China

by pythag011, Jul 10, 2009, 10:04 AM

Note: Chinese government censors facebook and other stuff

Currently in china.

Random problem (you might know the inspiration for this problem if you were at mop):

Let p be a prime that is 1 mod 10.

If $(p-1)!(1-\frac{1}{6}+\frac{1}{11}\ldots - \frac{1}{p-5})$ is congruent to $(p-1)!(\frac{1}{3}-\frac{1}{8}+\frac{1}{13}\ldots - \frac{1}{p-3})$ modulo p, then these numbers are also congruent to -1 times $(p-1)!(\frac{1}{2}-\frac{1}{7}+\frac{1}{12}\ldots - \frac{1}{p-4}).$

Also, how do you use linear algebra to solve differential equations?

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ummm.... china is thinking of censoring google.

by EggyLv.999, Jul 12, 2009, 3:21 AM

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"Chinese government censors facebook and other stuff"
rofl, its ture- a couple of my friends are also in china, and they cant get on facebook or youtube, and some cant even go on aim, they have to use meebo.....

by draops, Jul 15, 2009, 11:56 PM

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blah

Let $p=10k+11$ , and define

\[\begin{align*}
A_1
&=\frac{1}{1}-\frac{1}{6}\pm\cdots+\frac{1}{10k+1}-\frac{1}{10k+6}\\
&=\left(\frac{1}{1}+\frac{1}{6}+\cdots+\frac{1}{10k+6}\right)-\left(\frac{1}{3}+\frac{1}{8}+\cdots+\frac{1}{5k+3}\right),\\
A_2
&=\frac{1}{5}-\frac{1}{10}\pm\cdots+\frac{1}{10k+5}-\frac{1}{10k+10}\\
&=\left(\frac{1}{5}+\frac{1}{10}+\cdots+\frac{1}{10k+10}\right)-\left(\frac{1}{5}+\frac{1}{10}+\cdots+\frac{1}{5k+5}\right)\\
&\equiv-\left(\frac{1}{1}+\frac{1}{6}+\cdots+\frac{1}{5k+1}\right)\pmod{p},\\
B_1
&=\frac{1}{2}-\frac{1}{7}\pm\cdots+\frac{1}{10k+2}-\frac{1}{10k+7}\\
&=\left(\frac{1}{2}+\frac{1}{7}+\cdots+\frac{1}{10k+7}\right)-2\left(\frac{1}{7}+\frac{1}{17}+\cdots+\frac{1}{10k+7}\right)\\
&\equiv\left(\frac{1}{2}+\frac{1}{7}+\cdots+\frac{1}{10k+7}\right)+\left(\frac{1}{2}+\frac{1}{7}+\cdots+\frac{1}{5k+2}\right)\pmod{p},\\
B_2
&=\frac{1}{4}-\frac{1}{9}\pm\cdots+\frac{1}{10k+4}-\frac{1}{10k+9}\\
&=\left(\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{10k+9}\right)-2\left(\frac{1}{9}+\frac{1}{19}+\cdots+\frac{1}{10k+9}\right)\\
&\equiv\left(\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{10k+9}\right)+\left(\frac{1}{1}+\frac{1}{6}+\cdots+\frac{1}{5k+1}\right)\pmod{p},\\
C
&=\frac{1}{3}-\frac{1}{8}\pm\cdots+\frac{1}{10k+3}-\frac{1}{10k+8}\\
&=\left(\frac{1}{3}+\frac{1}{8}+\cdots+\frac{1}{10k+8}\right)-\left(\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{5k+4}\right)\\
&\equiv\left(\frac{1}{5k+7}+\frac{1}{5k+12}+\cdots+\frac{1}{10k+7}\right)\pmod{p}.
\end{align*}\]

We have $S\equiv A_1\equiv A_2\equiv C\pmod{p}$ and $T\equiv B_1\equiv B_2\pmod{p}$ , and we need to show that $S+T\equiv0\pmod{p}$ . Thus

\[\begin{align*}
T+S\equiv B_1+C
&\equiv2\left(\frac{1}{2}+\frac{1}{7}+\cdots+\frac{1}{10k+7}\right)\\
&\equiv-2\left(\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{10k+9}\right)\\
&\equiv-2(A_2+B_2)\\
&\equiv-2(S+T)\pmod{p}.\end{align*}\]

The result follows.

by math154, Dec 10, 2010, 1:05 AM

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easier

Let
$S_1=\binom{p}{1}+\binom{p}{6}+\cdots+\binom{p}{p-5}$ and define $S_2,\ldots,S_5$ similarly (the sums exclude $\binom{p}{0}$ and $\binom{p}{p}$ ). Since $S_1=S_5$ , we're given that
$S_1\equiv S_3\equiv S_5\pmod{p^2},$ and since $S_2=S_4$ while $S_1+\cdots+S_5=5(2^p-2)$ , it suffices to show that
$S_2\equiv2^p-2\pmod{p^2}$ for all $p=10k+1$ prime. If $\omega=e^{2\pi i/5}$ , then (the fact that $p\equiv1\pmod{5}$ is important in simplifying this)

\[\begin{align*}
S_2 &= 2^p+\sum_{j=1}^{4}\omega^j(1+\omega^{2j})^p\\
&= 2^p+\sum_{j=1}^{4}(\omega^j+\omega^{3j})^p\\
&=2^p-\sum_{j=1}^{4}\left(\frac{1}{1+\omega^{j}}\right)^p
\end{align*}\]

Now we will show that
$\sum_{j=1}^{4}\left(\frac{1}{1+\omega^{j}}\right)^p \equiv 2\pmod{p^2}$ for $p\equiv1\pmod{10}$ prime (in fact, it's true for all $p$ , I think). Indeed,

\[\begin{align*}
\sum_{j=1}^{4}\left(\frac{1}{1+\omega^{j}}\right)^p
&= \sum_{j=1}^{2}\left(\left(\frac{1}{1+\omega^{j}}\right)^p+\left(\frac{\omega^{j}}{1+\omega^{j}}\right)^p\right)\\
&= (1+\omega)^{p-1}+(1+\omega^2)^{p-1}\\
&= (1+\omega)^{p-1}+(1+\omega)^{1-p}\\
&= 2+\left((1+\omega)^{(p-1)/2}-\frac{1}{(1+\omega)^{(p-1)/2}}\right)^2.\end{align*}\]

Now note that the symmetric sums of $x^4+x^3+x^2+x+1$ in real life and in $\pmod{p}$ are the same. So doing everything with $g$ instead of $\omega$ and using FLT,
$\left((1+g)^{(p-1)/2}-\frac{1}{(1+g)^{(p-1)/2}}\right)^2\equiv0\pmod{p^2},$ as desired.

Edit:Proof of the thing that I said is true for all primes

We find that

\[\begin{align*}
X=\sum_{j=1}^{4}\left(\frac{1}{1+\omega^{j}}\right)^p
&= \sum_{j=1}^{2}\left(\left(\frac{1}{1+\omega^{j}}\right)^p+\left(\frac{\omega^{j}}{1+\omega^{j}}\right)^p\right)\\
&= \frac{1+\omega^p}{(1+\omega)^p}+\frac{(1+\omega)^p}{1+\omega^p}.
\end{align*}\]

Now note that this minus $2$ is
$\frac{[(1+\omega)^p-(1+\omega^p)]^2}{(1+\omega)^p(1+\omega^p)}=p^2\left(\frac{[(1+\omega)^p-(1+\omega^p)]}{p}\right)^2\frac{1}{(1+\omega)^p(1+\omega^p)}=p^2Y.$ Note that
$\frac{1}{1+\omega^j}=-\omega^j(1+\omega^{3j}).$ First, this formula tells us that $X$ , by a roots of unity filter, is an integer. Second, it implies that $1/(1+\omega^j)$ for $1\le j\le4$ is an algebraic integer, so $Y$ is an algebraic integer as well. But $Y=(X-2)/p^2$ must also be a rational number, so $Y\in\mathbb{Z}$ , as desired.

This post has been edited 2 times. Last edited by math154, Feb 22, 2011, 2:38 AM

by math154, Feb 22, 2011, 1:02 AM

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bump...
https://artofproblemsolving.com/community/c1352h993021
by Bryan0224, Nov 17, 2024, 1:13 PM
first shout of 2024
by QueenArwen, Nov 12, 2024, 12:04 PM
nice blog
by ryanbear, Aug 29, 2023, 6:35 PM
fantastic blog!!
by lneis1, Jul 8, 2021, 8:03 AM
what a pr0 blog
by centslordm, Jun 24, 2021, 4:34 PM
pythag011 orz
by tigerzhang, Jun 21, 2021, 5:39 PM
greetings
by Kanep, Oct 26, 2020, 12:38 PM
A very fine blog.
by Catoptrics, May 29, 2020, 11:42 PM
this is dead
by fukano_2, May 20, 2020, 10:35 PM
Found this
by Williamgolly, Nov 17, 2019, 8:47 PM
what a pr0 blog
by mathleticguyyy, Nov 9, 2019, 12:19 AM
$[asy]size(500); draw(polygon(7));[/asy]$
by FadingMoonlight, Jul 31, 2019, 11:11 AM
kala para na
by spy., May 4, 2018, 1:39 PM
So you've discovered this blog too? @archimedes15
by GeronimoStilton, Apr 5, 2018, 1:25 PM
What a dead blog...
by Archimedes15, Feb 19, 2018, 11:57 PM
bad physics. Relearn. Now.
by james4l, Aug 24, 2008, 7:15 PM
I already did.

With the new quote-simulation program (read my post on james4l's blog) I can slow my computer so much time goes backwards!

Just enter it for a thread with 1 trillion posts.

Time will flow baackwards.

Only 19.99! A 5000000 dollar value!

You can go backwards in time

Unfortunately, you'll have to stay there, because you can't go forward.
by pythag011, Aug 24, 2008, 1:32 AM
Nope it's 227 now. I should create some kind of anti-F5 button, so I can DROP a blog's views
by zephyredx, Aug 23, 2008, 6:33 PM
Whoever comes to this blog next will turn 225 views into 226
by pythag011, Aug 23, 2008, 3:34 AM
a.k.a. Mr. Ghost.
by zephyredx, Aug 22, 2008, 7:08 PM
BTW, if you go to microsoft word, double click n Mr.Green, and set brightness to 80% You get ghost Mr.Green.
by pythag011, Aug 20, 2008, 8:24 PM
Actually the beaver function is real...
by zephyredx, Aug 20, 2008, 2:24 AM
:D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D

SPAM!!!!!!!!!!!!!!
by pythag011, Aug 18, 2008, 5:24 PM
#1: 0 pts
comments: You have trivialized the problem by stating a direct reference to the solution of the problem.

#2: 0 pts
comments: irrelevant special case.

#3: 3 pts
comments: incoherent proof though the graders do applaud your use of a non-existent function

#4: 1 pts
comments: seriously, the joke's less funny second time through.

#5: 0 pts
comments: trivialization, not objectively stated the view of the contestant.

50% penalty for solving out of the time limit.

Your score is...

2 pts out of a possible 2000 pts
by james4l, Aug 18, 2008, 5:47 AM
LOL at your #2...
by zephyredx, Aug 18, 2008, 3:40 AM
1. By Perelman's proof, the Pionxare Conjecture is corect
2. No. The task of solving these problems has been proven to be NP, but not be P.
3. Apply an absolute zero Fourier Transform. Then, the velocity of every particle is 0. By the Fourier Untransforming Thereom, the velocity is f(R^3), where f is the Busy Beaver function (Zeb)
4. Yes. Consider the nth derivative of f(pi^e), where f is the busy beaver function. It must be an integer if pi^e is rational. But, it is positive and less than 1. Therefore, pi^e is irrational. Using the transcendality-irrationality bridge thereom, pi^e is transcendental.
5. This is a direct consequence of Zeb's conjecture.

How did you get into my blog?
by pythag011, Aug 18, 2008, 1:37 AM
IMO 2593 Problems:

Day 1:

You have 15 minutes.

#1 Prove the Poincare Conjecture.
#2 Is P=NP?
#3 Determine the velocity of every particle in Earth.
#4 Is pi*e a transcendental?
#5 Objectively show that the continuum hypothesis is true or false. Justify your response.
by james4l, Aug 18, 2008, 12:55 AM
Read the problem I just posted.
by pythag011, Aug 17, 2008, 12:38 AM
Peanut butter
by zephyredx, Aug 16, 2008, 10:53 PM

248 shouts

Random Trivial Problems

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by pythag011, Jul 10, 2009, 10:04 AM

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