_{n}$ so that $a_{n}\rightarrow\rho\ ''$[/color]

OK, you now have three versions of the proof. I think I'll take mlok's as my favorite. Note that all three of us are assuming as known that $\sqrt{2}$ is irrational. As I commented earlier, that's an arbitrary choice, and you could replace $\sqrt{2}$ in each of these proofs by any other number that you know is irrational. (In mlok's case, you specifically need a known irrational in $(0,1).$)

[quote="courtrigrad"]How do you show that between any two rational numbers $a, b$ there exists at least one irrational number, and consequently many?[/quote]
I know this problem has been solved and done in three ways, yet I just wanted to know if we could use the following argument to solve this problem: If $A$ is a countable set and $B$ an uncountable set, then $B-A$ is uncountable. Of course, I am implicitly assuming I know a few definitions beforehand.

[quote="Kent Merryfield"]The problem with that is that it doesn't address the condition that we were looking for irrational numbers in the interval $(a,b).$

Now, you could find a one-to-one function mapping that interval onto some standard set, like $(0,1).$ Then you would have an argument.[/quote]
I get it now. Thanks!


inequalities
calculus
calculus computations

Let $( a_n )^{\infty}_{n = 1}$ be a sequence of positive real numbers.

Is it true that
\[ \sum_{n = 1}^{\infty} a_n = + \infty \Rightarrow \sum_{n = 1}^{\infty} \frac{a_n}{1 + a_n} = + \infty  \]

It must be true.

Assume the contrary. Then $\sum_{n=1}^{+\infty}{\frac{a_{n}}{1+a_{n}}}<+\infty \Rightarrow \frac{a_{n}}{1+a_{n}} \rightarrow 0$. But this implies that $a_{n} \rightarrow 0$ (write it properly), and then $a_{n} \sim \frac{a_{n}}{1+a_{n}}$, meaning $\sum a_{n}$ must converge.

More rigorously, lets assume that $\sum_{n=1}^\infty a_n$ diverges and $\sum_{n=1}^\infty \frac{a_n}{1+a_n}$ converges. Then we know $\frac{a_n}{1+a_n} \to 0^+\ \Rightarrow\ 1 + \frac{1}{a_n} \to +\infty\ \Rightarrow\ a_n\to 0^+$.

Since $a_n$ and $\frac{a_n}{1+a_n}$ both tend to 0, we know that $\forall\epsilon > 0\ \exists n_0$ s.t. $n > n_0 \Rightarrow |a_n - \frac{a_n}{1+a_n}| < \epsilon$.

Now, take $n \ge m > n_0$ and consider that $\sum_{i=m}^n a_n - \sum_{i=m}^n \frac{a_n}{1+a_n} < (n-m)\cdot\epsilon$.

But $(n-m)\cdot\epsilon$ is finite, and since the sum of $\{a_n\}$ is divergent, its partial sums are also divergent.

Rearranging the inequality, we get $\sum_{i=m}^n a_n < \sum_{i=m}^n \frac{a_n}{1+a_n} + (n-m)\cdot\epsilon$, which implies that the partial sums of $\{\frac{a_n}{1+a_n}\}$ are divergent, a contradiction.

Yet an other way :

if $a_n$ is not bounded then there is a sub-sequence $a_{\phi(n)} \rightarrow +\infty$ but then $a_{\phi(n)}/(1+a_{\phi(n)}) \rightarrow 1$ and $\sum a_n/(1+a_n)$ is grossly divergent.

Else if $a_n < M$ then $a_n/(1+a_n) > a_n/(1+M)$ and conclusion follows.


Circles $C_{1}$ and $C_{2}$ are tangent to each other at $K$ and are tangent to circle $C$ at $M$ and $N$. External tangent of $C_{1}$ and $C_{2}$ intersect $C$ at $A$ and $B$. $AK$ and $BK$  intersect with circle $C$ at $E$ and $F$ respectively. If AB is diameter of $C$, prove that $EF$ and $MN$ and $OK$ are concurrent. ($O$ is center of circle $C$.)

First of all, we have to remark that the circumcircle of KEF has the same radius as C. This can be observed with the aid of an inversion of pole K and power -KA*KF. This inversion turns the line AB into the circle KEF and it leaves circle C alone (it's invariant). It turns circles C1 and C2 into 2 parallel lines which are both tangent to the images of C and AB, which are C and the circle KEF respectively. This means that 2 common tangents of C and KEF are parallel, so their radii are equal.

That was just an observation; no let's get to the problem: We use the same inversion further on. We have to show that the line KO and the circles KAB and KE'F' are concurrent, where X' is the image of X by the inversion. E' and F' are the points of tangency between C and 2 parallel lines, so they're the extremities of a diameter of C, and the conclusion follows easily: Let R be the radius of C. Then the powers of O with respect to the circles KAB and KE'F' are equal (both equal to R2), so O is on the radical axis of the 2 circles, so it's on KT where T is the other point where the 2 circles intersect (the first such point being K).

I hope there's nothing wrong with it...

Note that the condition revealed in the first paragraph, that the radii of KEF and C are equal, means that the angle <AKB is equal to 135 deg, or, in other words, OE is perpendicular to OF (this is another one of those useless observations :D).

Hate to bring this up again after 13 years, but I can't figure out what this problem's saying. How can $AB$ be the diameter of $C$ if $AB$, $C$, $C_1$, and $C_2$ are all tangent to each other?

Tiny diagram is attached- should give gist of problem's diagram.
(E, F are on circle)

Let $N'=KN\cap C, M'=MK\cap C, X=EF\cap MN, Y=MB\cap EN'$
Important lemma: $N'M'$ is diameter of circle $C$

Pascal's theorem on $M'N'EABM,$
$M'N'\cap AB=O$
$N'E\cap BM=Y$
$EA\cap MM'=K$ is collinear.

Pascal's theorem on $MNN'EFB,$
$MN\cap EF=X$
$M'N'\cap FB=K$
$N'E\cap BM=Y$ is collinear.

Thus, $(K,O,X,Y)$ is collinear.

Also, using pascal on other hexagon, $Z=FM'\cap NA$ is also in that line.


function
integration
trigonometry
calculus
derivative
calculus computations

a) Find  Fourier's tranformation of the function

$ f(x)\equal{}\left\{{\begin{array}{*{20}c}
{{1\minus{}\left|{\frac{x}{2}}\right|}\hspace{0.1cm},\hspace{0.2cm}|x|<2}\hfill\vspace{0.2cm}   \\
{{0}\hspace{0.1cm},\hspace{1.2cm}|x|>2}\hfill\\
\end{array}}\right.$

b) Using Parseval's theorem find  $ I\equal{}\displaystyle\int_{\minus{}\infty}^{\plus{}\infty}dy\,\frac{\sin^{4}{y}}{y^4}$

It's really just looking up the right formulae and plugging in. What do you have trouble with?

One thing you always have to do in communicating a Fourier transform problem is state what your conventions are - in particular, where does the $ 2\pi$ go? Here are my conventions:

$ \widehat{f}(s)\equal{}\int_{\minus{}\infty}^{\infty}e^{\minus{}2\pi isx}f(x)\,dx$

$ f(x)\equal{}\int_{\minus{}\infty}^{\infty}e^{2\pi isx}\widehat{f}(s)\,ds$

Plancherel (aka Parseval): $ \int_{\minus{}\infty}^{\infty}|f(x)|^2\,dx\equal{} \int_{\minus{}\infty}^{\infty}|\widehat{f}(s)|^2\,ds$

Derivatives: $ \widehat{f'}(s)\equal{}2\pi is\widehat{f}(s)$

That last formula is even more effective if you're willing to use weak (distributional) derivatives.

For the function in question:

$ f'(x)\equal{}\begin{cases}\frac12,&\minus{}2<x<0\\ \minus{}\frac12,&0<x<2\\0,&\text{otherwise}\end{cases}$

$ f''(x)\equal{}\frac12\delta(x\plus{}2)\minus{}\delta(x)\plus{}\frac12\delta(x\minus{}2).$

$ \widehat{f''}(s)\equal{}\frac12e^{\minus{}4\pi is}\minus{}1\plus{}\frac12e^{4\pi is}$

$ \equal{}\cos(4\pi s)\minus{}1$

Hence, $ \widehat{f}(s)\equal{}\frac{\widehat{f''}(s)}{\minus{}4\pi s^2}\equal{} \frac{1\minus{}\cos(4\pi s)}{4\pi^2 s^2}$

$ \equal{}\frac{2\sin^2(2\pi s)}{4\pi^2 s^2}\equal{}2\left(\frac{\sin(2\pi s)}{2\pi s}\right)^2$

Now we check Plancherel:

$ \int_{\minus{}\infty}^{\infty}|f(x)|^2\,dx\equal{}2\int_0^2\left(1\minus{}\frac x2\right)^2\,dx\equal{}\frac43$

Hence, $ \int_{\minus{}\infty}^{\infty}4\left(\frac{\sin(2\pi s)}{2\pi s}\right)^4 \,ds\equal{}\frac43$

Make the substitution $ y\equal{}2\pi s$ to get

$ \int_{\minus{}\infty}^{\infty}\frac{\sin^4y}{y^4}\,dy\equal{}\frac{2\pi}{3}$

After the substitution $ y=2\pi s $

$ \int_{-\infty}^\infty \left(\frac{\sin 2\pi s}{2\pi s}\right)^4\, ds=\frac13 $ becomes $ \int_{-\infty}^\infty \frac{\sin^4 y}{y^4}\, \frac{dy}{2\pi}=\frac13 $.

So you should multiply (not divide) both sides with $ 2\pi $ .


function
inequalities
number theory
prime numbers
number theory proposed

[quote="aham"]For any two consecutive powers of 2 , show that there exists a set of infinite primes such that
the sum of their reciprocals lies between the reciprocals of these two powers of 2.[/quote]
Fix $ k \ge 1$. We want to show that there exists a set $ P = \{p_0$, $ p_1$, $ \dots \}$ such that
\[ \frac {1}{2^{k + 1}} < \sum_{i = 0}^{\infty} \frac {1}{p_i} < \frac {1}{2^k}\]
holds true. We know from Bertrand's postulate that there exists a prime $ p_0$ such that $ 2^k < p_0 < 2^{k + 1}$. Let $ l$ be a positive integer such that
\[{ \frac {1}{2^k}} - \frac {1}{p_0} >\frac {1}{2^l}.\]
Then define a sequence of primes $ p_i$ such that $ p_i > 2^{l + i}$ for every $ i \ge 1$. Such primes clearly exist. Then we have
\[ \frac {1}{2^{k + 1}} < \frac {1}{p_0} < \frac {1}{p_0} + \sum_{i = 1}^{\infty} \frac {1}{p_i} < \frac {1}{p_0} + \sum_{i = 1}^{\infty} \frac {1}{2^{l + i}} = \frac {1}{p_0} + \frac {1}{2^l} < \frac {1}{2^k},\]
which is what we had to prove.

This is just a conjecture of mine is that the set
\[ S \equal{} \{ \sum_{j\equal{}1}^{k} \dfrac{1}{p_{i_j}} | \{k, i_1, i_2, ..., i_k\} \subset \mathbb{N} \}\]
(where $ 2 \equal{} p_1 < p_2 < ...$ is the set of prime numbers) is dense in $ (0, \plus{}\infty)$.

Mashimaru, your conjecture is correct.

[hide]We need only to prove that for any $ x>\epsilon > 0$ there exists a set of primes $ B$ such that $ x - \epsilon < \displaystyle\sum_{p\in B}{\frac {1}{p}} < x$.

Consider the set $ C$ of primes $ p$ such that $ \dfrac{1}{p} < \epsilon$. The function $ s(y) = \displaystyle\sum_{\begin{tabular}{c}p\in C \\
p\le y\end{tabular}}{\frac {1}{p}}$, grows from cero to infinity increasing by steps smaller than $ \epsilon$, so for some $ y$ we have $ x - \epsilon < s(y) < x$.

The same is truth exchanging the set of primes with any unbounded set $ A$ of positive reals numbers such that for any $ z > 0$, the function $ s_z(y) = s(y) - s(z)$ grows indefinitely (with the respectively $ s(y)$).[/hide]

Now, what about exact representions as sums of prime's reciprocals? Does anybodoy kwow something interesting to share?

:)

Dear [b]ElChapin[/b], I don't think such representation exists. Since if there does then we can find a formula for prime $ p$.

But of course, there are many approximate for the sum of reciprocals of primes since they play important role in proving properties involving inequality of Riemann's zeta function. May be we can Google it :lol:

[quote="FelixD"][quote="aham"]For any two consecutive powers of 2 , show that there exists a set of infinite primes such that
the sum of their reciprocals lies between the reciprocals of these two powers of 2.[/quote]
Fix $ k \ge 1$. We want to show that there exists a set $ P = \{p_0$, $ p_1$, $ \dots \}$ such that
\[ \frac {1}{2^{k + 1}} < \sum_{i = 0}^{\infty} \frac {1}{p_i} < \frac {1}{2^k}\]
holds true. We know from Bertrand's postulate that there exists a prime $ p_0$ such that $ 2^k < p_0 < 2^{k + 1}$. Let $ l$ be a positive integer such that
\[{ \frac {1}{2^k}} - \frac {1}{p_0} > \frac {1}{2^l}.\]
Then define a sequence of primes $ p_i$ such that $ p_i > 2^{l + i}$ for every $ i \ge 1$. Such primes clearly exist. Then we have
\[ \frac {1}{2^{k + 1}} < \frac {1}{p_0} < \frac {1}{p_0} + \sum_{i = 1}^{\infty} \frac {1}{p_i} < \frac {1}{p_0} + \sum_{i = 1}^{\infty} \frac {1}{2^{l + i}} = \frac {1}{p_0} + \frac {1}{2^l} < \frac {1}{2^k},\]
which is what we had to prove.[/quote]

Nice proof! :)
What i had in mind was to apply Bertrands postulate for every 2 consecutive powers of 2 starting from $ 2^k$  , invert the inequalities and sum them all up to obtain the required result.

Interestingly the sum of reciprocals of all primes diverges!


This is the thread for Mafia Game 12.  I will be choosing roles and sending them shortly after I post this.  There is no need to confirm your spot, since everyone has done this already.

[b]Players (15):[/b]
Klebian
samath
dts
DarkKnight
zeb
mark
chesspro
Peter VDD
yif man12
GoBraves
Pianoforte
apusey
noneoftheabove (replacing G-UNIT, who hasn't confirmed)
slamchess
dinoman

[b]Roles:[/b]

[i]Mafia (3)[/i]: Standard mafia role.  No day communication is allowed at all (not even during the first day).

[i]Thief (1)[/i]: The Thief is on the Mafia side, but the Thief and Mafia are unaware of each other's identities.  The Thief chooses one person each night to rob, and this person may not perform their special role that night.  Choosing the masons removes their ability to induct a new member, but not to communicate at night.  Choosing the mafia does nothing unless there is only one mafia remaining, in which case the mafia cannot kill that night.  The Thief shows up as innocent to inspection.

[i]Serial Killer (1)[/i]: The SK essentially functions as a single werewolf acting alone.  The SK shows up innocent to inspection.  The Doctor can prevent SK kills.

[i]Detective (1)[/i]: Standard Detective role.  The Detective searches only for the Mafia, not the Thief or Serial Killer.

[i]Doctor (1)[/i]: The Doctor can save one person (any person) per night.  This protects against being targeted by the Mafia OR SK.

[i]Magician (1)[/i]: This role is also known as the Bus Driver.  Chooses two people per night and "switches" them.  Anything that would happen to one now happens to the other.  This includes targeting by the doctor, mafia, thief, SK, and masons (for induction).  The Magician also can temporarily change the identity of the Officer and Miller (for one night only).  Note that the magician can cause the mafia to kill themselves.  The magician may not target him/herself unless the Thief is dead (since this can lead to a paradox).

[i]Masons (2)[/i]: Standard Mason role, with one big addition, the power of induction.  Once during the game, at night, the Masons can add a new member to their secret society.  Basically, this person retains their old role, but is told the identity of the masons and given the ability to communicate with them at night.  The masons should be careful not to induct a mafia or SK, as induction doesn't make the person innocent.  Also, bear in mind that the action of the Magician can screw up who you try to induct.  Masons may not communicate during the day.

[i]Citizens (3)[/i]: Standard Citizen role.

[i]Miller (1)[/i]: Citizen, but shows up guilty to inspection.  If targeted by the magician, the other targeted player becomes the Miller for that night only.  So it is possible (albeit unlikely) that, for example, the Doctor (or other innocent with a special role) will show up guilty to inspection.

[i]Officer (1)[/i]: The officer is a citizen, and is unaware of his/her own identity (like the Miller).  However, the Officer cannot be killed at night.  If targeted by the magician, the other targeted player becomes Officer for that night only.

We will begin with a day phase.  15 people alive, 8 needed for a lynch.  Like in game 11, you may (and should) establish a queue for people to target at night in case you miss a deadline.  I will limit nights to 24 hours, except for the first night.  Role PM's will go out shortly, and then you can start.

All role PM's have been sent.  To clarify one little thing, all the citizen PM's say "You may be the Miller or Officer".  That doesn't mean you are necessarily one of those.  Also, if you have any questions about your role or any other role, feel free to PM me.

Day 1 starts now.  15 people, 8 to lynch.  Good luck.

3 mafia? :) looking at the numbers in past games, this shouldn't be too hard...

let's see... the first one able to find (with proof) the 4 ending digits of $2001^{2001^{2001^{...}}}$ (2001 times) may make the first accusation :P

Peter, do you have the answer to that?

Yay for Mafia! Third consecutive game.

[quote="samath"]Peter, do you have the answer to that?[/quote]I have. :) obviously... it's not hard, just try it :P

I'm here.

A few questions for our moderator, recently reformed from the Mafia:

Is the Magician innocent or guilty?  I'm assuming innocent.

If the Magician, say, switches two people and the inspector inspects one of them, is he made aware of the switch?

If the Officer is targeted at night, are we told that he was targeted but the kill failed?

If the Officer or Miller is killed, are we told that he/she was a Citizen or Officer/Miller?

Peter VDD: Lemme see... Since $2001^{2001^{2001\ldots}}$ (2000 times) is trivially equivalent to 1 (mod 5) (since it must be equivalent to 1 (mod 10)), and, for integral $a$, $2001^a$ (mod 10000) cycles back to itself if $a$ is increased by 5 (see for yourself), the last four digits of your number must be 2001.

Do you need a more rigorous proof?

lol im here
this game had better go more quickly than game 10...

this is my first game so don't yell at me if i screw up something

Let's start already

[quote="dts"]A few questions for our moderator, recently reformed from the Mafia:

Is the Magician innocent or guilty?  I'm assuming innocent.

If the Magician, say, switches two people and the inspector inspects one of them, is he made aware of the switch?

If the Officer is targeted at night, are we told that he was targeted but the kill failed?

If the Officer or Miller is killed, are we told that he/she was a Citizen or Officer/Miller?[/quote]

[b]The magician is innocent!  Sorry I forgot to tell you this![/b]

If the inspector inspects A and the magician switches A and B, the inspector is given the answer as if A was inspected.  The switch does not affect this, since the Miller is enough to limit the power of the inspector.  The only time the magician would have any effect on the inspector is if A is inspected, A and B are switched, and B is the miller.  In that case, the response will be "guilty" no matter what, since A will be the miller for that night.

On failed kills, including attempted officer kills, you will just be told that the kill failed, not who was targeted.  Essentially, the officer operates as a second doctor that always saves him/herself.

If the officer or miller is killed, you will be told that a citizen died.

[quote="Peter VDD"]Ok dts, that's correct, you can accuse first, I'll second you. (unless you'd accuse me of course :D)[/quote]

Now, why do you assume I'm innocent just because I solved a problem?

Okay... (generates random number)... (looks down list)... I accuse noneoftheabove.

What are the odds he'll be Mafia twice in a row? :D

No idea. I'm not going to vote because we don't have enough evidence.

How are you supposed to have enough evidence before anything happens?

I also generated a random number.
Turns out, It fit none of the above also.
What are the chances of that!?!?
And by the way, the chance of him being mafia twice in a row is 1/5.
He's already been mafia last game, so this game it's 3/15.
Hehehe.
Since there's not too much evidence we're going to be gathering anyway, this being the first day, I second/AYE the accusation of Noneoftheabove.

You killed off the magician...but other than that good job sk.

But i never really was in the game lmao! :D

[quote="zeb"]For some reason, the mafia tried to kill PeterVDD [b]three nights in a row[/b]. Of course, by the third night, I was dead, so he died...[/quote]

We did?  :| I think we tried to kill Yif once...

Well, it's always either me or you they kill, zeb. :D

I had guessed that they would try to kill peter off once they knew I that knew that he was innocent...

[quote="Sly Si"]Yep, town wins.  Great game, everybody!  Sorry I haven't been around since Saturday.  So now that the game is over, here's all the stuff that you don't know (and some that you did):

[b]Roles:[/b]
Mafia: Samath, mark, DarkKnight
Thief: Klebian
Serial Killer: slamchess
Doctor: dts/la mafioso
Inspector: Peter VDD
Magician: zeb
Masons: chesspro, yif man, Peter VDD (inducted the same night he was killed)
Citizens: noneoftheabove, apusey, pianoforte
Miller: dinoman
Officer: GoBraves

Other stuff:
The mafia tried to kill Peter VDD three times.  The first two were thwarted by zeb.  Good job, zeb!
dts/la mafioso was saved every night by the doctor.
dts, mark, and GoBraves were inspected, in that order.
Klebian (thief) targeted apusey, apusey, and pianoforte, in that order.

I guess that's about it.[/quote]
Wait... did I target apusey two nights in a row?
I don't think so, let me check. I think I targeted GoBraves night 2.
But maybe not.

Heh, reading through the transcript of this was amusing, given that I knew la mafioso to be the doctor.

Peter VDD: Why did you suspect me?  ;)   Was it because I solved your problem?

[quote="Peter VDD"]Well, yes and no. I thought that a mafia would never repond to such a thing on turn 1, and I checked if my prediction was correct. And it was. :)
Second turn I found mark very suspicious and inspected him, third turn I doubted between GoBraves and Klebian. Wrong guess, but still good enough to win us the game.[/quote]

well u no 1 is guilty so if one is innocent the other is guilty so either way- rite choice :D

Um, Klebian would have showed up innocent to inspection.


algebra
polynomial
quadratics
system of equations
algebra proposed

Let $P(X,Y)=X^2+2aXY+Y^2$ be a real polynomial where $|a|\geq 1$. For a given positive integer $n$, $n\geq 2$ consider the system of equations:  \[ P(x_1,x_2) = P(x_2,x_3) = \ldots = P(x_{n-1},x_n) = P(x_n,x_1) = 0 . \] We call two solutions $(x_1,x_2,\ldots,x_n)$ and $(y_1,y_2,\ldots,y_n)$ of the system to be equivalent if there exists a real number $\lambda \neq 0$, $x_1=\lambda y_1$, $\ldots$, $x_n= \lambda y_n$. How many nonequivalent solutions does the system have?

[i]Mircea Becheanu[/i]

First note that if any of the $x_{i}$'s are $0$,so are all the other $x_{i}$'s.Thus there is one trivial  solution for all n where
$x_{i}=0$ for $i=0,1,2...n$.
Next ,suppose no $x_{i}$ is $0$.

Case $1$: $|a|>1$ and $n\geq 3$.
Then as $P(X,Y)=P(Y,X)$ we have
$P(x_{i},x_{i-1})=P(x_{i},x_{i+1})=0$ and so:

$x_{i-1}$ and $x_{i+1}$ are the roots of the quadratic expression in $t,P(t,x_{i})=0$ $-(1)$.

Hence,$x_{i-1}.x_{i+1}=(x_{i})^{2}$ .This implies that $x_{i}/x_{i+1}=k, k$ a real constant for all $i=0,1,...n$.By taking the product of all $x_{i}/x_{i+1}$ we get $k^{n}=1$,implying $k=1$ or $k=1,-1$ when $n$ is odd or even respectively.Also by $(1)$,

$x_{i-1}+x_{i+1}+2ax_{i}=0$

but as $|a|>1$ and none of the $x_{i}$'s are $0$, $k$ cannot be $1$ or $-1$.Hence there is no non trivial solution.

Case $2$: $a=-1$
The equation becomes $P(X,Y)=(X-Y)^{2}=0$.There is only one non trivial solution $x_{i}=p, p$ a non zero real constant.Also ,any other non trivial solution is equivalent to the above.

Case $3$: $a=1$
The equation is now $P(X,Y)=(X+Y)^{2}=0$.The only non trivial solution is for an even $n$,that is $x_{2i+1}=p,x_{2i}=-p, p$ a non zero real constant.Any other non trivial aolution is equivalent to the above.

Cheers
Bhargav

Ps:I haven't solved $n=2$ here but that is quite simple.

Observe that $ x_1^2 \plus{} 2ax_1x_2 \plus{} x_2^2 \equal{} 0 \Rightarrow x_2 \equal{} k x_1$, where $ k \in \{ \minus{} a \plus{} \sqrt {a^2 \minus{} 1}, \minus{} a \minus{} \sqrt {a^2 \minus{} 1} \}$

Therefore,
$ x_2 \equal{} k_1 x_1$
$ x_3 \equal{} k_2 x_2$
$ \cdots$
$ x_1 \equal{} k_n x_n$

where each $ k_i \in \{ \minus{} a \plus{} \sqrt {a^2 \minus{} 1}, \minus{} a \minus{} \sqrt {a^2 \minus{} 1} \}$

Therefore, $ k_1 k_2 \cdots k_n \equal{} 1$
$ \Rightarrow ( \minus{} a \plus{} \sqrt {a^2 \minus{} 1})^r ( \minus{} a \minus{} \sqrt {a^2 \minus{} 1} )^{n \minus{} r} \equal{} 1$, where $ r$ is the number of $ k_i$'s equal to $ \minus{} a \plus{} \sqrt {a^2 \minus{} 1}$
$ \Rightarrow ( \minus{} a \plus{} \sqrt {a^2 \minus{} 1})^ {2r \minus{} n} \equal{} 1$
$ \Rightarrow 2r \minus{} n \equal{} 0$, since $ \minus{} a \plus{} \sqrt {a^2 \minus{} 1}$ is real and not equal to $ \pm 1$
$ \Rightarrow r \equal{} n/2$

Therefore number of distinct solutions is exactly the number ways of choosing $ n/2$ of the $ k$'s to be $ \minus{} a \plus{} \sqrt {a^2 \minus{} 1}$ and the rest  $ \minus{} a \minus{} \sqrt {a^2 \minus{} 1}$
$ \equal{} \begin{cases} \binom{n}{n/2} \plus{}1 , \text{when n is even} \\
1, \text{when n is odd} \end{cases}$

The extra 1 is to count the $ (0, 0, \cdots, 0)$ solution.


geometry
circumcircle
modular arithmetic
ratio
conics
projective geometry
geometry unsolved

Let D be the midpoint of the arc BC opposite to A on the circumcircle of the triangle ABC and P a variable point inside the triangle. BP intersects the circle (ABC) at K, DK intersects AC at M, CP intersects the same circle at L, while DL cuts AB at N.

Prove that M, N and P are collinear and MN parallel to BC iff P lies on AP.

Remark: For MNP parallel to BC and P belonging to AD, I have a nice proof that K, as common to BP and DM and L, as common to CP and DN lie on (ABC).

Best regards,
sunken rock

Denote by $ X,Y$ the intersections of $ AB,CP$ and $ AC,BP$. We have:

$ (AXNB)\equal{}L(AXNB)\equal{}(ACDB)\equal{}K(ACDB)\equal{}(ACMY)$
$ \Rightarrow BY,CX,MN$ concurs, i.e. $ M,N,P$ are collinear.

Now suppose that $ P\in AD$, we have:

$ (AD,AC) \equiv (KB,KD) \pmod{\pi}$ so $ A,P,M,K$ are concyclic, since $ (MP,MA) \equiv (KP,KA) \equiv (CB,CA) \pmod{\pi}$, so $ MN$ is parallel to $ BC$, and we are done!

[quote="Mashimaru"][quote="babyandprettyboy"][quote="Mashimaru"]Denote by $ X,Y$ the intersections of $ AB,CP$ and $ AC,BP$. We have:

$ (AXNB) \equal{} L(AXNB) \equal{} (ACDB) \equal{} K(ACDB) \equal{} (ACMY)$
$ \Rightarrow BY,CX,MN$ concurs, i.e. $ M,N,P$ are collinear.

Now suppose that $ P\in AD$, we have:

$ (AD,AC) \equiv (KB,KD) \pmod{\pi}$ so $ A,P,M,K$ are concyclic, since $ (MP,MA) \equiv (KP,KA) \equiv (CB,CA) \pmod{\pi}$, so $ MN$ is parallel to $ BC$, and we are done![/quote]
The problem is obvious by Pascal theorem :)[/quote]

Yes, I just want to give a proof of Pascal's theorem using cross ratio. The newer part of the problem is just the second statement.

With cross ratio, we can still prove Pascal's theorem for any conic curve, as well as consider two straight lines as a conic. In this case, we have Pappus's theorem :lol:[/quote]
Very nice proof Mashimaru.
Here is a proof I found for the [b]Pascal theorem[/b]: (Using [b]Directed Angle[/b])
Let me restate again the [b]Pascal theorem[/b]:
[b]"Let $ A,B,C,D,E,E$ lie on circle $ (C)$, and they lie in any order. $ AB\cap DE\equiv G$, $ BC\cap EF\equiv H$, $ CD\cap FA\equiv K$ then $ G,H,K$ are collinear".[/b]
[i]Proof[/i]:
First I will prove $ (AB,DE) \plus{} (CD,FA) \plus{} (EF,BC)\equiv 0\pmod \pi$, $ (*)$
Indeed, it is easy to notice these following facts:
$ (AB,DE)\equiv (AB,BE) \plus{} (BE,DE)\equiv (AC,CE) \plus{} (BF,FD)\pmod \pi$, $ (1)$
$ (CD,FA)\equiv (CD,DA) \plus{} (DA,FA)\equiv (CE,EA) \plus{} (DB,BF)\pmod \pi$, $ (2)$
$ (EF,BC)\equiv (EF,FC) \plus{} (FC,BC)\equiv (EA,AC) \plus{} (FD,DB)\pmod \pi$, $ (3)$
From $ (1),(2),(3) \Longrightarrow (*)$.
Now, let $ \{Z\}\equiv (BDG)\cap (DFK)$, we claim to prove that $ B,F,Z,H$ are concyclic. Therefore, we just need to prove that $ (ZB,ZF)\equiv (HB,HF) \pmod \pi$. Obviously:
$ (ZB,ZF)\equiv (ZB,ZD) \plus{} (ZD,ZF)\equiv (GB,GD) \plus{} (KD,KF)\equiv (AB,DE) \plus{} (CD,FA)\equiv (BC,EF)\equiv (HB,HF)\pmod \pi$, due to $ (*)$.
Now we claim to prove $ H,Z,G$ are collinear. Hence, we just need $ (ZG,ZB)\equiv (ZH,ZB)\pmod\pi$.
Indeed, $ (ZG,ZB)\equiv (GD,DB)\equiv (ED,DB)\pmod\pi$. $ (DZ,ZB)\equiv (ED,DB)\pmod \pi$.
Similarly, $ K,Z,G$ are also collinear. Then, $ H,K\in GZ \Longrightarrow G,H,K$ are collinear.
Which ends our proof. :)
Hope there is no mistake. :(


1. Four men and four women are to be seated alternately at a round table. In how many ways can this be done ?

Pls explain your answer too.

2. How many numbers greater than 4000 can be formed from the figures 3,5,7,8,9? (repetitions not allowed).

3.What is the probability that the sum of any two different single-digit natural numbers is a prime number?

1
[hide]$(4 - 1)! = 6$ ways to arrange the men and the same for the women so $6 \cdot 6 = 36$[/hide]

2
[hide]Numbers can only start with 5,7,8 and 9. 24 arrangements for each number. $24 \cdot 4 = 96$[/hide]

3
[hide]$\frac {9!}{2!7!} = 36$ combinations of two.

1 can only reach 2, 3, 5, 7
2 can reach 5, 7, 11
3 can reach 7, 11
4 can reach 11, 13
5  can reach 11, 13
6 can reach 13
8 can reach 17

$4 + 3 + 2 + 2 + 2 + 1 + 1 = 15$

$\frac {15}{36} = \frac 5{12}$[/hide]

[quote="tarquin"]1

2
[hide]Numbers can only start with 5,7,8 and 9. 24 arrangements for each number. $24 \cdot 4 = 96$[/hide]

[/quote]
thats wrong, because you did not count the five digit numbers.
there can be $4*4*3*2=96$ four digit numbers, that are bigger than $4000$, also any five digit number written would be bigger than $4000$, thus you have to add $5!$ to it.
so the total is $96+120=216$ :)

good point amir! OA is 216.

tarquin, you may want to revisit answers to the other problems.

[quote="sheepwarrior"]isnt 1 just:
[hide]
4! * 4! or am i missing something??[/hide][/quote]
no, because its a CIRCULAR table :)

I probably did mess up since I did the problems fairly quickly. Just correct me if I'm wrong.

1. (4-1)! * 4! = 144.

2. OA is 2/5, but I don't know how they got it.


Let $x;y;z \in N^{*}$ and satisfying $2x^{x}+y^{y}=3z^{z}$.Prove that:$x=y=z$


superior algebra
superior algebra unsolved

The permutation consists of 6 cycles (1,2)(3,4)(5,6)(7,8)(9,10)(11,12).
The sol must consist of 3 cycles (a,c,b,d) where (a,b) and (c,d) or (d,c) form two of the original circles.
So we have $5.3.1.2^3=120$ solutions.


I recently posted this problem:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34529]Root Problem? A dog and a soldier formation[/url]

and sorry to bother again, but i did not stated that the dog´s speed was constant, so i was wrong  :( . and i think
the problem needs a new approach, help is really really appreciated.

Hmm, ive gone a little throught the problem,
 does the second solution from my last posts takes in account that the dogs speed is constant? it seems like it does.

Secondly, does sqrt(2) means,,      :sqrt: 2 ?

[quote="Oddgoo"]

Secondly, does sqrt(2) means,,      :sqrt: 2 ?[/quote]

yes


Points $ A_1,A_2,A_3$ are the midpoints of arcs $ BC,CA,AB$ of the circumcircle of triangle $ ABC$, and $ A_2,B_2,C_2$ are the contact points of the incircle with the sides of the triangle. Prove that the lines $ A_1A_2$; $ B_1B_2$ ;$ C_1C_2$ have a common point.

[quote="felixmann"]Points $ A_1,A_2,A_3$ are the midpoints of arcs $ BC,CA,AB$ of the circumcircle of triangle $ ABC$[/quote]

clearly here you mean $ A_1,B_1,C_1$ right?

[hide=" solution"]$ B_1C_1 \perp AA_1$ because $ \angle AB_1C_1 \equal{} \frac{\gamma}{2} \equal{} 90 \minus{} \frac{\beta}{2} \minus{} \frac{\alpha}{2} \equal{} 90 \minus{} \angle A_1AB_1$. Clearly also $ B_2C_2 \perp AA_1$ 'cause $ \triangle AB_2C_2$ is isosceles. Then $ B_1C_1 \parallel B_2C_2$ and cyclyc. So $ \triangle A_1B_1C_1$ and $ \triangle A_2B_2C_2$ are homothetic with a center on OI.[/hide]

[hide]Because of $ OA_{1}\parallel IA_{2}$ and $ OB_{1}\parallel IB_{2}$ and $ OC_{1}\parallel IC_{2},$ it is easy to show, By [b][size=100]Thales theorem[/size][/b], that the line segments $ A_{1}A_{2},\ B_{1}B_{2},\ C_{1}B_{2},$ intersect the line segment $ OI,$ at the same point so be it $ P,$ such that $ \frac {PI}{PO} \equal{} \frac {r}{R},$ where $ r \equal{} IA_{2} \equal{} IB_{2} \equal{} IC_{2}$ and  $ R \equal{} OA_{1} \equal{} OB_{1} \equal{} OC_{1}$ and so, the proof is completed.[/hide]

Kostas Vittas.


Solve this:
$x+\frac{2}{x}=y$
$y+\frac{2}{y}=z$
$z+\frac{2}{z}=x$

 :wink:

From (1) $sign(y)=sign(x)$, from (2) $sign(z)=sign(y)=sign(x)$. From (1)+(2)+(3) we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ - contradiction.

[hide="There is no real solution"]From each of the equations, $x,y,z$ must have the same sign.
But from the sum of the three equations, they can't have the same sign.[/hide]I don't know if it can be solved for complex solutions.
Rust beat me, of course.

[quote="Rust"]Complex solutions get from $3x^{6}+18x^{4}+24x^{2}+8=0$.[/quote]

can u solve?

Put $x^{2}=y$ and use Cardano's method  :wink:


a,b,c are natural numbers such that a^2+1 and b^2+1 are prime numbers. find all natural numbers satisfying (a^2+1)(b^2+1)=c^2+1.

is there any solution not using algebraic number theory?

It is easy
Solution 1:
Lemma 1:$ mn\equal{}x^2\plus{}1$ then exist $ a,b,c,d\in N$ such that
$ m\equal{}a^2\plus{}b^2,n\equal{}c^2\plus{}d^2,x\equal{}ad\plus{}bc,1\equal{}|ad\minus{}bc|$
Lemma 2:$ p\in P$,$ p\equiv 1 (mod 4)$ then it has unique represent in the form $ a^2\plus{}b^2$
Use 2 lemma easy to find out solution.
Solution 2:[hide="hint"]
$ a^2(b^2\plus{}1)\equal{}(c\minus{}b)(c\plus{}b)$[/hide]
Solution $ (a,b,c)\equal{}(2,3,7)$

can you post the detailed solution? thank you

I don't know what you mean in your problem

$ a,b,c$ are primes or $ a^2 \plus{} 1 , b^2 \plus{} 1$ are primes ?

If $ a^2 \plus{} 1 , b^2 \plus{} 1$ are primes then the solution of TTsphn is not correct: $ 3^2 \plus{} 1 \equal{} 10$ is not a prime.

[quote="TTsphn"]It is easy
Solution 1:
Lemma 1:$ mn \equal{} x^2 \plus{} 1$ then exist $ a,b,c,d\in N$ such that
$ m \equal{} a^2 \plus{} b^2,n \equal{} c^2 \plus{} d^2,x \equal{} ad \plus{} bc,1 \equal{} |ac \minus{} bd|$
Lemma 2:$ p\in P$,$ p\equiv 1 (mod 4)$ then it has unique represent in the form $ a^2 \plus{} b^2$
Use 2 lemma easy to find out solution.
Solution 2:[hide="hint"]
$ a^2(b^2 \plus{} 1) \equal{} (c \minus{} b)(c \plus{} b)$[/hide]
Solution $ (a,b,c) \equal{} (1,2,3)$[/quote]
I am sorry:
$ x^2\plus{}m^2\equal{}a^2\plus{}1$
$ y^2\plus{}n^2\equal{}y^2\plus{}1$
$ xn\plus{}ym\equal{}c$
$ 1\equal{}|xm\minus{}yn|\equal{}1$
Case 1:$ a\equal{}1,b\equal{}1$
then $ m\plus{}n\equal{}c$
        $ m\minus{}n\equal{}1$   so c is odd so $ (a^2\plus{}1)(b^2\plus{}1)\equiv 0(mod 2)$ so $ x\equal{}1$
Case 2:$ x\equal{}1,n\equal{}1$
$ my\plus{}1\equal{}c$
$ |m\minus{}y|\equal{}1$ so c odd we has same result.
Other case same method .So $ (a,b,c)\equal{}(1,2,3),(2,1,3)$

For solution 2 we suppose $ b>a$ then
$ b^2\plus{}1>c$ and  $ c>a^2$
$ a^2(b^2\plus{}1)\equal{}(c\minus{}b)(b\plus{}c)$
Easy to check $ b^2\plus{}1|b\plus{}c$
So $ k(b^2\plus{}1)\equal{}(b\plus{}c)$
If $ k>1$ then $ 2(b^2\plus{}1)<b\plus{}c<b^2\plus{}b\plus{}1$ so tradition.
So $ k\equal{}1$
$ a^2\equal{}c\minus{}b$
$ b^2\plus{}1\equal{}b\plus{}c$ so
$ 2b\equal{}(b^2\plus{}1)\minus{}a^2$ so
$ (b\minus{}1)^2\equal{}a^2$ so $ b\minus{}1\equal{}a$
because $ a^2\plus{}1,b^2\plus{}1$ is the prime then $ a\equal{}1,b\equal{}2$


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let a triangle ABC and a point P. Call A' the second intersection of the circle with center B from P and the circle with center C from P. Define B' and C' cyclically. Call O the circumcenter of A'B'C'. Prove that OP, AA', BB' cuncur iff AA', BB', CC' concur.

$ A',B',C'$ are obviously the reflections of $ P$ about $ BC,CA,AB,$ thus $ A,B,C$ become circumcenters of $ \triangle PB'C',$ $ \triangle PC'A'$ and $ \triangle PA'B'$ $ \Longrightarrow$ perpendicular lines dropped from $ A,B,C$ to $ B'C',C'A'$ and $ A'B'$ concur at $ O.$ Assume that $ AA',$ $BB',$ $CC'$ concur at a point $ Q.$ Then $ \triangle ABC$ and $ \triangle A'B'C'$ are perspective through $ Q$ and orthologic through the orthology centers $ P,O.$ By Sondat's theorem, $ P,Q,O$ are collinear $ \Longrightarrow$ $ AA',$ $BB',$ $CC',$ $OP$ concur. Conversely, assume that $ AA',$ $BB',$ $OP$ concur at a point $ Q.$ Let $ C'' \equiv CQ \cap PC'.$ Then $ \triangle ABC$ and $ \triangle A'B'C''$ are perspective through $ Q$ and orthologic, where $ P$ is one of the orthology centers. Thus by Sondat's theorem, it follows that the perpendicular line dropped from $ C$ to $ A'B'$ have to cut $ PQ$ at their second othology center, in other words, $ O$ is their second orthology center. Consequently, $ BO \perp A'C''$ and $ AO \perp B'C'',$ which implies that $ C'$ and $ C''$ are necessarily identical.


factorial
exponential
modular arithmetic
number theory
Divisibility
IMO Shortlist
Chinese Remainder Theorem

Find all positive integers $ n$ such that there exists a unique integer $ a$ such that $ 0\leq a < n!$ with the following property:
\[ n!\mid a^n \plus{} 1
\]

[i]Proposed by Carlos Caicedo, Colombia[/i]

If $n$ is even,$a^n+1=x^2+1,(x=a^{\frac{n}{2}})$
So 3 doesn't divide $a^n+1$ and we see 3 doesn't divide $n!$
So $n=2$.
But what if $n$ is odd,

We can use well-known (actually I never proved it with $a+1$ but rather $a-1$ but in this case I bet result is still valid and proof doesn't change much [induction])

if $p^b || a+1$ then $p^c || n$ is equivalent to $p^{b+c}||a^n+1$. Using that $v_p (n!)= \sum_{i=1} ^{\infty} [\frac{n}{p^i} ]$ we can compare $v_p(a^n+1)$ and $v_p(n!)$ and quickly bring whole problem to checking manually some small $n$.

No you cant since there are $\infty$ many $n$ with that property :D
Or how would you do it¿

I would fix $n$ and assume that there exist such an $a$, then use above (from which we get bound on $n$) -  is there really infinite number of solutions ?  :huh:  (can you show some infinite family of solutions)

Edit: I've seen my mistake - I thought $0 \leq a < n$  :blush:

Exactly the primes work :)
But I will not post a proof that they work since this spoils half of the problem.

[quote="Nima Ahmadi Pour"]Find all $n$ such that there exists a unique integer $a$ such that $0\leq a<n!$ with the following property:
\[ n!|a^n+1  \][/quote]
If $n!$ divides $a^n+1$ then $gcd(n,a) = 1$. But $a < n$. Hence $a = 0$ or $1$. Then $n!$ can be $1$ or $2$, so $n$ is $1$ or $2$.

Surely not, just try $n=3,5,7,...$ ;)
I think you missinterpreted some places of $!$'s.

Sorry :blush: it's $a<n!$
I 'll try to come up with a solution

Check that for $n=2,3$ there is unique $a$ satisfying problem. Now assume $n \geq 4$.
- If $n$ is even then $4|n!$. Hence $4|a^n+1$ and we get a contradiction.

- If $n$ is odd composite then there are at least two such $a$'s: Obviously for  $a=n!-1$ we have $n!|a^n+1$. Lets prove that if  $a=(n-1)!-1$ then $n!|a^n+1$. Note that
     \[a^n+1=((n-1)!-1)^n+1=((n-1)!)^2A+(n-1)!n-1+1=((n-1)!)^2A+n!\]
Because $n$ is composite $n!|((n-1)!)^2$ and so $n!|a^n+1$. Hence there are at least two such $a$'s.

- If $n$ is prime then there is unique such $a$: Let $n!=p!=r_1^{a_1}r_2^{a_2}....r_k^{a_k}$ be the prime factorization of $n!$ and let $(r,b)=(r_i,a_i)$ for some $i$ so that $r^b|a^p+1$. Then $r^b|a^{2p}-1$ and from Euler's theorem we also have $r^b|a^{r^{b-1}(r-1)}-1$. So $r^b|a^d-1$ where $d=\gcd(2p,r^{b-1}(r-1))|2$ . If $d=1$ then $a^p+1\equiv 2 \mod r^b$ which is obviously a contradiction. So $d=2$ and $r^b|a^2-1=(a-1)(a+1)$. If $r$ is odd then then $a\equiv 1 \mod r^b$ or $a\equiv -1 \mod r^b$. As above we must have $a \equiv -1 \mod r^b$. If $r=2$ then taking $a=2k+1$ gives $2^{b-2}|k(k+1)$. If $k\equiv 0 \mod 2^{b-2}$ then $a^p+1=(2k+1)^p+1\equiv 2 \mod 2^{b-1}$. Since $b \geq 3$ this is a contradiction. So $k\equiv -1 \mod 2^{b-2}$ and $a\equiv -1 \mod 2^{b-1}$. If $a\equiv 2^{b-1}-1 \mod 2^b$ then
\[a^p+1 \equiv (2^{b-1}-1)^p+1 \equiv 2^{b-1} \mod 2^b\]
This gives  $a\equiv -1 \mod 2^b$. Thus we proved that $a\equiv -1 \mod r_i^{a_i}$ for all $i$. From the Chinese remainder theorem this system of equations has unique solution in $[1, r_1^{a_1}r_2^{a_2}....r_k^{a_k})=[1, p!)$. Hence when $n$ is prime there is unique such $a$.

I have the same as [b]cefer[/b] for evens and primes, but my proof for odd composite is a little different.
[hide="Solution"]
Let $ n$ be an odd composite number.  Clearly, $ a\equal{}n!\minus{}1$ works.  Let $ p$ be a prime divisor of $ n$.  We consider all other $ a$ that satisfy the condition.  Since $ n$ is odd, we have that $ p$ is odd.  Letting $ p^x\parallel{}n$, $ p^l \parallel{} n!$, and $ p^xk\equal{}n$, we have that $ a^{p^xk}\equiv \minus{}1\bmod n!$.  We let $ a\equiv \minus{}1\bmod q^{e_q}$, where $ q$ is a prime divisor of $ n!$ not equal to $ p$ and $ q^{e_q}\parallel{}n!$.  Now, notice that $ a^{p^xk}\equiv \minus{}1 \bmod p^l$ entails that $ a^{p^xk}\equiv \minus{}1\bmod n!$ because we set $ a\equiv \minus{}1\bmod q^{e_q}$.  Since $ 3p\le n$, we have that $ l\ge x\plus{}2$.  Let $ g$ be a prime generator $ \bmod p^l$ (since $ p$ is an odd prime, it exists).  Letting $ a\equiv g^{p^{l\minus{}x\minus{}1}\cdot \frac{p\minus{}1}{2}}\bmod p^l$, we see that $ a^{2p^x}\equiv g^{\varphi (p^l)}\equiv 1\bmod p^l$, so $ a^{p^xk}\equiv \minus{}1\bmod p^l$, but $ a$ is not equivalent to $ \minus{}1\bmod p^l$.  Thus, $ a$ works for the conditions, but it is not $ n!\minus{}1$, so we have found a new solution, which means that odd, composite $ n$ does not work.  [/hide]

Note that $ n \equal{} 1, 2, 3$ are solutions. Suppose that $ n \geq 4$ is a solution, and let $ a$ be the unique integer. $ a$ and $ n!$ are relatively prime, since $ (a, n!) \leq (a^n, n!) \leq (a^n, a^n \plus{} 1) \equal{} 1$.

[b]I. [/b] [i]The equation $ x^n \equiv_{n!} 1$ does not have a solution in $ 2 \leq x \leq n!$.[/i]
[i]Proof.[/i] Suppose the contrary, and let $ 2 \leq x \leq n!$ be a solution to that equation. Then $ (ax)^n \equiv a^n x^n \equiv \minus{} 1 \pmod {n!},$ hence $ ax \equiv_{n!} a$ hence $ x \equiv_{n!} 1,$ contradiction.

[b]II.[/b] [i]$ n$ is prime.[/i]
[i]Proof.[/i] Suppose the contrary, and let $ q$ be the least prime divisor of $ n,$ so that $ 2q \leq n.$ Then $ q \,|\, \tfrac{n!}{q},$ and $ x \equal{} \tfrac{n!}{q} \plus{} 1$ satisfies $ x^n \minus{} 1 \equal{} \tfrac{n!}{p} (1 \plus{} x \plus{} \cdots \plus{} x^{n \minus{} 1})$ and also $ 1 \plus{} x \plus{} \cdots \plus{} x^{n \minus{} 1} \equiv_p 1 \plus{} 1 \plus{} \cdots \plus{} 1 \equiv_p 0,$ hence $ n!$ divides $ x^n \minus{} 1,$ contradiction.

[b]III.[/b] [i]All primes are solutions.[/i]
[i]Proof[/i] Let $ p$ be a prime. Then $ a \equal{} p! \minus{} 1$ is a solution to $ p! \,|\, a^p \plus{} 1.$ Suppose that $ b$ is another solution, then (because both are relatively prime to $ p!,$) $ b^{ \minus{} 1}a \bmod p!$ is a solution to $ x^p \equiv_{p!} 1,$ then $ (x, p!) \equal{} 1$ so if $ \delta$ is the order of $ x$ modulo $ p!$ it must be $ \delta \, |\, p$ and $ \delta \,|\, \phi(p!),$ and since $ (p, \phi(p!)) \equal{} 1,$ $ \delta \equal{} 1,$ so $ x \equal{} 1.$ Hence $ b^{ \minus{} 1} a \equal{} 1$ so $ a \equal{} b$ and we have that $ a$ is indeed unique.

[hide="Is this right?"]I claim that $ n$ satisfies the conditions given in the problem if and only if $ n$ is prime. (Note that $ n \neq 1$, since we cannot find an integer between 0 and 1!.) We first note that by the Chinese remainder theorem, it is necessary and sufficient for us to have $ a^n \equiv \minus{} 1 \pmod{q^k}$ to have exactly one solution for all primes $ q$ and positive integers $ k$ such that $ q^k | n!$.

$ n \equal{} 2$ clearly satisfies the condition. If $ a^n \equiv \minus{} 1 \pmod{n!}$ for any larger even $ n$, then $ (a^2)^{\frac {n}{2}} \equiv \minus{} 1 \pmod{4}$, as $ 4 | n!$; this is clearly impossible. Hence, $ n$ must be odd.

If $ q$ is some prime such that $ q | n$ and $ q^2 | n!$, then $ x^n \equiv \minus{} 1 \pmod{q^2}$ must have exactly one solution mod $ q^2$. But $ ( \minus{} 1)^n \equiv \minus{} 1 \pmod{q^2}$, and $ ( \minus{} 1 \plus{} q)^n \equiv ( \minus{} 1)^n \minus{} qn \plus{} q^2(\mbox{more stuff from binomially expanding}) \equiv \minus{} 1 \pmod{q^2}$ (since $ q | n$ and $ n$ is odd.) Since $ \minus{} 1$ and $ q \minus{} 1$ are distinct residue classes modulo $ q^2$, we have a contradiction. Therefore, if $ q | n$, then $ q^2 \not | n!$. But $ q^2 | n$ and $ q^2 \not | n!$ implies that $ q > \frac {n}{2}$. On the other hand, since $ q | n$, if $ q \neq n$, then $ n \geq 2q > n$, a contradiction, so $ n \equal{} q$. Therefore, if $ n$ satisfies the problem's conditions, it must be prime.

I shall now show that all primes $ n \equal{} p$ satisfy the given conditions. We have that $ p^1$ fully divides $ p!$, and that $ x^p \equiv x \equiv \minus{} 1 \pmod{p}$ has exactly one solution mod $ p$. For any $ q | p!$, $ q \neq p$, I claim that the equation $ x^p \equiv \minus{} 1 \pmod{q}$ has just the solution $ x \equiv \minus{} 1 \pmod{q}$. Let $ \omega$ be a primitive root mod $ q$, and let $ \omega^a \equiv x \pmod{q}$, then $ \omega^{ap} \equiv \omega^{\frac {q \minus{} 1}{2}} \pmod{q}$, so $ ap \equiv \frac {q \minus{} 1}{2} \pmod{q \minus{} 1}$, that is, $ q \minus{} 1 | 2ap$. $ p > q$, so $ q \minus{} 1 | 2a$, that is, $ \frac {q \minus{} 1}{2} | a$. Hence, $ \omega^a \equiv \pm 1 \pmod{q}$. In addition, as $ q \neq 2$, $ x \equiv \omega^a \not \equiv 1 \pmod{q}$, so $ x \equiv \minus{} 1 \pmod{q}$, as desired.

Now, since $ p( \minus{} 1)^{p \minus{} 1} \not \equiv 0 \pmod{q}$, by Hensel's lemma, the congruence $ x^p \equiv \minus{} 1 \pmod{q^k}$ has exactly one solution for all positive integers $ k$, which completes our proof. [/hide]

[hide="Solution with help from dinoboy and yugrey"]
Note that $n$ is odd if $n$ is not 2 because $a^n+1$ then is a sum of two squares and no primes $p\equiv 3\pmod{4}$ can divide it.

Let $n$ be a composite odd integer. Let $p|n$ and $p$ a prime. Then note that $(\frac{n!}{p}-1)^n+1\equiv 0\pmod{\frac{n!}{p}}$. However, since $v_p((\frac{n!}{p}-1)^n+1^n)=v_p(\frac{n!}{p})+v_p(n)\ge v_p(n!)$, we have

\[(\frac{n!}{p}-1)^n+1\equiv 0\pmod{n!}\]

Thus $a=\frac{n!}{p}-1$ is a solution. However, since $a=n!-1$ is a solution, $a$ is not unique and $n$ cannot be composite.

Let $n$ be an odd prime and let $q$ be a prime less than $n$, with $q^k|n!$ with $k$ as maximal. Since $a^{2n}\equiv 1\pmod{q^k}$ and $a^{q^{k-1}(q-1)}\equiv 1\pmod{q^k}$ and $q,q-1$ are relatively prime to $n$ because $n$ is a larger prime, we have $a\equiv -1\pmod{q^k}$. Thus $a\equiv -1\pmod{n!}$, and $a$ is unique. Thus our solutions are primes.
[/hide]

Easy problem. This is my [hide="solution"]solution.

First, take some $n$ that works. Then for every $p \le n$ a prime, say $p^x || n!$, and let there be $f(p)$ solutions $a$ in $\mathbb{Z}_{p^x}$ of the equation $a^n+1=0$. Then, because of Chinese Remainder Theorem, there are $\prod_{p \le n} f(p)$ solutions $a$ in $\mathbb{Z}_{n!}$ of the equation $a^n+1=0$. But the problem says there is only one solution. Therefore $f(p)=1$ for all $p \le n$. So there is exactly one solution $a$ in $\mathbb{Z}_{p^x}$ of the equation $a^n+1=0$, where $p^x || n!$ and $p \le n$.

Consider $g$ a primitive root modulo $p^x$. Then $\{g^0, g^1, ..., g^{p^{x-1}(p-1)}\}$ is a reduced residue system modulo $p^x$. If $p^x | a^n+1$, clearly $a$ must be coprime to $p$. So the solution $a$ must belong to that set, and it must occur exactly once. Now take the set $\{g^{0n}, g^{1n}, ..., g^{p^{x-1}(p-1)n}\}$. The number $a^n$ must occur exactly once. But it is easy to see it must occur a multiple of $gcd(n, p^{x-1}(p-1))$ number of times (possibly 0). Therefore $gcd(n, \phi(p^x))=1$ for any $p^x || n$.

But if $n$ isn't prime, we can take $q \le \sqrt{n}$ such that $q | n$ and therefore $q | (n, \phi{q^x})$ since $q ^2 | n!$. Contradiction. Therefore $n$ must be prime.

To check that all primes work, simply follow the proof above. The condition is equivalent to $gcd(n, \phi(p^x))=1$ for any $p^x || n$, because then the set $\{g^{0n}, g^{1n}, ..., g^{p^{x-1}(p-1)n}\}$ is a reduces residue set, and so $-1$ appears exactly once. If $p \le n$ this is clear. If $p=n$, $x=1$ and so $gcd(n,n-1)=1$. [/hide]

ONE MUST NOT FORGET: Also note that if $n=1$, the only possible $a$ is $a=0$ since $0 \le 0 \textless 1$ and it works, so $n=1$ works. [quote="Zhero"]Note that $ n \neq 1$, since we cannot find an integer between 0 and 1!.[/quote] That is wrong since $n=1$ is possible.

So we are done and the answer is [hide="answer"] answer: $n$ prime or $n=1$.[/hide]

The answer is all primes $n$.

Clearly $a=1$ is the unique value for $n=2$, so we can assume $n>2$.

Claim: If $n=p$ is prime, then the unique value of $a$ is $p! - 1$.
Proof: Since $p$ is odd, we have $(p! - 1)^p \equiv -1\pmod{p!}$, which implies $a=p! - 1$ works. Now we show it is unique.

We will show $p!\mid a+1$, which clearly finishes. We have \[p! \mid a^p + 1 ,\] Let $q$ be a prime such that $2\le q\le p$.

We have $a^p \equiv -1\pmod q$ and $a^{2p}\equiv 1\pmod q$.

 Thus $\mathrm{ord}_q(a)  $ divides $2p$ but not $p$, so it must be either $2$ or $2p$. However, $\mathrm{ord}_q(a)$ also divides $q-1$, so it cannot be $2p$.

 So $q$ divides $a^2 - 1 = (a-1)(a+1)$, but not $a-1$, which implies $q\mid a+1$. Now, if $q\ne p$, by LTE, \[\nu_q(a^p + 1) = \nu_q(a+1) \]Since $a^p + 1$ and $a + 1$ have the same exponent of each prime less than $p$, we have \[(p-1)!\mid a+1\] Since $p\mid a+1$, we get $p!\mid a+1$, as desired.
$\square$

If $n$ is even, since $n>2$, we get that $a^n \equiv 3\pmod 4$, contradiction because $a^n$ is a perfect square.

Now assume $n$ is odd and composite. We claim that $a=n! - 1$ and $a=(n-1)! - 1$ are both valid, which will show $a$ is not unique.

Notice that \[(n! - 1)^n\equiv -1\pmod n!,\] so $a=n! - 1$ works. It remains to show $a=(n-1)! - 1$ works.

We wish to show \[ ((n-1)! - 1)^n \equiv -1\pmod{n!}\]

Consider the binomial expansion of $((n-1)! - 1)^n$. Since $n! \mid ( (n-1)!)^2$, the only terms that are not divisible by $n!$ is $-1$ (since $n(n-1)! = n!$). Thus \[((n-1)! - 1)^n \equiv -1\pmod{n!}\]

The answer is when $n$ is prime.

[b]Lemma 1.[/b] Even $n>2$ don't work.
[b]Proof.[/b] Suppose otherwise. Then $a^n$ is a square that is $-1$ mod $n!$. But $-1$ is not a quadratic residue mod $3$, which divides $n!$, contradiction.

[b]Claim.[/b] Composite numbers $n$ don't work.
[b]Proof.[/b] Suppose otherwise. If $n$ is even, we are done by lemma 1. Otherwise, let $n=pm$ for a prime $p$ and an integer $m>1$. Suppose that $k=\nu_p(n!)$. Then
\[k\ge\nu_p(p\cdot 2p\cdot \dots\cdot mp)\ge m\ge 2.\]
Therefore, $p\mid\phi(p^k)$, so $\gcd(\phi(p^k),n)>1$.
This implies that there exists a residue $x$ mod $p^k$ such that $x^n\equiv 1\pmod{p^k}$ but $p^k\nmid x-1$. But then $a=n!-1$ and $a=x(n!-1)$ are both solutions, contradiction.

[b]Lemma 2.[/b] If $n$ is an odd prime, $n$-th powers that are relatively prime to $n!$, mod $n!$, are unique.
[b]Proof.[/b] Consider every prime power divisor $p^k$ of $n!$. Clearly $\gcd\left(\phi\left(p^k\right),n\right)=1$.

[b]Claim.[/b] Prime numbers $n$ do work.
[b]Proof.[/b] Obviously $n=2$ works. Otherwise, $a=n!-1$ works, and by lemma 1, that is unique.

The answer is primes only. Obviously $n$ must be odd for Fermat Christmas reasons.

Suppose $n$ is composite; then pick $a = (n-1)! - 1$. Observe that for any $p \mid n!$, we must have $p \mid (n-1)!$ as well, so by LTE $$\nu_p(((n-1)!-1)^n + 1) = \nu_p((n-1)!) + \nu_p(n) = \nu_p(n!).$$ On the other hand, any prime $p \nmid n!$ does not cause any problems. However, $a = n!-1$ tautologically works, so this contradicts uniqueness.

Now if $n=p$ is prime, then for any $q \leq p$ prime, $$q \mid a^p + 1 \iff q \mid a^{\gcd(p, q-1)} + 1 = a + 1,$$ so in particular we must have $a \equiv -1 \pmod q$ for all such $q$. On the other hand, it is easy to check by LTE that such primes must appear with multiplicity exactly the same as they do in $n!$, so it follows only $a = n!-1$ works here.

Felt easy for an N4?

Only primes. To show these work, we construct $a$. For each prime $q\leq p$, let $k=\nu_q(p!)$. Then we have $a^{2p}\equiv -1 \pmod{q^k}.$ Note that $\text{ord}_{q^k}(a) \mid \gcd(2p, q^{k-1}(q-1)) = 2$ (since if $q=p$ then $k=1$). Since it is not 1, it must be $2$, and this implies $a\equiv -1\pmod{q^k}$ for $q>2$. For $q=2,$ it is easy to see that $\nu_2(a^p+1) = \nu_2(a+1)$ and this implies the desired result. By CRT, we have $a\equiv -1\pmod{p!},$ and this is unique as desired.\\
    For even composites, it fails by mod 4. Consider an odd $n$. We claim $a$ is not unique. First, note that $n!-1$ works. We claim $(n-1)!-1$ works as well. Consider a prime $p\leq n$. Then $\nu_p(((n-1)!-1)^n +1) = \nu_p(n) + \nu_p((n-1)!) = \nu_p(n!)$, as desired.

The answer is all primes. If $n>2$ and is even, then $4\mid n!$ and so $a^n+1\equiv 0\pmod 4$ but since $a^n$ is a perfect square this is impossible. If $n$ is odd then $(n!-1)^n+1\equiv 0\pmod n!$. If $p\mid n$ for some prime $p<n$ then it is clear to see by binomial expansion that
\[\left(\frac{n!}{p}-1\right)^n+1\] is divisible by $n!$, because $n!\mid (\tfrac{n!}{p})^2$ and the only terms that aren't divisible by that is $n\cdot \tfrac{n!}{p}$ which is clearly divisible by $n!$. We now prove that all primes work. It is easy to check the result for $n=2$ and $3$ so assume $n=5$. If $n$ is an odd prime, then let $n!\mid a^n+1$. Then for all odd $p\le n$, we have $p\mid (-a)^n-1$ and $(-a)^{p-1}-1$. Let $g=\text{ord}_p(-a)$ then $g\mid n$ and $g\mid p-1$. Since $n$ is prime and $p-1<n$, $g=1$. Thus, $p\mid a+1$. However, then we have
\[\frac{a^n+1}{a+1}=a^{n-1}-a^{n-2}+a^{n-3}+\dots + 1\equiv n\pmod p\]
It follows that if $p<n$ then $p\nmid \tfrac{a^n+1}{a+1}$. Thus, for all $p<n$, $\nu_p(a^n+1)=\nu_p(a+1)$ so $(n-1)!\mid a+1$ but since we already know $p\mid a+1$ for all $p\le n$, we have $n!\mid a+1$, which has unique solution in $a$.

lol this took me so long for no reason, i had the heuristic down and the idea of orders and $v_p$ but took me like 1.5-2 hours :( also why is the prime case harder than the composite case

We claim that the answer is all primes and 1. Obviously n=1 has unique solution $a=0$, n=2 has unique solution $a=1$, and n=3 has solution a=5. From here on now assume $n>3$. Note that if n is even, we need $4|n!|a^n+1$, but since $a^n$ is a perfect square $a^n+1$ can only be 1 or 2 mod 4, contradiction.

[b]Case 1.[/b] n is composite.
Note that $a=(n-1)!-1$ works, because taking mod $k<n$ (and noting that $k|(n-1)!$) yields $$((n-1)!-1)^n+1\equiv ((n-1)!)^n+(-1)^{odd}+1\equiv 0\pmod k,$$ (it's well known that $n|(n-1)!$,) so for all composite n $a^n+1$ is divisible by all numbers less than n and including n, since it's divisible by n's divisors (1).
On the other hand, $a=n!-1$ obviously works (taking mod n! is just -1+1=0). But this implies that there is not a unique a, since there are (at least) two numbers that work, contradiction.

[b]Case 2.[/b] n is prime.
Clearly the solution $a=n!-1$ still works. Taking a prime $p\le n$ and a number $q=v_p(n!)$ (motivated to do this by CRT) we have that $$p^q|a^{2n}-1\iff ord_{p^q}a=r|\gcd(2n,p^{q-1}(p-1))=2\iff a^2\equiv 1\pmod {p^q}.$$ (This is still 2 even if p=n since it follows that q=1 and so its just gcd(2n,p-1).)
However, if $a\equiv 1\pmod {p^q}$, this is impossible by just plugging it in which yields $a^n+1\equiv 2\ne 0\pmod {p^q}$, so it's not divisible by $p^q$ (since $p^q\ne 2$ for $n>3$) which is a factor of $n!$, contradiction.
Hence for all primes p, $a\equiv -1\pmod {p^q}$. By CRT we know there is a distinct root modulo n! (the lcm of all of them), and since $a<n!$ there's only one value of a that satifies this. $\blacksquare$

[hide = (some would call it a sketch, i call it a: ) solution]
We'll show that the only answer is $n = 1$ and $n$ prime. It's obvious $n = 1, 2$ work, so assume $n > 2$ now. If $n$ were even, $3 \mid a^{n}+1$ would give a contradiction, thus $n$ is odd and $a = n!-1$ must be the unique solution. Let's show that $n$ being prime is both sufficient and necessary. First the necessity. If $n$ was not prime, there would be a prime $p$ with $p^2 \leq n$ and $p \mid n$. Let $k = \nu_p(n!)$. Observe that if some $a$ satisfies $p^k \mid a^p+1$ we must have $p^k \mid a+1$ as otherwise by CRT on $p^k$ and $n!/p^k$ we can find $a \ne n!-1$ with the desired properties. But this is false by Hensel's, the congruence $a^p+1 \equiv 0 \pmod {p^k}$ has at least $k-1 \geq 2$ solutions. It therefore follows that $n$ is prime. Now the sufficiency. If $q \mid a^p+1$ for some $q$, we have $q \mid a^{2p}-1$ and so the order of $a$ mod $q$ divides $2p$. It also divides $q-1$, which results in the order being $2$, implying $q \mid a+1$. After this we can force $n! \mid a+1$ by using LTE over each prime power $q^r$ of $n!$. Observe that for $q = n$ we have $r = 1$ which saves us from trouble.
[/hide]

We claim that all prime $\boxed{p}$ work.

We first show that this is true. We can see that $a = p!-1$ works. We show that this is unique. Consider a prime $q$ such that $q \leq p$ or $q |p!$. We can see that $a^p \equiv 1 \pmod{q}$ or $a^{2p} \equiv 1 \pmod{q}$ This means that $ord_q (a) | \gcd(2p, q-1) = 2$ since $q \leq p$. This means that we either have $a^2 \equiv 1 \pmod{q}$ or $a \equiv 1 \pmod{q}$ the latter is impossible because then $a^{p} \equiv 1 \pmod{q}$. Note that the second case implies that $a\equiv -1 \pmod{q}$ because otherwise we have no way to obtain $a^x \equiv -1 \pmod{q}$

Now we show that all others do not work.

We can obviously see that $a = n!-1$ works as before. We claim that $(n-1)! -1 $ also works. From LTE we get that $v_p(((n-1)! -1)^n +1) = v_p(n-1!) +v_p(n) = v_p(n!)$. $\blacksquare$ (note that this is different than when n was prime because all primes $p\leq n$ have a positve value of $v_p((n-1)!)$

We claim that the answer is when $n$ is prime.\\

The question is just asking for which $n$ is there a unique residue solution $a$ to $$a^n\equiv -1\pmod{n!}.$$ By taking mod 4, this has no solutions when $n$ is even for $n\geq 4$ since $-1$ is not a quadratic residue mod 4, so note that $n=2$ works and assume $n$ odd from now on.\\

When $n$ is prime, say $n=p$, consider any prime $q\leq p$. Let $k$ be the largest positive integer such that $q^k\mid p!.$ By Chinese Remainder Theorem, it suffices to show that $$a^p\equiv -1\pmod{q^k}$$ has a unique solution for $a\pmod{q^k}$ for all primes $q\leq p.$ Let $r$ be a primitive root mod $q^k$, since all odd prime powers have a primitive root. Then, this becomes $$a^p\equiv r^{\frac{(q-1)q^{k-1}}{2}}\pmod{q^k}.$$ Clearly, if $a$ is not relatively prime to $q$, there are no solutions in that case, so suppose that $a\equiv r^s\pmod{q^k}$. Then, this becomes $$r^{ps}\equiv r^{\frac{(q-1)q^{k-1}}{2}}\pmod{q^k}.$$ Since $r$ is a primitive root, this is equivalent to $$ps\equiv \frac{(q-1)q^{k-1}}{2}\pmod{(q-1)q^{k-1}}.$$ Since by definition $q\leq p$, $p$ is relatively prime to $(q-1)q^{k-1}$, hence there exists a unique residue $s$ that satisfies this, hence proven.\\

Now, consider if $n$ is not prime. Write $n=pk$ for a prime $p$ and an integer $k\geq 2.$ Then, let $s$ be the largest positive integer such that $p^s$ divides $(pk)!$. In particular, note that $s\geq 2$ since the factor has at the very least a $p$ and a $2p$.\\

Now, consider $$a^{pk}\equiv -1\pmod{p^s}.$$ Again, letting $r$ be a primitive root mod $p^s$. this is $$a^{pk}\equiv r^{\frac{(p-1)p^{s-1}}{2}}\pmod{p^s}$$ If $a=r^v$, this is $$r^{vpk}\equiv r^{\frac{(p-1)p^{s-1}}{2}}\pmod{p^s}$$ so $$vpk\equiv \frac{(p-1)p^{s-1}}{2}\pmod{(p-1)p^{s-1}}.$$ However, the left side, the right side, and the modulus are all divisible by $p$, which means that there cannot be a unique residue $v$ that works, contradiction, so we are done.

The answers are $n=1,n=2$ and $n$ prime.

First settle trivial case of $n$ even. If $n$ is even, -1 must be a quadratic residue mod all primes $<n$. So then if $n\geq 3$ 2 is a quadratic residue mod 3 which is wrong.

Now we have $n$ is odd. If $n$ is not prime and not 1, then it can be split into $pr$ where $p$ is prime and $r\geq 2$. Hence we have $2p\le n$, so $v_p(n!)\geq 2$. Let $k=v_p(n!)$. Now since such a solution is unique, and there is at least one solution($a=p!-1$), then if there is more than 1 solution to $a^n\equiv -1$ mod $p^k$ in the range $[0,p^k-1]$ then there are more than 2 solutions(by CRT).

Let us choose a primitive root $g$ of $p^k$. Then $g^{\frac{p^k-p^{k-1}}{2}}\equiv -1 \mod{p}$, and we only need to show that $xn \equiv \frac{p^k-p^{k-1}}{2} \mod{p^k-p^{k-1}}$ has more than 1 solution, which is obvious.

Now if $n$ is prime, it works. Firstly if there is only one solution for each $p$ for $xn \equiv \frac{p^k-p^{k-1}}{2} \mod{p^k-p^{k-1}}$ if $k=v_p(n)$ we are done. This is because by CRT there will only be one solution mod $n!$. So, this is obvious for primes $<n$ since $n$ is coprime to $p^k-p^{k-1}$(coprime to $p$ and $p-1$). Clearly for $p=n$, $n$ is coprime with $n-1$ so we are done, since $v_n(n!)=1$.

One can use Hensel to generate another solution.

The answer is all primes and $1$ which is trivial. For evens $>2$ we have $n! \equiv 0 \pmod 4,$ so $a^n \equiv 3 \pmod 4$ but is a square, impossible. We check $n=2$ works.

If $n$ is odd and is a prime $p$ then the order of $a \pmod p$ must divide $2p$ but not $p$ so it must be either $2$ or $2p.$ But it must also divide $\varphi(p!)$ so it is not $2p.$ Thus $a^2 \equiv 1 \pmod p$ so $a^p \equiv a \pmod p$ so $a \equiv -1 \pmod p$ and $a=p!-1$ is the only solution which works.

If $n$ is odd and composite, we take $a=(n-1)!-1$ and since $n \mid (n-1)!$ we have $n! \mid (n-1)!^2$ so when we expand all terms of $a^n$ using the binomial formula all terms must be divisible by $n!$ except for the last two, but the second to last term is $(n-1)! \cdot n = n!$ so it is also divisible by $n!$ so $a^n \equiv -1 \pmod{n!}$ and $n! \mid a^n+1.$ However we can also just take $a=n!-1$ so these don't work and we are done.

 :coolspeak:
$\textcolor{blue}{\text{Claim } n \hspace{0.2cm}\text{even doesn´t work}}$
We would have for some $p\mid n!$ prime $p\mid a^n+1$ so by Fermat Christmas Theorem $p=2$ or $p\equiv 1\pmod 4$

$\textcolor{blue}{\text{Claim } n \hspace{0.2cm}\text{odd and composite doesn´t work}}$
As $n$ odd $n\mid (n!-1)^n-1$ so $a=n!-1$ works
[hide=One way to found a contradiction]
We claim $a=\frac{n!}{p}-1$ satisfies the divisibility, where $p\mid n \hspace{0.2cm}\text{prime}$, thus $p^2\mid n!$, by Lifting the Exponent
$$\nu_p\left(\left(\frac{n!}{p}-1\right)^n+1\right)=\nu_p\left(\frac{n!}{p}\right)+\nu_p(n)=\nu_p(n!)-1+\nu_p(n)\geq \nu_p(n!)$$
And, as $\frac{n!}{p}\mid \left(\frac{n!}{p}-1\right)^n+1$, then $n!\mid \left(\frac{n!}{p}-1\right)^n+1$, so there is not a unique $a$
[/hide]

[hide=Another way]
We claim $a=(n-1)!-1$ satisfies the divisibility. Let $p\mid n! \hspace{0.1cm}$ prime so $p\mid (n-1)!$, so by Lifting the Exponent
$$\nu_p\left(((n-1)!-1)^n+1\right))=\nu_p\left((n-1)!\right)+\nu_p(n)=\nu_p(n!)$$
so $n!\mid ((n-1)!-1)^n+1$, so there is not a unique $a$
[/hide]
$\textcolor{blue}{\text{Claim } p \hspace{0.2cm}\text{prime work}}$
Let $q\mid p! \hspace{0.1cm}$ prime, so $a^{2p}\equiv 1\pmod q$ and $a^{q-1}\equiv 1\pmod q$ so $\operatorname{Ord}_q(a)\mid 2\gcd\left(p,\frac{q-1}{2}\right)$ so $\operatorname{Ord}_q(a)=2$ thus $q\mid a+1$, then
$$\nu_q\left(a^p+1\right)=\nu_q(a+1)+\nu_q(p)=\nu_q(a+1)$$
So $p!\mid a+1$ then $a=p!-1$ is the unique solution here

The answer are the $\boxed{\text{primes}}$. If $n$ is even and greater than $2$, we have

\[a^n \equiv -1 \pmod{n!} \iff a^n \equiv -1 \pmod{4},\]

which is impossible.

Manually check that $n=2$ is valid, so assume $n$ is odd henceforth.

If $n$ is prime, we have

\[a^n \equiv -1 \pmod{n!}\]
\[\implies a^{2n} \equiv 1 \pmod{n!}.\]

Hence, the order of $a$ modulo $n!$ is divisible by $2n$ but not $n$. This means it is either $2$ or $2n$. Consider a prime $p \mid n!$.

We have

\[a^{p-1} \equiv 1 \pmod{p},\]

so the order of $a$ modulo $n!$ is $\gcd(p-1, 2n) = 2$. This means that

\[a^n \equiv a \equiv -1 \pmod{p!},\]

fixing $a=p!-1$.

If $n$ is odd and composite, $a=n!-1$ still is valid, but we claim $(n-1)!-1$ will also be valid, violating the uniqueness of $a$. Note that

\[\nu_{p}(((n-1)!-1)^n+1) = \nu_{p}((n-1)!) + \nu_{p}(n) = \nu_p(n!). \]

We are done.

We claim the answer is $\boxed{\text{all primes}}$.
[list]
[*] For all primes $n=p$, we claim the desired unique integer is $p!-1$, which clearly works. To prove this, take an arbitrary prime $q<p$, and note we require
\[a^p \equiv -1 \pmod q \implies \operatorname{ord}_qa \mid 2p, q-1 \implies \operatorname{ord}_qa = 2 \implies q \mid a+1.\]
We conclude using LTE to get
\[v_q(p!) \leq v_q(a^p+1) = v_q(a+1) \implies p! \mid a+1 \implies a=p!-1.\]
[*] For all evens other than 2, notice that if $m$ works, $n!-m$ also works (and clearly $\tfrac{n!}{2}$ always fails), giving us two distinct values. For all odds, we simply take $n!-1$ and $(n-1)!-1$. $\blacksquare$
[/list]


[quote="Elio (n)"]I saw ailleur this sum

$ 1 \plus{} 10 \plus{} 10^{2} \plus{} 10^{3} \plus{} ...................... \equal{} \minus{} \frac {1}{9}$

How do you take this?[/quote]

Consider it as an GP where $ a \equal{} 1$ and $ r$ = 10. Try and calculate the sum to infinity (assume that it converges (which it doesn't))

$ S_\infty \equal{} \frac {a}{1 \minus{} r}$, for $ |r|<1$

[quote="Elio (n)"]I saw ailleur this sum

$ 1 \plus{} 10 \plus{} 10^{2} \plus{} 10^{3} \plus{} ...................... \equal{} \minus{} \frac {1}{9}$

How do you take this?[/quote]

This is false, but maybe this is how it's shown:

Let $ s \equal{} 1 \plus{} 10 \plus{} 10^2\plus{}\dots \equal{} 1 \plus{} 10(1 \plus{} 10 \plus{} 10^2 \plus{} \dots) \equal{} 1 \plus{} 10s$. Thus $ 9s \equal{} \minus{} 1\implies s \equal{} \minus{} \frac {1}{9}$.

This infinite sum has no value unless a finite number of terms are specified.

[quote="Elio (n)"]I saw ailleur this sum

$ 1 \plus{} 10 \plus{} 10^{2} \plus{} 10^{3} \plus{} ...................... \equal{} \minus{} \frac {1}{9}$

How do you take this?[/quote]

$ \sum _{x \equal{} 0} ^\infty 10^x$

$ \mid r \mid \ge 1$, so there is no sum.

may be you mean this:

$ 10^{\minus{}1} \plus{} 10^{\minus{}2} \plus{} 10^{\minus{}3} \plus{} ...................... \equal{}  \frac {1}{9}$

Set $ H_n\equal{}\sum_{k\equal{}1}^{n}{\frac{1}{10^k}}$

Note that:

$ 10H_n\minus{}H_n\equal{}1\minus{}\frac{1}{10^n}$, with this kind of thought you can "evaluate" many of this sums.

 :)

No, he meant that he saw a proof for this obviously false fact and wanted to know what it was.

ooooooooooooooooooooooooooooooooooooooooo  :blush:  jiji thanks....

 :)


geometry
geometric transformation
reflection
parallelogram
trigonometry
ratio
AMC

Given $\triangle ABC$, $M$ is on $AB$ and $P$ is on $CA$ such that $AM=MB$, $CP: PA=4: 3$. $CM$ and $BP$ meet at $Q$. If $AQ=CQ=4$ and $BQ=5$, find the area of $\triangle ABC$.

[hide="Hint!!!"]Extend $AQ$, use Ceva, and you'll see something interesting with the ratios.  :) [/hide]

I got a very messy solution involving triangle area relationships. I drew the perpendicular to side $AC$ from $Q$ and found $AC$, then I used $A=1/2 (b*h)$
[hide="I got"]
$85/4$ ?
[/hide]

That's not correct... The answer is an integer.

Corrected..I hope I didn't make any more errors this time
[hide="solution"]
Extend $Q$ to meet $BC$ at $S$.  Let $[APQ] = 3a$. Then $[PQC] = 4a$
Since $AM=MB$, $[BQC]=3a+4a=7a$.
$[AQB]/[BQC] = 3/4$. Therefore, $[AMQ] = [MQB] = 21a/8$.
Using the relationship between areas and bases, we find $QP = 20/7$.
Draw the perpendicular from $Q$ to $AC$, meeting $AC$ at $R$. Since $AQ=QC$, $AR=RC = 3.5x$
Using stewarts theorem with $QP$ as the cevian we obtain ${x = 4\sqrt{2}/7}$
We can now find the area of the triangle $PQC$ and find the value of $a$. When you do this, $a=8/7$.
Therefore, the total area of the triangle $ABC$ is $8/7(77/4) = 22$ [/hide]

[hide="another solution"]
$C$, $Q$, and $M$ are obviously collinear, so using Menelaus' Theorem with $\triangle CMA$ gives

\[ \frac{CQ}{MQ}\cdot \frac{MB}{AB}\cdot \frac{AP}{CP}=1 \]

and since $CQ=4$, $\frac{MB}{AB}=\frac 12$, and $\frac{AP}{CP}=\frac 34$, $MQ=\frac 32$.

Now let $Q^\prime$ be the reflection of $Q$ over $M$ so that $AQBQ^\prime$ is a parallelogram. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides (easy to prove using Law of Cosines), so if $AM=MB=a$, then $2(4^2+5^2)=3^2+(2a)^2\Rightarrow 2a=\sqrt{73}$.

Using Heron's Formula and using difference of squares, \[ [ABQ]=\sqrt{\frac{9+\sqrt{73}}{2}\cdot \frac{9-\sqrt{73}}{2}\cdot \frac{1+\sqrt{73}}{2}\cdot \frac{-1+\sqrt{73}}{2}}=\sqrt{2\cdot 18}=6 \]

If $h_X$ is the distance from $X$ to $AB$, then $\frac{[ABC]}{[ABQ]}=\frac{h_C}{h_Q}=\frac{CM}{QM}=\frac{11}{2}$ and solving, $[ABC]=\boxed{22}$.
[/hide]

[hide="My solution"]
My solution is the same as scorpius119's solution until the part where he constructed a parallelogram and started to calculate the area of the parallelogram.
Once you construct the parallelogram, you can find that $MQ'=\frac{3}{2}$ (which means $QQ'=3$) and $AQ'=BQ=5$.
Therefore, $\triangle AQQ'$ is a 3-4-5 triangle, so $AQ\perp MC$.

Hence the area of $\triangle ABC$ is $2\cdot \triangle AMC=2\cdot \frac{1}{2}\cdot 4\cdot \frac{11}{2}=\boxed{22}$

[/hide]

use mass points, you get $PM=\frac{3}{2}$,

$PM$ is a median, using the median formulas, $AB^2+3^2=2(4^2+5^2)\implies AB=\sqrt{73}$,

by heron, $[APB]=6$,

then looking at the length ratios, $\frac{[ABC]}{[APB]}=\frac{11}{3}\implies [ABC]=\boxed{22}$

note: this is somewhat similar to an aime #15 from some time ago

[quote="Altheman"]note: this is somewhat similar to an aime #15 from some time ago[/quote]If you're talking about Aime 1989#15, they're similar, but different. :D

I solved this with mass points as well. :)


The solution to $ \minus{}4 < 2(x \minus{} 1) < 8$ is expressed in the form $ a < x < b$. Find the value of $ a \plus{} b$.

Let's solve the inequality for $ x$ first. This is fairly simple; follow the standard procedures.

$ \minus{}4 < 2(x\minus{}1) < 8$
$ \minus{}2<x\minus{}1<4$
$ \minus{}1<x<5$
By comparing this with the give $ a<x<b$, we see that $ a\equal{}\minus{}1$ and $ b\equal{}5$, so $ a\plus{}b\equal{}\minus{}1\plus{}5\equal{}\boxed{4}$.


LaTeX
combinatorics unsolved
combinatorics

[url]http://www.dpmms.cam.ac.uk/~wtg10/enthusiast1.pdf[/url]

Question 5. This probably isnt that hard but im stumped.

A piece of advice: I, as a forum user, am much more likely to skip over a post if it would require me to load another website in order to see a problem when there are so many problems here on this forum for which I do not have to wait for a .pdf to load.  If you were to type the text of the question in, you would probably be more likely to get a result.

Also, the question number 5 (about cycles in $\mathbb{Z}^{3}$) is not a question of analysis, so it is not suprising it has received no answers in the analysis forum.

you make 2 good points well. although its here now and i dont know how the code stuff works so i cant type in the question.

"The code" is called Latex.  It's worth learning because it makes it much easier to type mathematics so that other people can read it.  There is a tutorial around this site somewhere (maybe at the link labelled "Latex" on the sidebar) and also a forum where you can practice Latexing things up.

[quote]5. Does there exist a cycle in $\mathbb{Z}^{3}$ (i.e., a path consisting of line segments going from
lattice point to neighbouring lattice point and ending up where it started - a lattice point
being a point with integer coordinates) such that none of the projections in the $x, y$ and
$z$ directions contains a cycle?[/quote]

And with that, I'll move it to someplace that isn't the analysis forum.

What does 'none of the projections in the x, y and z directions contains a cycle' mean? I don't understand, can you give an concrete example?

i interpreted it as, for example, that the projection in the x-direction is what the cylce looks like on the y-z plane if you squashed it flat in the x direction


Hi everyone!

Here's the list of guys from the Republic of Moldova:
1. Plamadeala Eugeniu  a.k.a. AndersoN (me!) (HM at IPhO2004, Bronze at IPhO2005)
2. Vanovschi Vladimir (Bronze at IPhO2005)
3. Patrinica Artiom
4. Vicol Dorian
5. Burovenco Igor

As I've heard, we're probably leaving on Friday.

I'm going too  :lol:.

The french team:
1/ Florian Kahn
2/ Igor Kortchemski (me  :) )
3/ Suzanne Lanéry
4/ Swann Piatecki
5/ Arthur Stril

Great!  :)
When did the final selection end? How long have you been preparing :?:

We had a preparation in our respective schools (2 hour per week since October for my school). In the beginning of April we had a 3 hour theoretical test (the 2005 test is available here in french: [url=http://www.obspm.fr/~lunap/SaE/IPhO/QCM05.pdf]part 1[/url], [url=http://www.obspm.fr/~lunap/SaE/IPhO/ex05.pdf]part 2[/url], [url=http://www.obspm.fr/~lunap/SaE/IPhO/pb05.pdf]part 3[/url]).

The 14 best students were selected for an experimental camp which lasted one week in May. At the end there was an experimental test (3 hours), and the team was formed by the results of the two tests. Finally, we have a two day training session on Wednesday and Thursday before departing  :) .

How were you prepared?

In the previous years I used to go to special physics lessons in my school. About 6 hours per week.
But, this school year I have studied physics only in the second semester, from March to June.

The selection was part of the Republican Physics Olympiad and consisted of a theoretical examination (5 problems; 3 hours). After that, we mostly studied by ourselves, sometimes doing experiments at the Moldova State University, where our team leader teaches.

Ok, the serbian team is as follows:

1. Delic Uros
2. Nikolic Branko
3. Radicevic Djordje (me!)
4. Jankovic Marija
5. Pavlovic Marko

Our selection system was very strange this year -- at least, in comparison to all the previous years when the team was chosen on a single National Olympiad, comprising of a theoretical and an experimental part. We had the first part of the Olympiad in April (4 theory + 1 exp. data interpretation in four hours); the first 15 proceeded to the next part, held in May, which consisted of four more theoretical problems in four hours. The first 10 proceeded onto the third part which took place a day after the second part and was made up of two experiments in four hours. The first seven went on a two week preparation programme, and the fourth part of the Olympiad was held immediately after these preparations (in the beginning of June); it consisted of two theoretical and one experimental problem (four hours for work, again). After the team was announced, we continued getting more problems for practice by e-mail... and that's about that. Before the preparations started in June, we were all left to prepare within our high schools.

Interesting selection. I'd say it's VERY selective :D
What about those problems that you got through e-mails? Did you get solutions for them too? :)

I had National High School Examinations (Baccalaureatte) till 18 June, which was tiresome and time wasting.
I have heard Russian IPhO ppl saying that IPhO team members are exempted of national exams, they directly get the marks that they have at each subject in the first semester. This is not a bad way to save time for preparations and minimize stress for these ppl.

What about experiment preparation?

[quote="AndersoN"]This is not a bad way to save time for preparations and minimize stress for these ppl.[/quote]

I think that selection camps / exams are the best way to prepare & get to know others. I've never learnt more than on the high levels of team selections. (Incomparably more than in any school-related preparation.)

Anyway, the Slovak Team is quite young / new:

Jakub Imriska,
Tomas Bzdusek (Gold at EUSO),
Marcela Hrda,
Peter Peresini (Gold+Bronze at IOI:)
and someone else I don't know.

[but I hope they'll do well anyway:)]

Well, the concept of our selection was great, but we had to traditionally mess a few things up while organizing it. For example, we didn't have enough professors available to hold theory problems, so we ended up doing a week and a half of lab work... I think it wasn't bad, I learned a great deal, but I'd be much happier if we had more theory than experiments... And unfortunately the preparations were held in Belgrade (the capital city) because of the experiment part, so our overall productivity was rather low -- of seven students to attend the preparations, four were from Belgrade and it's just not the same when you have to go on some lectures and then can trot right back home and forget almost everything... (I think so). Fortunately, we still don't have national exams at any level of high school, so this made things a lot easier -- for instance, the schools themselves are rather lenient, so there was no problem for us to skip entire weeks of classes and still have good grades.

BTW, we're going to Singapore from Zurich on Friday evening (22:30, I believe).

And no, I'm waiting for solutions, today was the deadline for sending in our own solutions...

[b]Here's the Vietnamese Team[/b]

1. Nguyễn Đăng Phương (Third Prize at VPhO 2005, Second Prize at VPhO 2006, Silver at APhO 2006)

2. Trần Xuân Quý (Third Prize at VPhO 2005, Second Prize at VPhO 2006, Bronze at APhO 2006)

3. Phạm Tuấn Hiệp (Second Prize at VPhO 2006)

4. Phan Hữu Thành (Second Prize at VPhO 2006)

5. Bùi Đức Thắng (Third Prize at VPhO 2006)

[img]http://i34.photobucket.com/albums/d128/tungcly/DoituyenQT1.jpg[/img]

Ok, does anybody want to say something about the past IPhO? It seems strange that no one posted any kind of comment here (so far)...

Well, concerning the entire Olympiad program, I considered it to have been quite efficiently organized -- notwithstanding that the target population for a great lot of events was 10 years or so (Sentosa beach games and the Musical Fountain, anyone?). With all those effects, they could've pulled out quite a show :( As for the competition itself, I thought that there was [b]way[/b] too much optics... What happened to classical physics? I hadn't even had the opportunity to use Newton's laws :o Why so? :huh:

Oh, @[b]AndersoN[/b]: I hadn't figured out who you are until the closing ceremony, and after that I couldn't find you anywhere... Were you flying back via Istanbul?

Post pictures!

This competition seemed more difficult to you that the previous one?

you know where I can find the results by country ?

Djole you have medal of silver!!

 :lol:

[quote="Djole"]
Oh, @[b]AndersoN[/b]: I hadn't figured out who you are until the closing ceremony, and after that I couldn't find you anywhere... Were you flying back via Istanbul?[/quote]
Yep. Istanbul was my way.  I didn't manage to figure out who YOU are at all. :(  I only managed to speak to Igor from the French team.
Congrats on your medals! Really nice.

I totally screwed up... because of several reasons. One of which is... as you said, wayyy too much optics.
Anyway, life doesn't end here.  See y'all around.

Ok, I wrote a message twice and, due to various unpredictable reasons, got it erased :mad: I'll be brief, for now:

@[b]AndersoN[/b]: I thought that you've seen me, because our team was seated right next to the Moldovans in that sports hall prior to the astronomy night. I couldn't approach your team because I spent the entire time there talking to my leaders about the results (some 10 minutes), and after that we departed towards the hockey field and I lost sight of you.

Tnx for the congrats! I still can't believe that I managed to get a silver on such problems (especially the one with the digital camera)... This Olympiad is too crazy :)

Here are the photos of the french team: http://shadowmare.free.fr/photos .

And the first 27 countries (out of 250, [b]before moderation[/b]):

1. China (213.85)
2. Indonesia (195.55)
3. USA (193.35)
4. South Koria (189.55)
5. Taiwan (181.8 )
6. Azerbaijan (165.1)
7. Iran (162.8 )
8. Hungary (162.5)
9. Russia (157.4)
10. Germany (155.15)
11. Thailand (153.15)
12. Romania (140.7)
13. Ukraine (139.35)
14. Singapore (139.05)
15. India (138 )
16. Australia (134.95)
17. Canada (<134.3)
18. Hong Kong (132.3)
19. Turkey (130.05)
20. France (130)
21. Czech Republic (127.9)
22. Bulgaria (119.75)
23. Israel (118.4)
24. Japan (118.55)
25. United Kingdom (117.3)
26. Bielorussia (117.2)
27. Vietnam (115.8 )

Heh, I thought I'd never see that list... Thanks for making it available ;) Too bad it's not valid anymore (a lot of people had made more points after the moderation...)

Well, I am very happy that Indonesia could get the [b]absolute winner[/b] of IPhO 2006. May the achievement could be achieved by Indonesia this year, and forever.
This is the team of Indonesia 2006:
Jonathan Pradana Mailoa
Pangus Ho
Andy Octavian Latif
M. Firmansyah Kasim
and I forget this one...

2007 IPhO national team of Indonesia
M. Firmansyah Kasim
Musawwadah Mukhtar
David Halim
Yosua Michael Maranatha
ughh, I also forget this one.....[/b]


Angela, Bob, and Charlie have a total of $ \$35$ in one-dollar bills. Bob has $ \frac34$ as much as Angela and Charlie together. How many dollar bills does Bob have?

Let $ A$, $ B$, and $ C$ be the amounts of each person (by first initial).  We have $ A\plus{}B\plus{}C\equal{}35$ and $ B\equal{}\frac{3}{4} (A\plus{}C)$.  Multiplying each side of our second equation by $ \frac {4}{3}$, we get $ \frac {4}{3}B\equal{}A\plus{}C$.  Entering that into our first fraction, we get $ \frac {4}{3}B\plus{}B\equal{}35$.  We then have $ \frac {7}{3}B\equal{}35$, so we multiply each side by $ \frac {3}{7}$ and get $ B\equal{}\boxed {15}$


Delete this, because nobody takes a joke.

I thought this was supposed to be the FUN factory. But since nobody can take a joke, I will just delete it.

What's the joke? It's not funny at all. It is mostly annoying.

[quote="mathelete"]Nobody understanda what a joke is.[/quote]

You don't know what a joke is.  Apparently, you don't know what a [b]ban[/b] is either.


function
real analysis
real analysis solved

Let  $f$ and  $g$  $2$ continous and convexe functions defined over $ [0..1] $ such that $sup(f,g)\ge 0$. Prove that there exist  $x\ge 0$  such that $x*f+(1-x)*g\ge 0 $ .We can generalize the result for n functions and show that exists a linear combination of them (with the sum 1) that is bigger(or equal) than 0.

It's clear that $h_t=t\cdot f+(1-t)\cdot g$ is also a convex function for $t\in[0,1]$. Given a convex function $h_t$, it's easy to see that the set on which it takes negative values is an interval $I_t$. These intervals ($I_0,I_1$) are open and disjoint for $h_0=g$ and for $h_1=f$. Take $x_t$ to be the midpoint of this interval for $h_t$. It's easy to see that $x_t$ is a continuous function of $t$ (assuming that the conclusion is wrong and there is no $t\in[0,1]$ s.t. $h_t$ takes only non-negative values), so there must be some $x_{t_0}$ in which $f$ and $g$ take non-negative values (because $x_t$ goes continuously from $x_0$ to $x_1$, so it must pass through a point which separates $I_0$ and $I_1$), but this yields a contradiction.


If x & y are positive numbers such that x^3 + y^3 +1/27 = xy. Find the value of 1/x? :huh:  :)

The value of $ x$ is clearly not unique.  Are you sure you have the condition written correctly?

Actually, I think there doesn't exist such $ (x,y)$

[hide]
The given implies $ x\plus{}y\plus{}\frac{1}{3}\equal{}0$, which is impossible since $ x, y$ are both positive.[/hide]

Where did you get that from? $ x \equal{} y \equal{} 1/3$ works.

Oops...I'm wrong. I misused this identity:

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} \frac {1}{2}(a \plus{} b \plus{} c)((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)$

Alright, then this problem has an unique solution as you have said, $ \frac{1}{x} \equal{} 3$.

[hide="AM-GM..."]
$ x^3\plus{}y^3\plus{}\frac{1}{27} \ge xy$... equality iff $ x^3\equal{}y^3\equal{}\frac{1}{27}$[/hide]


Let point $ P$ be inside the square $ ABCD$ such than $ PA\equal{}3$, $ PB\equal{}4$, and $ PD\equal{}5$. Find the length of $ PC$ and the side length of the square.

[[British Flag Theorem]] (not really Pre-Olympiad, but .. ).

thanks, but lets say that instead of a square, we had a triangle. for example:

An equilateral triangle $ ABC$ has an interior point  $ P$. If $ PA\equal{}3$, $ PB\equal{}4$, and $ PC\equal{}5$, what is $ AB$?

For the equilateral triangle problem, rotate the triangle $ 60^{\circ}$ clockwise about a vertex, notice the construction of a right triangle, and make use of the cosine law.


[quote="tranthanhnam"]i think "=" <=.
x= 0.8012
i can't solve this prolem[/quote]
?????? :?
If I wasn't wrong then $x=\frac{\sqrt[4]{5}}{\sqrt{3}}$

Tranthanhnam: Not fair, this is problem of "Mathematics of Young".
I think $x = \frac{\sqrt{3}}{2}$

I agree with b2.  This is my proof:

Let $D=\sqrt{x+x^3}+\sqrt{x-x^3}$.  For $x\in[0,1]$, $D\ge0$, so to maximise $D$ it is sufficient to maximise $D^2=2x(1+\sqrt{1-x^4})$.

Differentiating this, and letting $u=\sqrt{1-x^4}$, we get $\frac{d}{dx}D^2=2+2u-\frac{2x^4}{u}$.

$\frac{d}{dx}D^2=0\Leftrightarrow u+u^2-2x^4=3u^2+u-2=0\Leftrightarrow u=-1$ or $\frac{2}{3}$.

Obviously $x\in[0,1]\Rightarrow u\in[0,1]\Rightarrow u=\frac{2}{3}\Rightarrow x=\sqrt[4]{1-\frac{2}{3}^2}=\frac{\sqrt[4]{5}}{\sqrt{3}}$.  It is easy to confirm that this turning point is in fact the maximum.


modular arithmetic
function
number theory

Jo is celebrating 2009 by going to the Fall Out Boy concert, but his dad wont let him go unless he gets a math problem right. Help him go to the concert by finding the tens digit of $ 2009^{2009}$

( i know i already asked something like this, but there was a pattern in the last one, and i dont see a pattern this time)

You want to find the residue of $ 2009^{2009}$ in modulo 100.

$ 2009^{2009}\equal{}9^{2009}\equiv{x}\pmod{100}$ where x is the residue. By Euler's Totient theorem: $ 9^{\phi(100)}\equal{}9^{40}\equiv1\pmod{100}\implies 9^{2000}\equiv{1}\pmod{100}$. This leads to finding the ten's digit of $ 9^9$, which is 8.

Because of this: http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem

[quote="DaReaper"]okay. i read that and still have no idea where the phi comes in and how you got the 40...  :huh:[/quote]

Look up the phi function. They'll probably explain it waaay better than I will. They as in AoPSWiki.

[quote="DaReaper"]how did you get $ 2009^{2009}$ as being equal to $ 9^{2009}$?[/quote]

no; they are congruent mod 100 because $ 2009 \equiv 9 \pmod{100}$. (Or at least I think that's what you're asking)

but then after that you apparently are confused by the phi function (aka Euler's totient.) You might want to learn some number theory first... Fermat's Little Theorem and Euler's Theorem are pretty useful to know.


calculus integration LaTeX algebra polynomial

I rewrote it it LaTeX :)

$ \int_0^{\pi/2} \frac {\sin(5x)}{\sin(x)} dx$

expand the integrand to get a polynomial in cos(x)

@ modularmarc101  :thanks   :D

@Aashay THE ROKR :nice solution :first:

hello
$ I \equal{}\int_{0}^{\pi/2}\frac{\sin(nx)}{\sin(x)}dx$
if
$ n$ is odd $ \plus{}ve$ integer.
$ I\equal{}\frac{\pi}{2}$ only.


function
floor function
logarithms
algebra theorems
algebra

How do you prove that:

If  $f(x)$, $f: \mathbb{Z^{+}}\rightarrow \mathbb{Z^{+}}$ is an increasing function, such that$f(m)f(n)=f(mn)$, then $f(x)=x^{c}$, where $c$ is a constant.

I think shobber misread $f(x)=c^{x}$ instead of $f(x)=x^{c}$.

So we also have that:

If  $f(x)$, $f: \mathbb{Z^{+}}\rightarrow \mathbb{Z^{+}}$ is an increasing function, such that$f(m+n)=f(mn)$, then $f(x)=c^{x}$, where $c$ is a constant. :?:

No. :P

If  $f(x)$, $f: \mathbb{Z^{+}}\rightarrow \mathbb{Z^{+}}$ is an increasing function, such that $f(m+n)=f(mn)$, then  $f(x)=c$, where $c\in\mathbb{Z^{+}}$. If we want a strictly increasing function, there's no solution.

If  $f(x)$, $f: \mathbb{Z^{+}}\rightarrow \mathbb{Z^{+}}$ is an increasing function, such that $f(m+n)=f(m)f(n)$, then $f(x)=c^{x}$, where $c\in\mathbb{Z^{+}}$. If we want a strictly increasing function, then we must have $c>1$.

Both statements are proved taking $n=1$ and then straightforward induction.

The original problem is more difficult.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55870 .

But the case given here is easier since it has strong multiplicativity:
Let $f(2)=v=2^{t}$. Then for any positive integer $n$ let $k=\left\lfloor \log_{2}(n)\right\rfloor$.
Thus $2^{k}\leq n <2^{k+1}$, so $v^{k}\leq f(n) \leq v^{k+1}\implies k \leq \log_{v}(f(n)) \leq k+1$.
Now lets change $n$ into $n^{s}$:
$k \leq \log_{v}(f(n^{s}))=\log_{v}(f(n)^{s})=s \cdot \log_{v}(f(n)) \leq k+1$ (note that this $k$ depends on $s$), equivalent to: \[\frac{k}{s}\leq \log_{v}(f(n)) \leq \frac{k+1}{s}\] So we know that $\log(f(n))$ is from some interval of length $\frac{1}{s}$. But letting $s$ increase this shows that there is only one value for $\log_{v}(f(n))$ that works, so also only one value for $f(n)$. But we know one working value: $f(n)=n^{t}$ (easy to check that it works).
this it has only one solution for each $t$, all of them being like claimed.

Let f: R ->R>0 | f(x+y)=f(x)*f(y).
 Is f(x)=c^x a unique solution?

No, counterexamples are archieved the same way than for Cauchy's equation.


advanced fields
advanced fields unsolved

For a sequence $ (\alpha_j)$ which is dense in $ [0,1]$ and linear operator $ T: l^2 \to l^2, y\equal{}Tx, x\equal{}(\xi_j), y\equal{}(\eta_j), \eta_j\equal{}\alpha_j\xi_j$, find $ \sigma_p(T)$ and $ \sigma(T)$.

I started by applying the operator $ T_\lambda(x)\equal{}(T\minus{}\lambda(I))x\equal{}0$, which yields $ (\alpha_j\minus{}\lambda)\xi_j\equal{}0$.  I'm not entirely sure where to go from there, or if that is even the correct approach.  Any input is appreciated.

First you should try to find the point spectrum $ \sigma_p(T)$ (= "eigenvalues"), you are already half way there.
After you are done, try to use your result and some elementary fact about spectra of bounded operators to conclude that $ \sigma(T) \supset [0,1]$.
And finally: What would a possible inverse mapping of $ T \minus{} \lambda$ for $ \lambda \notin [0,1]$ look like? Is it well defined?


search
inequalities proposed
inequalities

Let $ a,b,c$ positive reals such that:

$ \frac{1}{a\plus{}b\plus{}1}\plus{}\frac{1}{b\plus{}c\plus{}1}\plus{}\frac{1}{c\plus{}a\plus{}1} \ge 1$

Then prove that:

$ a\plus{}b\plus{}c \ge ab\plus{}bc \plus{} ac$


Actually, that's wrong.  If you start with $ \frac{1}{10}$, then your next term will just add to that and so you'll be greater than what you want.

[hide="Answer to given series"]Notice that $ \frac{1}{10}$ is $ 0.1$
$ \frac{1}{10^2}$ is $ 0.01$
$ \frac{1}{10^3}$ is $ 0.001$
and so on.

Adding the terms of this series together gives $ 0.11111111111...$, which is equivalent to $ \frac{1}{9}$.[/hide]

[hide="Solution"]Let:
\[ S \equal{} {\left(\frac {1}{10}\right)}^1 \plus{} {\left(\frac {1}{10}\right)}^2 \plus{} {\left(\frac {1}{10}\right)}^3 \plus{} \cdots
\]
Then:
\[ \frac {S}{10} \equal{} {\left(\frac {1}{10}\right)}^2 \plus{} {\left(\frac {1}{10}\right)}^3 \plus{} {\left(\frac {1}{10}\right)}^4 \plus{} \cdots
\]
Subtracting the second equation from the first equation, we have:
\[ S \minus{} \frac {S}{10} \equal{} \frac {1}{10} \\
\Rightarrow 10S \minus{} S \equal{} 1 \\
\Rightarrow 9S \equal{} 1 \\
\Rightarrow \boxed{S \equal{} \frac {1}{9}}
\]In this way, we can also derive the general form for the sum of an infinite geometric series with first term $ a$ and common ratio $ r$:
\[ S\equal{}\frac{a}{1\minus{}r}\]In this problem, $ a\equal{}\frac{1}{10}$ and $ r\equal{}\frac{1}{10}$.
[/hide]


I had been browsing the net, but I can't find a good and understandable proof of the centroid theorem? Any idea?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=centroid&t=87177[/url]

i posted a few solutions at the bottom, along with a few other people's

[quote="mathnerd314"]What does the centroid theorem say? (I'm not familiar with the name)[/quote]

it states that the three medians of a triangle are concurrent.. and a lot more...

The concurrent thing we can just use a special case of Ceva's...

as for the 2:1 ratio, I think you can use area ratio to proof that.

But can we simplify it? I mean, can we use some fundamentals to explain without using Ceva's theorem?

you can easily prove that the medians are concurrent with coordinate geometry and vectors, also you can use area ratios (which is essentially ceva)


prove that every positive rational number can be written in the form
$\frac{a^2+b^3}{c^5+d^7}$
where $(a,b,c,d\in \mathbb{N}^+)$

http://www.mathlinks.ro/Forum/viewtopic.php?t=79523
http://www.mathlinks.ro/Forum/viewtopic.php?t=26785 (not exactly the same, but nice generalisation mentioned)
http://www.mathlinks.ro/Forum/viewtopic.php?t=24088
http://www.mathlinks.ro/Forum/viewtopic.php?t=6130


The odds of losing a certain card game is 7 to 13, which means that on average there are 7 losses for every 13 wins. What is the probability of losing the game? Express your answer as a percent.

So we have $ \frac{7}{20}\equal{}\boxed{35\%}$.


$ f(t)\equal{}\cos t_1 \cos t_2 ... \cos t_n$, then show that $ (\tan t_1 \plus{} \tan t_2 \plus{} \tan t_n \plus{} .... \plus{} \tan t_n)\equal{} \minus{}(\frac {f'(t)}{f(t)})$

Probably take 1/f(x) first and then take log on both sides :)
Then differentiate to get the answer :wink:

[quote="hell_ever"]$ f(t) \equal{} \cos t_1 \cos t_2 ... \cos t_n$, then show that $ (\tan t_1 \plus{} \tan t_2 \plus{} \tan t_n \plus{} .... \plus{} \tan t_n) \equal{} \minus{} (\frac {f'(t)}{f(t)})$[/quote]

Use uvwxyzabcd.......rule.  :rotfl:  :rotfl:  :rotfl:

Basically fix all other terms and differenciate one term alone.

You get f'(x) = Sigma (-sin(ti)Pi(cosj))
now divide by f(x)

If you want the proof for uvwxyzabcd....... rule use log as rokesh said./

Shre take 1/f(x) and take log and then diff. u will get the ans so directly. :D


When a matrix $ A$ has been reduced to a matrix $ U$, say U has 4 columns but only 2 pivots, and columns 2 and 3 have both their first two terms nonzero. Can I pick either one as the pivot column to generate the basis of $ A$?

Your question makes no sense. In particular, the phrase "generate the basis for $ A$" makes no sense. What are you talking about here? The row space? The column space? The null space?