algebra
polynomial
binomial theorem
number theory
number theory unsolved

Hi,

i have to prove that: if p is a prime number, then a^p+b^p = (a+b)^p modp.

"=" is not equality but congruence.

Any suggestion?

A very good idea is to use Newton's binomial theorem
http://en.wikipedia.org/wiki/Binomial_theorem
You can also use fermat's little theorem
http://en.wikipedia.org/wiki/Fermat's_little_theorem

I won't post the solution because I think it is pretty clear what to de next :)

Daniel

Using Newton's binomial theorem we have $ (a\plus{}b)^p\equal{}a^p\plus{}C^1_pa^{p\minus{}1}b\plus{}...\plus{}C^{p\minus{}1}_pab^{p\minus{}1}\plus{}b^{p}$ and from the fact that $ p$ divides $ C^1_p,C^2_p...C^{p\minus{}1}_p$(for $ p$ prime), results that $ (a\plus{}b)^p \equiv a^p\plus{}b^p(mod p)$.

Fermat's little theorem only shows that this is true when $ a, b$ are integers, but in fact this identity is true formally.

Hello t0rajir0u. Can we also talk about fractions here ? Could u please explain me ?

A solution with Fermat's little theorem.
We have $ a^p\equiv a \mod p$   and also  $ b^p\equiv b \mod p$. Adding these two we get $ a^p \plus{} b^p\equiv a\plus{}b \mod p$.

But $ (a \plus{}b)^p\equiv a\plus{}b \mod p$ so $ a^p \plus{} b^p\equiv (a\plus{}b)^p \mod p$
This approach tho, as t0rajir0u stated holds for positive integers $ a,b$.

[quote="gordon_c"]But what would suppose to mean $ a \equiv b (mod p)$ with $ a,b$ fractions and $ p$ an integer?[/quote]

For any real $ a$, write $ a\equal{}kp\plus{}r$ so that $ k$ is an integer and $ r$ is the smallest non-negative real and say $ r$ is remainder of $ a$ mod $ p$ (I don't know its real name indeed :) ). Then $ a\equal{}b(mod p) \Leftrightarrow$ their remainders mod $ p$ are equal.
P.S. Actually, $ p$ is not necessarily an integer.

stephencheng, this result is false if you define remainders that way.  What I mean is that this is true as a [i]polynomial[/i] identity.  We say that a polynomial is congruent to $ 0 \bmod p$ if all of its coefficients are.

In any case, the appropriate "fractional" definition is as follows:  $ \frac {a}{x} \equiv \frac {b}{y} \bmod p$ if and only if $ p | ay \minus{} bx$.  We should restrict $ p \not | x, y$ or else we run into problems.  Since the inverses of $ x, y$ exist $ \bmod p$, this case actually [i]can[/i] be covered by Fermat's little theorem, but again, it's not what I meant.

I read the fermat's little theorem. So if i use it i have that ->   a^p-1 = 1modp

How we prove that a^p=amodp ???

I don't know, maybe it is dummy question!! But can someone explain it to me?

thanks

Just multiply $ a^{p\minus{}1} \equiv 1(modp)$ with a.

Actually, FTL states[b] that [/b] $ a^p\equiv a\mod p$ for positive integer $ a$ and prime $ p$.

And [b]if[/b] $ (a,p)\equal{}1$ $ \Rightarrow$ $ a^{p\minus{}1}\equiv 1\mod p$


If you're a Mathcounts participant going to the IL State Mathcounts Competition on March 7, please post here!

I'm going, so that's why I made this post. :D GL suma_milli who has chapter on valentines day :D , Stargirl240 whose chapter's is on Feb. 21, and jgjgjmath on the 21st. :D

Er right this is probably a little late.. but I'm going to be wearing a white polo shirt that has mathcounts and a knight on it as a logo.

i'm going to be wearing a striped shirt and jeans....

and i'll be eying ravi... :agent:


For $a;b;c \in R$ and satisfying $x \leq y \leq z$.Let $a;b;c$ satisfying $x \leq a;b;c \leq y$ and $a+b+c=x+y+z$.Prove that:
$a^{2}+b^{2}+c^{2}\leq x^{2}+y^{2}+z^{2}$


We want to prove that:

$\sum_{k=1}^{p-1} \frac{1}{k^3} \equiv 0 \mod{p}$
It's easy to see that $\frac{1}{k^3} \equiv k^{p-4} \mod{p}$ (from Fermat's Little Theorem).
So:
$\sum_{k=1}^{p-1} \frac{1}{k^3} \equiv \sum_{k=1}^{p-1} k^{p-4} \mod{p}$
But:
$\sum_{k=1}^{p-1} k^{p-4} = \sum_{k=1}^{\frac{p-1}{2}} k^{p-4} + (p-k)^{p-4}$
And $k^{p-4} + (p-k)^{p-4}$ is obviously divisible by $p$, because $p-4$ is odd, so the $k^{p-4}$ term disappears and it's also easy to verify that other terms are divisible by $p$...

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 (this simple extends to any power not divisible by $p-1$)


Is it possible to find 100 positive integers less than 25000, making them the sum of the two different from each other?

clarify zee eenglish, please! do you mean "do there exist 100 DISTINCT positive integers less than 25000 such that all parwise sums are different?"?

for example
a[1]=1
a[2]=1
a[3]=1
a[4]=2
a[5]=3
a[6]=5
a[7]=8
a[8]=13
a[9]=21
a[10]=30

you first chose 2 number,then chose another 2,the sum of the first 2 and the sum of the second 2 is always different.

is it clear?

If you construct the sequence greedily, the answer is almost, but not quite:  see the extension [url=http://www.research.att.com/~njas/sequences/b011185.txt]b011185[/url] associated to [url=http://www.research.att.com/~njas/sequences/A011185]A011185[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  This doesn't definitively prove that the answer is "no," of course -- it might be possible to sacrifice a few smaller terms to fit in several larger terms.


In terms of pi, what is the radius of a circle whose circumference is 5?

[hide]
if the circumference is 5 and $C=\pi \cdot d$ then $\frac{5}{\pi}$ is the diameter and half of that must be the radius so the answer is:

$\boxed{\frac{\frac{5}{\pi}}{2}}$[/hide]

[quote="math92"]
$\boxed{\frac{\frac{5}{\pi}}{2}}$[/quote]

Simplifying Math92's answer:

[hide]$\frac{\frac{5}{\pi}}{2}$

$\frac{5}{\pi}*\frac{1}{2}$

$\frac{5}{2\pi}$[/hide]


Ya..I dont think its a right triangle because if you used pythagorean theorm AD is sqr65, but if you used angle bisector theorm, AD is 8..

[quote="now a ranger"]In triangle $ABD$, $BD$ is the hypotenuse, and $AB=4$ and $AC$ is the angel bisector of Angle $A$. $BC=3$, and $CD=6$. Find the length of Angle Bisector $AC$.[/quote]
[hide="we can use"]
Angle bisector theorem[/hide]
[hide="and a piece of pie"]
stewart's[/hide]

[quote="cincodemayo5590"][quote="kohjhsd"]Does the word "hypotenuse" mean it's a right triangle? (I think it doesn't in this problem since then the Angle Bisecor Theorem won't work.)[/quote]

I think that here, it just means "the longest side".

[hide="Hint"]
A man and his dad put a bomb in the sink...  :D [/hide][/quote]
Acutally you dont have to use stewats, you can just use angle bisector theorm (for finding AC), as AC^2+(BC)(CD)=(AB)(AD)

[quote="GlenthemanN"][quote]Acutally you dont have to use stewats, you can just use angle bisector theorm (for finding AC), as AC^2+(BC)(CD)=(AB)(AD)[/quote]
ACtually we cant, as in this problem AC is the midpoint. Angle bisector just concerns the two sides of the triangle and the other sides which are cut in half.

In this example the triangle is ABD, not ABC, thus AC is the bisector, we have side d with measure of 4, side a with a measure of 9 split into m and n with measures 3 and 6 respectively.., we cannot use stewarts to find the length of the bisector until we know the length of side b.

did anyone get an answer?

I got [hide="99 bottles of beer on the wall.."]$\sqrt{14}$[/hide][/quote]
I dont get your reasoning..AC is the angle bisector so we can use that formula and by the fromula AC^2+(BC)(CD)=(AB)(AD), we gegt AC^2+(3)(6)=(4)(8)-by the angle bisector formula AD is 8., so AC^2=32-18, so AC^2=14, so AC=$\sqrt{14}$, you can use stewarts if you want..but thats a little longer

where do you get AC^2+(BC)(CD)=(AB)(AD) from?

its pretty much stewarts..

[quote="GlenthemanN"]you are using stewarts as well as angle bisector..
angle bisector theory doesnt have anything to do with AC^2.. angle bisector states that:
for triangle ABC with D on BC such that AD is the bisector of angle BAC, then $\frac{DB}{AB}= \frac{DC}{AC}$.[/quote]

thats the angle bisector formula..the angle bisector theorm states that the AC^2+(BC)(CD)=(AD)(AB) and that is NOT stewerts.
I mightve gotten the formula and therom thing switched around but I know that works-if you dont believe me try it on anothe problme.

ohh got the two confused! eghgh
but this page http://www.tjhsst.edu/~tmildorf/math/GeometryI.pdf states that wat I posted was the angle bisector theorem

either way im going to search for your forumla now, I do believe you. It just seems like stewarts adjusted as we know stewarts is
$ad^{2}+amn = mb^{2}+nc^{2}$ and you state that $d^{2}+mn =cb$

[quote="GlenthemanN"]ohh got the two confused! eghgh
but this page http://www.tjhsst.edu/~tmildorf/math/GeometryI.pdf states that wat I posted was the angle bisector theorem

either way im going to search for your forumla now[/quote]

Its in Art of Problemsolving Vol. 2 if you have it..
That formula onlny works for when it is the angle bisector, stewarts works for any thing that divdes the opposite side

well I'll have to remember it, thanks for helpin me clear that up :)


calculus
integration
trigonometry
LaTeX
real analysis
real analysis unsolved

Prove that $ \int_0^{\frac{\pi}{4}} arctan \sqrt{\frac{cos2\theta}{2cos^2\theta}} d\theta \equal{} \frac{\pi^2}{24}$

No idea for this problem ? My friend suggests to use power series method, but I don't have any idea how to simplify the series..

Yes, it's really hard integral. Nobody posted solution here (it was a year ago):
http://www.mathlinks.ro/viewtopic.php?t=215976

Can you show the identity

$ \int_{0}^{\pi /4}\cot^{\minus{}1}\sqrt{\frac{\cos 2x}{2\cos^{2}x}}dx\equal{}2\int_{0}^{\pi /4}\tan^{\minus{}1}\sqrt{\frac{\cos 2x}{2\cos^{2}x}}dx$

Even i don t get that. i mean  the identity

$ \int_{0}^{\pi /4}\cot^{\minus{}1}\sqrt{\frac{\cos 2x}{2\cos^{2}x}}dx\equal{}2\int_{0}^{\pi /4}\tan^{\minus{}1}\sqrt{\frac{\cos 2x}{2\cos^{2}x}}dx$

Please can u show me the working :)

Think harder, look at the problem this way,

$ \int_{0}^{\pi /4}\left( 2\tan ^{\minus{}1}\sqrt{\frac{\cos 2x}{2\cos ^{2}x}}\minus{}\cot ^{\minus{}1}\sqrt{\frac{\cos 2x}{2\cos ^{2}x}}\right) dx$

$ \equal{}\int_{0}^{\pi /4}\tan ^{\minus{}1}\frac{\sqrt{2}\cos 3x}{\left( 2\cos 2x\plus{}3\right) \sqrt{\cos 2x}}dx\equal{}?$

@ Indochina

please help me i don get it still

Hi vijaymenon. I have the following:

Using $ \tan^{ - 1} {x} + \tan^{ - 1}{y} = \tan^{ - 1}\left(\frac {x + y}{1 - xy}\right)$ and $ 2\tan^{ - 1} x = \tan^{ - 1}x + \tan^{ - 1} x = \tan^{ - 1}\left(\frac {2x}{1 - x^{2}}\right)$,

$ 2\tan^{ - 1}\sqrt {\frac {\cos 2x}{2cos^{2} x}} - \cot^{ - 1}\sqrt {\frac {\cos 2x}{2\cos^{2}x}}$

$ = \tan^{ - 1}\frac {2\sqrt {\frac {\cos 2x}{2\cos^{2}x}}}{1 - \frac {\cos 2x}{2\cos^{2}x}} - \tan^{ - 1}\sqrt {\frac {2\cos^{2}x}{\cos 2x}}$

$ = \tan^{ - 1}\frac {\sqrt {2\cos 2x}}{\cos x}\times\frac {2\cos^{2}x}{2\cos^{2}x - \cos 2x} - \tan^{ - 1}\sqrt {\frac {2\cos^{2}x}{\cos 2x}}$

$ = \tan^{ - 1}\frac {2\cos x\sqrt {2\cos 2x}}{1} - \tan^{ - 1}\sqrt {\frac {2\cos^{2}x}{\cos 2x}}$

$ = ...$,

[using $ \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left( \frac{x-y}{1+xy}\right)$, multiplying above and below by $ \frac {\sqrt {\cos 2x}}{\sqrt {\cos 2x}}$, using $ \cos 2x = \cos^{2}x - 1$ and $ \cos 3x = 4\cos^{3}x - 3\cos x$]

$ = \tan^{ - 1}\frac {\sqrt {2}\cos x(4\cos^{2}x - 3)}{\sqrt {\cos 2x}(2\cos 2x + 3)}$

$ = \tan^{ - 1}\frac {\sqrt {2}\cos 3x}{\sqrt {\cos 2x}(2\cos 2x + 3)}$

Now (I think) we need to show that

$ \int_{0}^{\pi/4}\, \tan^{ - 1}\left(\frac {\sqrt {2}\cos 3x}{(2\cos 2x + 3)\sqrt {\cos 2x}}\right)\,dx = 0$.

I can write out the 4 lines which should be where the dots ... are, but I guess you don't need them and I'm VERY slow at $ \text{\LaTeX}$.

Hope this helps you.

@ANDREW

should we do it by parts the next step     ??

i deduce it to find $\int_{0}^{\infty }\frac{(1-x^{2})\arctan x^{2}}{(1+4x^{2}+x^{4})}dx$.
the next step?

I calculated it in this way.
$\int_{0}^{\frac{\pi }{2}}\arctan\sqrt{\frac{1+\cos \theta }{\cos \theta }}d\theta=\int_{0}^{\infty }\frac{1}{1+x^{2}}\arctan \sqrt{\frac{1+2x+x^{2}}{2x}}dx=\frac{\pi ^{2}}{4}+\frac{1}{\sqrt{2}}\int_{0}^{\infty }\frac{(1-x)\arctan x}{(1+4x+x^{2})\sqrt{x}}dx=\frac{\pi ^{2}}{4}-\frac{\sqrt{2}}{2\sqrt{3}}\frac{\sqrt{6}-3\sqrt{2}}{24}\pi ^{2}-\frac{\sqrt{2}}{2\sqrt{3}}\frac{\sqrt{6}+3\sqrt{2}}{24}\pi ^{2}=\frac{\pi ^{2}}{6}$
so given definite integral=$\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\arctan \sqrt{\frac{\cos \theta }{1+\cos \theta }}d\theta =\frac{\pi ^{2}}{24}$

Can anybody modify this proof
to show one of the [url=http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R%28x,cos,arccos%29]Coxeter Integrals[/url].

In fact, I deduced $\int_{0}^{\infty }\frac{\arctan x^{2}}{x^{2}+k}=\frac{\pi }{\sqrt{k}}(\arctan (1+\sqrt{2k})-\frac{\pi }{4})$.

[quote="Doe John"]How did you get the equation

$ \int_{0}^{\frac{\pi }{2}}\arctan\sqrt{\frac{1+\cos\theta }{\cos\theta }}d\theta=\int_{0}^{\infty }\frac{1}{1+x^{2}}\arctan\sqrt{\frac{1+2x+x^{2}}{2x}}dx $ ?

Substitution $ \theta=\arctan x $ gives something else.[/quote]

the well-known omnipotant formula! But this is first time when I find it is appled in such case.


This has been pasted in the "High School " division.

Maybe this has been posted before.
"On a remote Norwegian mountain top, there is a huge checkerboard, 1000 squares wide and 1000 squares long, surrounded by steep cliffs to the north, south, east, and west. Each square is marked with an arrow pointing in one of the eight compass directions, so (with some possible exception of some squares on the edges) each square has an arrow pointing to one of its eight nearest neighbors. The arrows on squares sharing an edge differ by at most 45°. A lemming is placed randomly on one of the squares, and it jumps from square to square following the arrows. Prove that the poor creature will eventually plunge from a cliff to its death."

Lemma: There must be a loop or else he will fall off. Proof: There are a finite amount of squares, if at some point he encounters a square he has been on before then he is in a loop. Done.
Now if we were to fill up the board with loops, there would always be some open square (I'm assuming that we are proving that there is some square that kills the lemming... otherwise there is an easy counterexample).
There always is an open square because if you consider the corners, the only way to include a corner makes one out of the two pieces next to it unusable...

I'm pretty sure that we're supposed to prove the statement as written.  Perhaps you could post your counterexample?

Also, assuming that every square leads to a loop does not imply that the the grid is composed of loops.

In response to 2nd line: doesn't it imply though that he never falls off the board?

And for a counterexample: Just put a loops of any size and there is a chance that he starts out on the loop and never falls off... anyway i think i read the question wrong on this part (i.e. its implied for every square...)

[quote="me@home"]In response to 2nd line: doesn't it imply though that he never falls off the board?[/quote]

If you're asking "Doesn't having the checkerboard being entirely composed of loops imply that he never falls off the board?", then the answer is yes, but that's the wrong direction.  What you assumed is "If he never falls off the board, then the board must be composed of loops", and that isn't true in general.

[quote]
And for a counterexample: Just put a loops of any size and there is a chance that he starts out on the loop and never falls off... anyway i think i read the question wrong on this part (i.e. its implied for every square...)[/quote]

Right, so I assume we're trying to prove that there cannot be a loop of any size on the board.

Sorry, I don't see how that isn't true... how could a board keep him on forever and not have a loop? After a finite amount of steps isn't he forced to come upon a square he's already been on???

Oh sure, there has to be a loop... but then you say:

[quote]Now if we were to fill up the board with loops, there would always be some open square [/quote]

What's wrong with that?

I am not going to be completely rigorous, but I am sure the idea works.

We say that "switching" the board is rotating all its arrows 90 degrees (clockwise or counterclockwise). We say that a set of squares is a "sad cycle" if these squares form a cycle after switching the board in some way. Also, it is easy to see that after a switching of the board, the property of "no more than 45 degrees' difference" is conserved.

Suppose that the lemming never falls to his death. Then there must be a cycle. It is easy to see that the cycle cannot "cross" itself, that is, if we joined the centers of adjacent squares in the cycle we would get a non-intersecting broken line. Thus we can speak of squares being "inside" and "outside" the cycle. Now, switch the board so that all the arrows on the cycle now point inside the cycle. It is easy to see that after the switch, no square inside the cycle can point to a square outside the cycle or on the cycle itself. That is, if the lemming is placed inside the cycle and left to jump following the arrows, it will never get out. Then there must be a cycle in here. It is actually a sad cycle in the original board.

Then we switch the board again and look inside the new cycle. By the same method we can find either a cycle or a sad cycle inside it, and so on. We can always find smaller and smaller cycles, unless the last one we find has no interior squares (otherwise we can repeat the process). But again, it is easy to see  ( :P ) that a cycle with no interior squares will have two adjacent squares with angles differing by more than 45 degrees.


LaTeX
algorithm
number theory
relatively prime

salam
$ x^{2}\plus{}y^{2}\minus{}xy \equal{} 727$
Soale asoonie,vali be nazare man ghashange!
[color=red][Moderator Edit: LaTeX Formula corrected][/color]

moadele ye ebarate daraje dovvom bar hasbe $ x$ hast :
$ x^{2}\minus{}xy\plus{}(y^{2}\minus{}727)\equal{}0$ . pas bayad $ \Delta\equal{}z^{2}$ ke $ z$ adade sahihe .
pas : $ y^{2}\minus{}4(y^{2}\minus{}727)\equal{}z^{2}$ ke inam hast $ z^{2}\plus{}3y^{2}\equal{}2908$ . hala maloome ke in moadele javabash too adade sahih motenahie . Dge dar ovordane javabash ba khodetoon :) .

[quote="khashi70"]moadele ye ebarate daraje dovvom bar hasbe $ x$ hast :
$ x^{2}\minus{}xy\plus{}(y^{2}\minus{}727) \equal{} 0$ . pas bayad $ \Delta \equal{} z^{2}$ ke $ z$ adade sahihe .
pas : $ y^{2}\minus{}4(y^{2}\minus{}727) \equal{} z^{2}$ ke inam hast $ z^{2}\plus{}3y^{2}\equal{} 2908$ . hala maloome ke in moadele javabash too adade sahih motenahie . Dge dar ovordane javabash ba khodetoon :) .[/quote]
toolanish nakon  :D  in rahe hal sade tare : moadele ro bebari be $ mod$ $ 5$ bad maloom mishe ke masalan $ x\equal{}5a\plus{}1$ $ y\equal{}5b\plus{}3$ jaygozin ke koni dobare hamin karo bokoni javabesh saritar dar miad  :blush:
hala ino hal konid(shod marathone Diophantine :roll: ) :
$ p^{2}\minus{}p\plus{}1 \equal{} a^{3}$ where $ p$ is prime and $ a$ is natural

are marathone diophantine kheli hal mide na khashi??? :wink:  delam barat tang shode...!!!! in soale $ 2$ balkan 2005ee / soloution :  http://www.mathlinks.ro/resources.php?c=1&cid=19&year=2005&p=224998
[url]

aya adade $ m\in\mathbb N$ vojoood darad be torike moadele $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{abc}\equal{}\frac{m}{a\plus{}b\plus{}c}$ baraye $ a,b,c$ dar adaaade $ \mathbb N$ binahayat javab dashte bashad??[/url]

[quote="sepehr ahmadzadeh"]aya adade $ m\in\mathbb N$ vojoood darad be torike moadele $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{abc}\equal{}\frac{m}{a\plus{}b\plus{}c}$ baraye $ a,b,c$ dar adaaade $ \mathbb N$ binahayat javab dashte bashad??[/quote]
are,$ m\equal{}12$  :wink:

rahe halle man vase in soal,ye meghdar motefavete,va az namosavi estefade kardam!
albate momkene joob dashte bashe:D
hala farda rahamo tozih midam

[quote="sepehr ahmadzadeh"]khob babak plz tozih bede begoo chera be $ 12$ residi??? faghat $ 12$ ?? va chera $ 12$ mishe? tanx
ye soal dg kheli bahale
Let $ a,b,c$ be positive integers such that $ a$ and $ b$ are relatively prime and $ c$ is relatively
prime either to $ a$ or $ b$. Prove that there exist infinitely many triples $ (x,y,z)$ of distinct positive integers such that
$ x^{a}\plus{}y^{b}\equal{} z^{c}$    soale kheli ghashangiee az shortlist $ 1997$[/quote]
$ X$ va $ Y$ ro betore delkhah entekhab konid(tori ke mosavi nabashan va $ (X,Y) \equal{} 1$ [kar az mohkam kari eib nemikone]) va farz konid $ X^{a}\plus{}Y^{b}\equal{} Z$ . chon $ (c,ab) \equal{} 1$ pas adadi manande $ d$ vojood dare ke
$ dab\equiv\minus{}1 (mod$ $ c)$ hala darim : $ Z^{abd}(X^{a}\plus{}Y^{b}) \equal{} Z^{abd\plus{}1}$  dar natije darim $ (X\times Z^{bd})^{a}\plus{}(Y\times Z^{ad})^{b}\equal{} (Z^{{{abd\plus{}1}\over c}})^{c}$ baghiash ham ke dige badihiye  :)

[quote="sepehr ahmadzadeh"]
Let $ a,b,c$ be positive integers such that $ a$ and $ b$ are relatively prime and $ c$ is relatively
prime either to $ a$ or $ b$. Prove that there exist infinitely many triples $ (x,y,z)$ of distinct positive integers such that
$ x^{a}\plus{}y^{b}\equal{} z^{c}$    soale kheli ghashangiee az shortlist $ 1997$[/quote]
aval az hame soalo tarjome mikonim,chon inja forume iranias va dalili nadare be zaboone digei benevisim:
[b]farz konid $ a,b,c$ aadade sahih va mosbati bashand ke $ a,b$ nesbat be ham aval bashando hamchenin $ c$ nesbat be yeki az $ a,b$ aval bashad.sabet konid binahayat setaie $ (x,y,z)$ az aadade sahihe mosbat va motefavet vojood darad ke $ x^{a}\plus{}y^{b}\equal{} z^{c}$[/b]
rahe hallesh ine:
[hide="rahe hal"]
bedoone az dast dadane kolliat farz konid $ (c,a) \equal{} 1$.amma $ (a,b) \equal{} 1$ pas $ (a,bc) \equal{} 1$...
khob hala eddea mikonim ke adadaye mosbat va sahihe $ k,t$ vojood daran ke $ k.a \equal{} 1\plus{}t.bc$...
[hide="esbate eddea"]
bebinid aval az hame chon $ (a,bc) \equal{} 1$ pas $ m,n$ peida mishe ke $ m.a\plus{}n.bc \equal{} 1$...
hala age $ m$ ya $ n$ barabare $ 0$ bashe ke masale halle (ba yekhorde deghat)...
hamchenin chon $ m.a\plus{}n.bc \equal{} 1$ va $ a,b,c$ mosbato sahihano $ m,n$ ham sahihan,pas hatman yekishoon bayad manfi basheo oonyeki mosbat...
khob age $ n$ manfi bashe ke oon moghe $ \minus{}n$ mosbate va darim $ m.a \equal{} 1\plus{}(\minus{}n).bc$ yeni hamoon chizi ke mikhastim.
hamchenin age $ m$ manfi bashe oon moghe ye adade $ l$ ro dar nazar begirid ke $ m\plus{}l.bc > 0$ oon moghe be tasavie zir deghat konid:
$ (m\plus{}l.bc).a\plus{}(n\minus{}l.a).bc \equal{} 1$ hala too in tasavi zaribe $ a$moon mosbat shode...
pas aadade sahiho mosbate $ k,t$ vojood daran ke $ k.a \equal{} 1\plus{}t.bc$
--------------------------------------------------------
[/hide]
khob pas farz konid dashte bashim $ k.a \equal{} 1\plus{}t.bc$ ke $ k,t$ sahih va mosbatan...
khob hala be tasavie zir deghat konid:

$ (r^{c}\minus{}s^{b})^{ka}\equal{} (r^{c}\minus{}s^{b})^{1\plus{}tbc}\equal{} (r^{c}\minus{}s^{b})(r^{c}\minus{}s^{b})^{tbc}$

$ \equal{} r^{c}(r^{c}\minus{}s^{b})^{tbc}\minus{}s^{b}(r^{c}\minus{}s^{b})^{tbc}$

$ \equal{} [r.(r^{c}\minus{}s^{b})^{tb}]^{c}\minus{}[s.(r^{c}\minus{}s^{b})^{tc}]^{b}$

pas darim:

$ [(r^{c}\minus{}s^{b})^{k}]^{a}\plus{}[s.(r^{c}\minus{}s^{b})^{tc}]^{b}\equal{} [r.(r^{c}\minus{}s^{b})^{tb}]^{c}$

khob hala kafie gharar bedim:

$ x \equal{} (r^{c}\minus{}s^{b})^{k}$

$ y \equal{} s.(r^{c}\minus{}s^{b})^{tc}$

$ z \equal{} r.(r^{c}\minus{}s^{b})^{tb}$

yeni darim $ x^{a}\plus{}y^{b}\equal{} z^{c}$...

khob hala faghat ye kare koochik mimoone oonam ineke sabet konim binahayat ta $ x,y,z$e matloob vojood dare...
yeni kafie binahayat ta $ r,s$ peida konim ke $ x,y,z$emoon sahiho mosbat beshano motafavet...
khob vase inkaram kafie $ r,s$ sahiho mosbat bashano $ r^{c}> s^{b}$ bashe...[/hide]

ok doroste :)   khob hala bademoon nemiad ye chandta soal bezarin maham fekr konim hamash man bezaram???hala age diophantine nabood bood che behtar vali tarjihan nazarye adad bashe ta mabahese dg..

in ham ye soal az balkan
x^y - y^x = xy^2 - 19
tamame adade avale x va y ra biabid ke dar moadeleye fogh sedgh konand

albate age ye topic ham vase soale hendese bezanim,kheili khoob mishe :wink:

[quote="Toxic_Frog"]in yeki kheili ghashang nist  :D  : $ \varphi (mn) \equal{}\varphi (m)\plus{}\varphi (n)$[/quote]
khob farz konid masalan $ m\equal{}p_{1}^{a_{1}}...p_{k}^{a_{k}}$

oon moghe darim:

$ \varphi (m)\equal{}p_{1}^{a_{1}\minus{}1}...p_{k}^{a_{k}\minus{}1}(p_{1}\minus{}1)...(p_{k}\minus{}1)$

khob hala adade $ mn$ be forme $ p_{1}^{b_{1}}...p_{k}^{b_{k}}.W$ hastesh ke $ (W,p_{i})\equal{}1$ hamchenin chon $ m\mid mn$ pas bettab $ b_{i}\geq a_{i}$...

khob darim:

$ \varphi (mn)\equal{}p_{1}^{b_{1}\minus{}1}...p_{k}^{b_{k}\minus{}1}(p_{1}\minus{}1)...(p_{k}\minus{}1).W'$

ke khode $ W'$ be forme $ q_{1}^{c_{1}\minus{}1}...q_{t}^{c_{t}\minus{}1}(q_{1}\minus{}1)...(q_{t}\minus{}1)$ hastesh...

khob hala ba moghayese $ \varphi (m),\varphi (mn)$ va tavajoh be rabete $ b_{i}>a_{i}$ naitje mishe ke $ \varphi (m)\mid\varphi (mn)$...

khob hala ba tavajoh be tasavie $ \varphi (mn)\equal{}\varphi (m)\plus{}\varphi (n)$ natije mishe ke $ \varphi (m)\mid\varphi (n)$...

be tarighe moshabeh mishe sabet kard ke $ \varphi (n)\mid\varphi (m)$ pas $ \varphi (m)\equal{}\varphi (n)$...

khob hala ye ghazie maroof hast ke mige age farz konim $ d\equal{}(m,n)$ oon moghe $ \varphi (d).\varphi (mn)\equal{}\varphi (m).\varphi (n).d$

(esbatesham ziad sakht nist,kafie dotarafo be $ mn$ taghsim konido az in nokte estefade konid ke $ \frac{\varphi (m)}{m}\equal{}(1\minus{}\frac{1}{p_{1}})...(1\minus{}\frac{1}{p_{k}})$ va sade konid...)

khob hala ba tavjoh be inke $ \varphi (m)\equal{}\varphi (n)$ natije mishe ke:

$ \varphi (mn)\equal{}\frac{d.\varphi (n)^{2}}{\varphi (d)}$

khob pas moadelamoon be shekle zir dar miad:

$ \frac{d.\varphi (n)^{2}}{\varphi (d)}\equal{}\varphi (m)\plus{}\varphi (n)\equal{}2\varphi (n)$

ke age sadash konim mishe:

$ \frac{d}{\varphi (d)}\equal{}\frac{2}{\varphi (n)}$

bebinid samte raste in tasavi yekhorde koochike,yeni age $ \varphi (n)>2$ bashe oon moghe kasremoon az $ 1$ kamtar misheo oon moghe bayad dashte bashim $ d<\varphi (d)$ ke inam ghalate,cherak $ \varphi (d)$ tedade aadade [u]koochektar[/u] az $ d$e ke nesbat be oon avvalan,pas bettab $ \varphi (d)<d$...

pas bayad dashte bahsim $ \varphi (n)\leq 2$ yeni $ \varphi (n)\equal{}1$ ya $ \varphi (n)\equal{}2$...

amma dashtim:

$ \varphi (n)\equal{}q_{1}^{\alpha_{1}\minus{}1}...q_{s}^{\alpha_{s}\minus{}1}(q_{1}\minus{}1)...(q_{s}\minus{}1)$

ke ba tavvajoh be tasavie $ \varphi (n)\equal{}1\text{or}2$ natije mishe ke ya $ s\equal{}1,q_{1}\equal{}2,\alpha_{1}\equal{}1$ yeni $ n\equal{}2,\varphi (n)\equal{}1$ ya inke $ s\equal{}1,q_{1}\equal{}2,\alpha_{1}\equal{}2$ yeni $ n\equal{}4,\varphi (n)\equal{}2$ ya $ s\equal{}1,q_{1}\equal{}3,\alpha_{1}\equal{}1$ yeni $ n\equal{}3,\varphi (n)\equal{}2$ ya inke $ s\equal{}2,q_{1}\equal{}2,q_{2}\equal{}3,\alpha_{1}\equal{}1,\alpha_{2}\equal{}1$ yeni $ n\equal{}6,\varphi (n)\equal{}2$

pas $ n\in\{2,3,4,6\}$...
ke fek konam $ n\equal{}6$ gheire ghabele ghabool bashe...

[quote="goodgirl_2278"]in ham ye soal az balkan
x^y - y^x = xy^2 - 19
tamame adade avale x va y ra biabid ke dar moadeleye fogh sedgh konand
[/quote]
soorate soal ine: $ x^{y}\minus{}y^{x}\equal{} xy^{2}\minus{}19$,ya ine $ x^{y}\minus{}y^{x}\equal{} (xy)^{2}\minus{}19$ :?:
dar har soorat rahe halleshoon taghriban yekkie,faghat dovomie javab nadare...
rahe halle avvalie ine:
midoonim $ x\mid x^{y},x\mid xy^{2}$ pas $ x\mid y^{x}\minus{}19$ amma chon $ x$ adade avvlae pas tebghe ferma $ x\mid y^{x}\minus{}y$ pas bayad dashte bashim $ x\mid y\minus{}19$ hamchenin be tarighe moshabeh natije mishe ke $ y\mid x\plus{}19$...

khob hala bebinid age $ y\minus{}19\geq 0$ oon moghe khahim dasht:

$ x\mid y\minus{}19\Rightarrow x\leq y\minus{}19\Rightarrow 19\leq y\minus{}x$

az tarafi darim:

$ y\mid x\plus{}19\Rightarrow y\minus{}x\leq 19$...

pas $ y\minus{}x \equal{} 19$ yeni yeki az $ x,y$ zoje va oonyeki farde pas $ x \equal{} 2$ yeni $ y \equal{} 21$ ke tanaghoze chon $ y$ bayad avval bashe...

pas farze $ y\minus{}19\geq 0$emoon ghalate yeni bayad dashte bashim $ y < 19$...

pas darim:

$ x\mid 19\minus{}y\Rightarrow x\leq 19\minus{}y\Rightarrow x < 19$...

pas $ x,y < 19$ khob hala dge baghiash yekhorde jagozarie,aval az hame deghat konid ke $ x \equal{} y$ javab nadare...
hala kafie gharar bedim $ y \equal{} 17,13,11,7,5,3,2$ ke natijash az rabete $ x\mid 19\minus{}y$ ine:

$ (y,x) \equal{} (2,17),(3,2),(5,7),(5,2),(7,3),(7,2),(11,2),(13,3),(13,2),(17,2)$

amma midoonim $ y\mid x\plus{}19$ pas mimoone javabaye zir:

$ (y,x) \equal{} (2,17),(3,2),(7,2)$

ke dar halate $ (y,x) \equal{} (2,17)$ samte chape tasavie hokmemoon manfi mishe...

pas mimoone javabaye zir:

$ (y,x) \equal{} (3,2),(7,2)$ ke jofteshoon javaban...

[quote="sepehr ahmadzadeh"]khob babak plz tozih bede begoo chera be $ 12$ residi??? faghat $ 12$ ?? va chera $ 12$ mishe? tanx [/quote]
shayad rahesh maskhare be nazar biad vali behar hal ye rahe dge:
bara inke begam chera be $ 12$ residam,kamel migam ke chetori fek kardam...
avval az hame biaid az sharre $ c$ khalas shim,man khodam gharar dadam $ c \equal{} a\plus{}b$...
(vali dalili nadare in behtarin rah bashe)
yeni moadelamoon mishe:

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{a\plus{}b}\plus{}\frac{1}{ab(a\plus{}b)}\equal{}\frac{m}{2(a\plus{}b)}$

hala dotarafo dar $ (a\plus{}b)$ zarb konid mishe:

$ \frac{a\plus{}b}{a}\plus{}\frac{a\plus{}b}{b}\plus{}1\plus{}\frac{1}{ab}\equal{}\frac{m}{2}$

$ \Rightarrow 3\plus{}\frac{a^{2}\plus{}b^{2}\plus{}1}{ab}\equal{}\frac{m}{2}$

$ \Rightarrow\frac{a^{2}\plus{}b^{2}\plus{}1\minus{}(\frac{m}{2}\minus{}3)ab}{ab}\equal{} 0$

yeni bayad dashte bashim:

$ a^{2}\plus{}b^{2}\plus{}1\minus{}\frac{m\minus{}6}{2}.ab \equal{} 0$

khob pas kafie binahayat ta $ a,b$e sahih peida konam ke rabete bala bargharar bashe...
khob farz konid $ (a,b)$ too rabete bala sedgh kone,deghat konid ke in rabete ye moadele daraje $ 2$ bar hasbe $ a$ hastesh:

$ a^{2}\minus{}(\frac{m\minus{}6}{2}.b)a\plus{}(b^{2}\plus{}1) \equal{} 0$

ya be ebarati:

$ f(x) \equal{} x^{2}\minus{}(\frac{m\minus{}6}{2}.b)x\plus{}(b^{2}\plus{}1) \equal{} 0$

khob hala in moadele dota rishe dare masalan $ x_{1},x_{2}$...

khob midoonim $ x_{1}\plus{}x_{2}\equal{}\frac{m\minus{}6}{2}b$...

hamchenin hamoontori ke goftim $ a$ yeki az rishehas pas darim:

$ x_{2}\equal{}\frac{m\minus{}6}{2}.b\minus{}a$

yeni dar vaghe $ (b,\frac{m\minus{}6}{2}.b\minus{}a)$ ham oon tasavi ro sefr mikone...
hala dobare hamin karo bokonid yeni dobare ye moadele tashkil bedid ba $ b$e jadidemoon va dobare,va dobare,va dobare...
hamchenin deghat konid ke $ x_{1}.x_{2}\equal{}b^{2}\plus{}1>0$ pas $ x_{1},x_{2}$ ham alamatan,yeni age yekishoon mosbat bashe,oon yeki ham mosbate...
pas ba in ravesh mishe binahayat ta $ a,b$ sakht,amma deghat konid ke $ a,b$ bayad sahih va mosbat bashand pas bayad dashte bashim:

$ 2\mid m\minus{}6$

va $ m\minus{}6 > 0$

masalan $ m \equal{} 12$ ke tasavimoon ya be ebarati tabamoon mishe $ a^{2}\plus{}b^{2}\minus{}3ab\plus{}1 \equal{} 0$

ya masalan $ m \equal{} 18$ ke tasavimoon ya be ebarati tabamoon mishe $ a^{2}\plus{}b^{2}\minus{}9ab\plus{}1 \equal{} 0$...

ya kheili javabaye dge...

khob ba tavajoh be algorithme bala,age $ (a_{n},b_{n},c_{n})$ javab bashan oon moghe $ (a_{n\plus{}1},b_{n\plus{}1},c_{n\plus{}1})$ ham javaban ke dar an:

$ a_{n\plus{}1}\equal{} b_{n}$

$ b_{n\plus{}1}\equal{}\frac{m\minus{}6}{2}.b_{n}\minus{}a_{n}$

$ c_{n\plus{}1}\equal{} a_{n\plus{}1}\plus{}b_{n\plus{}1}$

yeni masalan too halate $ m \equal{} 12$ darim $ b_{n\plus{}1}\equal{} 3b_{n}\minus{}a_{n}$...

dge fek nakonam chizi vase tozih dadan moonde bashe,aha dalile inke oon bala neveshtam $ m \equal{} 12$ be jaye inke benevisam masalan $ m \equal{} 18$ in bood ke ye javabe avvaliei yeni ye $ (a_{0},b_{0},c_{0})$ bayad peida konim ke bad az roo oon badiaro besazim dge,avvalin javabi ke be zehne adam miad ine $ (a,b) \equal{} (1,1)$ pas $ c \equal{} 2$ ke age jagozari konim mishe $ m \equal{} 12$...  :D

salam!
be monasebate voroodam be mathlinks,1 soal mizaram:

(2x+1)^2 + (y-1)^2 = 1386 + 12xy

migam bacheha felan bikhiale in soal beshin!:-/ :(

kochektarin adade tabiie $ t$ ra biabid betorike adade sahihe $ x_{1},x_{2},\dots,x_{t}$ vojood dashte bashad ke $ \sum_{i\equal{}1}^{t}x_{i}^{3}\equal{}2002^{2002}$

chera bikhial shim ?:D
avval moadelaro in shekli benevisin :
$ 4x^{2}\plus{}x(4\minus{}12y)\plus{}(y^{2}\minus{}2y\minus{}1384)\equal{}0$ .
hala ye moadeleye daraje do bar hasbe $ x$ hast pas bayad $ \Delta \equal{} z^{2}$ bashe ke natije mide : $ 16\plus{}144y^{2}\minus{}96y\minus{}16(y^{2}\minus{}2y\minus{}1384) \equal{} z^{2}$
ke age sadash konin mishe : $ 128y^{2}\minus{}64y\plus{}22128\equal{}z^{2}$ , hala maloome ke $ 16$ samte chapo ad mikone pas bayad $ z\equal{}4t$ ke age gharar bedin va sade konin mishe :
$ 8y^{2}\minus{}4y\plus{}1383\equal{}t^{2}$ . hala maloome ke samte chap be shekle $ 4a\plus{}3$ hast amma morabae hich adadi $ 4a\plus{}3$ nist . pas moadele javab nadare :)

Babak ye soali , fekr konam $ m = 12$ too oon soale javabe masalan age $ m = 18$ oonvaght moadelat mishe :
$ a^{2}+b^{2}+1-9ab = 0$ , hala farz kon $ a\geqq$ dar in soorat age $ a$ too in moadele sedgh kone $ \frac{b^{2}+1}{a}$ ham sedgh mikone amma agar $ b < a$ oon moghe in rishe az $ a$ kamtare hala mishe noozool zado az in joor kara bad be in natije miresin ke bayad too ye marhale dashte bashim $ a=b$ . amma age $ a=b$ natije mide :
$ 2a^{2}+1-9a^{2}= 0$ ke maloome risheye tabiyi nadare !!
dar vaghe dar halate kolli agar $ a^{2}+b^{2}+1-mab = 0$ va $ a,b > 0$ oon moghe $ m = 3$ :)

[quote="sepehr ahmadzadeh"]kochektarin adade tabiie $ t$ ra biabid betorike adade sahihe $ x_{1},x_{2},\dots,x_{t}$ vojood dashte bashad ke $ \sum_{i \equal{} 1}^{t}x_{i}^{3}\equal{} 2002^{2002}$[/quote]
biaid be peimane $ 9$ negah konid,samte rast hastesh $ 4$,pas samte chapam bayad $ 4$ bashe...
amma midoonim $ x^{3}\equiv\minus{}1,0,1 (\bmod 9)$
pas haddeaghal $ 4$ ta mokaab bayad ba ham jam she ta be hange $ 9$ adade $ 4$ro tolid kone...

$ 2002^{2002}\equal{} 2002^{2001}(1\plus{}1\plus{}1000\plus{}1000)$

[quote="khashi70"]Babak ye soali , fekr konam $ m = 12$ too oon soale javabe masalan age $ m = 18$ oonvaght moadelat mishe :
$ a^{2}+b^{2}+1-9ab = 0$ , hala farz kon $ a\geqq$ dar in soorat age $ a$ too in moadele sedgh kone $ \frac{b^{2}+1}{a}$ ham sedgh mikone amma agar $ b < a$ oon moghe in rishe az $ a$ kamtare hala mishe noozool zado az in joor kara bad be tanaghoz mirese . pas bayad hatman $ a = b$ ke natije mide :
$ 2a^{2}+1-9a^{2}= 0$ ke maloome risheye tabiyi nadare !!
dar vaghe dar halate kolli agar $ a^{2}+b^{2}+1-mab = 0$ va $ a,b > 0$ oon moghe $ m = 3$ :)[/quote]
emmmmmm,na bebin,avval az hame inke man ye ghalate emlai daram too postam oonam ineke $ \frac{m-6}{2}$ be ezae $ m=18$ mishe $ 6$ na $ 9$...
dar har sorat in chize mohemmi nist,khob bia farz kon $ a<b$ yeni mikhaim javabamoon too har marhale be in shekl bashe $ (a_{n},b_{n},c_{n})$ ke $ a_{n}<b_{n}$
khob bebin age $ a_{n}<b_{n}$ bashe oon moghe $ a_{n+1}<b_{n+1}$ chera ke $ a_{n+1}=b_{n}$ va hamoontori ke khodet gofti $ b_{n+1}=\frac{b_{n}^{2}+1}{a_{n}}$ pas kafie sabet konim:
$ b_{n}<\frac{b_{n}^{2}+1}{a_{n}}$
$ \Rightarrow a_{n}b_{n}<b_{n}^{2}+1$
ke inam be dalile $ a_{n}<b_{n}$ doroste...
pas dar vaghe mishe javabe avvaliaro tori tain kard ke hame javaba $ a$shoon az $ b$shoon kamtar bashe,pas fek nemikonam $ m=18$ moredi dashte bashe...  :maybe:

Khob bebin bayad ye javabe avvalie masalan $ (a_{1},b_{1})$ bashe ke beshe ba rabete bazgashti vasash binahayat ta javab dar ovord .

ow,hagh ba toe...
pas faghat $ m\equal{}12$ sedgh mikone,be hamoon dalili ke gofti,yeni moadele $ a^{2}\plus{}b^{2}\minus{}mab\plus{}1\equal{}0$ be ezae $ m\not\equal{}3$ javabe sahih va mosbat nadare...

[quote="Toxic_Frog"]hala in yeki ro peida konid : kochektarin adade tabiii ke natoonim beforme $ {2^{a}\minus{}2^{b}\over 2^{c}\minus{}2^{d}}$ benvisim (russia 2005)[/quote]
kheyli asoon mishe sabet kard ke $ 1,2,...,10$ ro mishe nevesht . hala sabet mikonim ke $ 11$ ro nemishe nevesht .
avval tavvajoh konin mitoonim farz konim ke yeki az in adada sefran do halat darim $ b \equal{} 0$ ya $ d \equal{} 0$
age $ b \equal{} 0$ natije mide $ 2^{a}\minus{}1 \equal{} 11(2^{c}\minus{}2^{d})$ ke inam natije mide $ d \equal{} 0$ oonyeki halatesham hamintorie , yani $ b \equal{} d \equal{} 0$
pas darim $ 2^{a}\minus{}1 \equal{} 11(2^{c}\minus{}1)$ , hala age $ c > 1$ dar in soorat $ a > 1$ va hala ba peymaneye $ 4$ be tanaghoz mirese
pas $ c \equal{} 1$ va dar natije $ 2^{a}\minus{}1 \equal{} 11$ ke maloome javab nadare :)

[quote="goodgirl_2278"]aghaye khashi70,halle soale ario,joob dare!
akharesh mishe
8y^2-4y+1385=t^2[/quote]
age akharesh injoori ham bashe samte rast $ mod$ $ 9$  $ \in\left\{{\left.{1,4,0,7}\right\}}\right.$   va samte chap  $ mod$ $ 9$  $ \in\left\{{\left.{3,5,8,2}\right\}}\right.$   ke bazam neshoon mide moadele bedoone javabe   :oops:

Are rast migin , hagh ba shomast :) , va mamnoon az sina vase rahe hallesh .
hala ham ke bahse china shode tooye marathone hendese biayn inam ye soale nazarie az china :
hameye dotayihaye tabiyi mesle $ (a,n)$ ro peyda konid ke $ n|(a\plus{}1)^{n}\minus{}a^{n}$ . soale khoobie roosh fekr konin .

[quote="khashi70"]
hameye dotayihaye tabiyi mesle $ (a,n)$ ro peyda konid ke $ n|(a\plus{}1)^{n}\minus{}a^{n}$ . soale khoobie roosh fekr konin .[/quote]
$ n\equal{}1$...

Are doroste babak , hala in yekio hal konin :
hameye dotayihaye tabiyi mesle $ (m,n)$ ro peyda konid ke $ mn\minus{}1|m^{2}\plus{}n^{2}$ :)


function
limit
real analysis
real analysis unsolved

Consider the function $f(x) = \frac{ax^{2}+bx+c}{px^{2}+qx+r}$.If $q^{2}-4pr < 0$, is it always true that $f(x)$ does not have an asymptitic behavior at any point????


function
complex analysis
complex analysis unsolved

Show that every function f which is analytic in a symmetric region G can be written in the form
of f1+if2 where f1 and f2 are analytic functions in G which are real on the real axis.

hope iam not disturbing anyone

thx

This can not be correct as stated above, e.g. $f(x)=i$.

You are right, i agree with u . For this reason i posted this problem but it is found in Ahlfor's book for complex analysis as i wrote it

$f(z)=i$ (constant function) is no problem for this theorem - just let $f_1(z)=0$ and $f_2(z)=1.$

The theorem is true.

Let $f_1(z)=\frac{f(z)+\overline{f(\overline{z})}}2$

And let $f_2(z)=\frac{f(z)-\overline{f(\overline{z})}}{2i}$

The main thing we need to show is that for $f$ holomorphic, then $\overline{f(\overline{z})}$ is also holomorphic.

For $x\in\mathbb{R},\,f_1(x)=\text{Re}(f(x))$ and $f_2(x)=\text{Im}(f(x)).$ Both are real.


Four distinct integers $ a$, $ b$, $ c$ and $ d$ have the property that when added in pairs, the sums 16, 19, 20, 21, 22, and 25 are obtained. What are the four integers?

WE know a+b=16 and c+d=25, and we know the difference between either a and b or c and d is 1, and this must be c and d because they add to an odd number. So, c=12 and d=13. Then we can figure out that the values are a=7, b=9, c=12, and d=13.

You know wuts hilarious?? I followed the instructions with those same numbers and they said I was wrong. In the answer, they didnt put a space  or a comma. unbelievable right?


Support
geometry
MATHCOUNTS
analytic geometry

What do you think about the word mathlete? I remember when I used to participate in math competitions, back more than a lustrum ago (actually, almost two) I used to dislike it. I mean, I didn't think a participant in a math competition should be called that way (the word is too amorphus, at least that is my opinion), nevertheless I have seen that some people here in the forum use it to refer to themselves.

I personally, really dislike it, what about you?

hmmm...personally, i don't care if i'm in a math competition and they call us all mathletes. the word isn't degrading us, isn't it? then again, being called mathletes doesn't mean anything. heck, i'm a mathlete just cause i do math. if somebody called you an athlete, it means you participate in sports, etc.

i know i've used it for lack of a better word but i tend to dislike the term. it might apply more if schools financed and actually paid attention to their bright students and athletic teams. i go to Summerville High where the heart and soul of the school is the football team with the super old coach who won 500 games, among other sports teams, and where it's like pulling teeth to get the school to pay for the amc b. so, i don't feel that "math" and "-lete" should go to together.  (i should have gone to the governer's school of science and mathematics when i had the chance)

Why not? If you compete in math, than that is like an athlete, so call math competitors mathletes.

Dislike. No creativity. Think of something better. Mathlete, psh...

I hate it when ppl compare math to athletics

[quote="joml88"]Hmm...David, you just said "it might apply more if schools financed and actually paid attention to their bright students and [b]athletic[/b] teams" and then went on to condemn the support that sports team get as excessive.  From my perspective, our school (Summerville) gives us adequate support and funding.  I don't know what the majority of schools are like but I'm fairly certain that we are very lucky to be on Summerville's.  It is true that the sports teams, specifically football, get the majority of the support.  But that's because football brings in revenue and people can actually go and have fun watching a football game.

Finally, why on earth would you want to go to the Gov School  :P ?[/quote]

because we have to get sponsors. the school didn't buy the hoodies, a steel corporation did.  and brinson is going to try to get more sponsors so we can more stuff (particullary more aops books).

I really do not care either way, and mathlete is a short (albeit corny) way of calling someone a math competitor.

[quote="filletwho"]Howabout Mather?[/quote]

How about nerd...jk!(Sorry if that offends anybody)

*gasps* noooo... I think you emotionally scarred me for life. JK :D . When it's that obvious, should I say jk? owell. I think nerd is a good term, however. builds self-confidence in your math skills by setting you apart from others. Geek works for me too.

[quote="filletwho"]You're right, we aren't the stereotypical nerd. We're problem solvers! Btw, I'm in track, but I skipped it for math club. :)[/quote]

Im in mathcounts, and im in football and i have practice every tuesday (sameday as mathcounts) and i've only been to one mathcounts meeting, but since only 4 ppl tried out for mathcounts(including me) i got a free entry.

[quote="GoBraves"][quote="plokoon51"]hmmm...nerd applies to some people who are good at math, but most people who are good at math seem to do something else (etc. sports) from personal experience, most nerds are really not that smart. One kid i "know" is a 8th grader who reads books on analytic geometry and such but he isn't any better at math than us freshman.

i wouldn't consider myself a nerd but my friends say i am and i just take that "title" without embarassment. its just a word, like mathlete[/quote]

You dont have to be smart to be a nerd....you just have to act like your smart...and read a bunch of books that nobody in your grade will understand, even if you dont know what it means.[/quote]

Yup, totally agree. Or just act like a smart-ass, regardless of what you read or whatever.

[quote="plokoon51"][quote="filletwho"]You're right, we aren't the stereotypical nerd. We're problem solvers! Btw, I'm in track, but I skipped it for math club. :)[/quote]

problem solvers are not bound to math. problem solvers are bound to problems, and the universe is full of problems. therefore, problem solvers will always have a purpose.

btw, i do track too, but our math club is lame and we meet like once a month...[/quote]

I would try to solve other problems, but looking at the problems of some ppl around me just depresses me. And I was helping out at the middle school math club. ( I was actually hoping they would have a pizza party.

Well, I rest my point. After reading all the post, it seems that there are people that, as me, dislike the term, and the other people that accept it just because of the lack of a better term, but nobody is "proud" to be considered or called a "Mathete". I agree that it is not offensive, it is just that I don't like it.

im proud of being a mathlete. i lift proverbial math weights everyday, and probably put as much time into it as most recognized athletes, so i dont think mathlete is a bad term. unless of course you recognize some of the social stigma it entails...

[quote="tekno10m"]Random fact:

"Mathlete" is a registered trademark of MATHCOUNTS. On all of their publications, they put the (R) symbol after "Mathlete" every time they write it.[/quote]

If we take the definition of "Mathlete" as someone who participates in Mathcounts specifically, I would be proud of being a Mathlete. If we said that all the students who didn't get past the school round were not Mathlets, then I'd be prouder.

I think we should change it to Math1337 :lol:

Well, i think it is ok for me.
Perhaps because i speak Chinese in my daily life so i cannot think of any other more creative words to describe math competitors.

[quote="tekno10m"]Random fact:

"Mathlete" is a registered trademark of MATHCOUNTS. On all of their publications, they put the (R) symbol after "Mathlete" every time they write it.[/quote]
The funny thing is that "mathlete" was a reasonably common term even 25 years ago, before MathCounts existed.

[quote="filletwho"]Howabout AoPSer?[/quote]

that's probably trademarked as well.  :D  :lol:  :P

[quote="diophantus"][quote="miraculouspostmaster"]IT'S CORNY!!!!!!
Mathlete?!?!?!?!

(then again, in my experience, people who love math are kind of corny :D )[/quote]
Hmmm.  Funny place to express that opinion. :blush:[/quote]

I'm included in this group of people, of course.

[quote="miraculouspostmaster"]I'm included in this group of people, of course.[/quote]

who isn't that is on this forum?

hehe...I find paladins idea amusing Math133t...

Its just a term...I frankly don't care.  For as long as I don't call myself or others one I am fine.  Think of Mathlete as an outsiders term for the people of math.

Yet I cannot help but feel how dumb it makes us sound... :(

[quote="Farazy"]
Yet I cannot help but feel how dumb it makes us sound... :([/quote]
That's insignificant.  I guarantee you that if you were not in mathcounts, then by high school you won't even remember the term.  If you were in mathcounts, then you will realize that the term does not collectively identify stupid people.

[quote="Farazy"]hehe...I find paladins idea amusing Math133t...[/quote]

yay, i got support for one of my dumb ideas!

i'm not even on my school's math team and i still get called that, i even try to hide the math books i read from my family, it's like heroin:
1-you do it once and you're all of a sudden an addict
2-you try to hide it underneath your bed
3-you eventually accept what you are and yet do not try to get help (math team, acadec, etc.)

well, the last one's only half true

acadec rules!!

I resent the term "nerd" and it's not because I'm ashamed of loving math. It's because some people tend to generalize people who like math and compete at it as unathletic, antisocial, mouth-breathing, dictionary-reading geeks. And I don't abide by that, and everybody who knows me knows that.

I think that mathlete is a pretty good term, given the structure of today's public schools (at least around where I live). Their athletics programs are really the core of their interaction with the community; how good a school's football team is seems to determine how people in the community view a school. I think that this is terrible, and people ought to look at the academics in a school, but this is how it is. Anything which puts math team in the same group as atheletics (such as the word "mathlete") is helping to get community support, as well as support from the administration of the school. However, I live in Georgia, and I realize that in more educated parts of the world, people probably care more about the academic side of schools.

Well, I also live in Georgia, and I can tell you that besides the few guys that have a Football schoolarship, the rest of the people is accepted here at GATech mainly because of their academic performance more than because of their sport performance, and inside here, people look at schools mainly because of their accademic achivements.

Sports are great, they are the main entretainment of the mass basically everywhere in the world (in the US football, and baseball in the summer that is too hot to play anything else, in the rest of the world, soccer!), but I still dislike a lot the term Mathlete, not because it is bad or good, just because in my opinion is an ugly word!

what a funny  :blush: word!! i have never heard it before


category theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

1.Are mod-R and mod-Mn(R) isomorphic for n>1?
2.Define the center of a category C to be the class of natural transformations of the identity functor 1c to 1c.Let C=R-mod and let c be an element of the center of R. For any M∈ob R-mod let ηm(c) denote the map x--->cx.. x∈M. Show that ηc: M--->ηm(c) is in the center of C and every element of the center of C has this form. Show that c--->η(c) is a bijection and hence that the center of R-mod is a set.
I have show that ηc: M--->ηm(c) is in the center of C,but I cannot proof every element of the center of C has this form.Who can help me ,thanks~~

[b]1.[/b] i don't know that, but these categories are equivalent, i.e. there are functors in both directions, whose composites are isomorphic to the identity functor.

if $M$ is a $R$-module, we make $F(M) : = M^{n}$ to a $M_{n}(R)$-module via matrix multiplication. $F$ yields a functor $R$-mod $\to M_{n}(R)$-mod. let's show that every $M_{n}(R)$-module arises in this form, up to isomorphism.

consider the set $C_{i}$ of matrices, that have nonzero entries only in the $i.$ row. this is a right-ideal of $M_{n}(R)$. let's denote $\delta_{i}j$ the matrix, which is $1$ at $(i,j)$ and elsewhere $0$. then

$\delta_{11}+...+\delta_{nn}= I$
$\delta_{ii}\delta_{jj}= \left\{\begin{array}{cc}0 & i \neq j \\ \delta_{ii}& i = j \end{array}\right.$
$C_{i}= \delta_{ii}M_{n}(R)$

now let $M$ be a $M_{n}(R)$-module. for every $m \in M$ we have $m = \delta_{11}m+...+\delta_{nn}m$, so it follows $M= M_{1}+...+M_{n}$ if we put $M_{i}= \delta_{ii}M_{i}$. via $rm : = (rI)m$, $M$ and thus also $M_{i}$ is also a $R$-module - that's because of $M_{i}= \delta_{ii}M = \delta_{ii}M_{n}(R) M = C_{i}M_{n}(R)$. now, the direct sum $M = M_{1}+...+M_{n}$ is direct, since $\delta_{11}m_{1}+...+\delta_{nn}m_{n}= 0$ implies $\delta_{ii}m_{i}= 0$ after left-multiplication with $\delta_{ii}$.

thus, $M$ is a $R$-module which is the direct sum of the $R$-modules $M_{i}$. but these modules are all isomorphic: if $P_{i}j$ is the permutation matrix according to the transposition $(ij)$, we have $P^{2}=I$ and $P_{ij}C_{i}= C_{j}$, $P_{ij}C_{j}= C_{i}$. then $M_{i}\to M_{j}, m \mapsto P_{ij}m$ and $M_{j}\to M_{i}, m \mapsto P_{ij}m$ are inverse $R$-module-isomorphisms.

we have proven that for every $i$, we have $R$-module-isomorphisms

$\alpha : M_{i}^{n}\to M , x \mapsto \sum_{j=1}^{n}P_{ij}x_{j}$
$\beta : M \to M_{i}^{n}, m \mapsto (P_{ji}\delta_{jj}m)_{j}= (\delta_{ij}m)_{j}$

but $M_{i}^{n}$ is also a $M_{n}(R)$-module, and an easy calculation shows, that $\alpha,\beta$ are even $M_{n}(R)$-module-isomorphisms.

finally, define the functor $G_{i}$ from $M_{n}(R)$-mod to $R$-mod by $M \to C_{i}M$. the above shows $M \cong F(G_{i}(M))$ via $m \mapsto (\delta_{ij}m)_{j}$. naturality is checked easily, so that $1 \cong F \circ G_{i}$. besides $G_{i}(F(M)) = \{ (0,...,m,...,0) : m \in M\}$ is canonical isomorphic to $M$, so that $1 \cong G_{i}\circ F$.

this establishes an equivalence of categories $M_{n}(R)$-mod $\equiv$ $R$-mod for every $i$.

[b]2. [/b] assume $\tau$ is a natural transformation $1_{C}\to 1_{C}$, where $C=R$-mod. expecially, we have a map $\tau_{R}: R \to R$, which is determined by $c : = \tau_{R}(1)$, since then $\tau_{R}(r)=c r$. now, if $M$ is a arbitrary $R$-module, and pick $x \in M$. we want to show $\tau_{M}(x)=c x$. consider the homomorphism $\sigma : R \to M$ which is given by $r \to rx$. naturality of $\tau$ yields $\tau_{M}(\sigma(1))=\sigma(\tau_{R}(1))$, hence the desired result $\tau_{M}(x)=c x$.

this even shows that the center of $R$-mod is isomorphic to $R$. actually if $R$ is not commutative, it is isomorphic to the center of $R$.

How can you proof that c=Tr(1) is in the center of R?

just check what makes $\tau_{R}$ to be an homomorphism


summer program
Mathcamp
Stanford
college
geometry
function
USAMTS

Breaking up the silence.

Dead forum = bad.

Probably not, I might go in high school though...

i thought we already had a forum thingy on this... but w/e.
i loved amsp. by far the best 3 weeks of my life. but.... ive gotta move on. so no .im not coming backkk. prolly a diff. mathcamp or debate/forensics camp & tennis camp. but no amsp '09 :'(
how 'bout u leoxnlin?

I think a dead forum should be normal. Except for AMY talk. But anyways, I'm probably not coming back. Liked it, though.

[quote="dannyhamtx"]MOP

[hide="."]The message is too small. Please make the message longer before submitting.[/hide][/quote]

Hoping to go to MOP as well...

I almost wish I was a 9th grader.

Oh well.

[quote="dannyhamtx"]MOP
[/quote]

I hope so too. Dearly hope so.

[quote="shelly32494"]i thought we already had a forum thingy on this... but w/e.
[/quote]

um... oops?  :oops:

[quote="pythag011"]I almost wish I was a 9th grader. [/quote]

No... you... don't.

You're so lucky.

lol. dont wry leoxnlin it doesnt rly matter.
omg. pythag011 you DO NOT want to be in high school. NONONONONONO. enjoy middle school while u cannnn. srsly.

haha i think david will enjoy high school. he'll be able to take harder classes and not be stuck in stupid middle school classes athat are wayyytoo easy for him xP

im probably not gonna return either.
i dont wanna go to a math camp...
its either violin camp-_- or stanford debate camp
or some random science research internship bleghh

or intern for state government or city hall or something that willl get me good rec letters for my college apps LOL

[quote="1=2"]@pythag011: take middle school to improve your proof writing skills so you can be Evan O'Dorney(mathematics-wise)[/quote]

I'm not doing anything except random problems. Writing down a few every week, but I still fail at writing...

Oh well...

[quote="1=2"]Do you have any nearby friends that are at your caliber in mathematics? If you do, you could practice with them.[/quote]

Yes, I do...

I'm happy that I'm not living in a state where nobody is interested in math...hmm....

EDIT: Also, how do you get good at geometry?

DO geometry!

And if a problem is hard, persist. The solution can be attempted by very simple tactics easier than the angle bisector theorem.

[quote="BOGTRO"][quote="leoxnlin"]Bobby Shen is the eighth-grader best known as the two-time first place Texas MATHCOUNTS individual winner who placed thirteenth and first in the written competition in sixth and seventh grade respectively, but lost to Darryl Wu in seventh grade in an oral match (placing him second). An AoPSer.[/quote]

what is his username?[/quote]

That reminds me; at Six Flags, the Texas nats team started trying to convince me that Bobby Shen's username was stppo<number>, that he doesn't post at all in mathcounts forum, that he posts a lot in Olympiad Algebra forum, and that he got a 15 on a practice AIME.

[quote="BOGTRO"]1913. [/quote]
lol im higher than you but thats damn good for a 6th grader  :P

The chess tournament was fun.  The funniest part was that I ended up beating 3 people who were about 300 points lower than me and losing to 2 who were about 700 higher then me... so all the database registered was that I finished 3 and 2 against people whose ratings averaged 150 points higher than mine...

Congratulations to everyone who did well in the tournament since you are all 10 times better than me...

[quote="BOGTRO"][quote="leoxnlin"]Bobby Shen is the eighth-grader best known as the two-time first place Texas MATHCOUNTS individual winner who placed thirteenth and first in the written competition in sixth and seventh grade respectively, but lost to Darryl Wu in seventh grade in an oral match (placing him second). An AoPSer.[/quote]

what is his username?[/quote]

few things: bobby shen got 1st, 1st, and 1st* in national mathcounts in 7th grade, in written, masters, and cd.

his username is dannyhamtx.

*lost to darryl wu 4-3.

I thought his username was Mewto55555...

Bobby Shen's username is neither Mewto55555 nor dannyhamtx. I'm not sure if he would want his identity to be revealed. I though everyone knew... I guess only Texans in or just out of middle school would know...

I know...

Incidentally, only certain people have been told who he is... dannyhamtx tried to keep it a secret from me a long time ago...

These people include me and the texas team memmers... heres a hint (once you figure out the hint, you will realize who he is)

The texas team members told me that he was sttpo123.

OK... he might not want to have his identity reveaked

surprisingly, a lot knows about it

such as high school students from his region
MC kids
those who use clues he gave us to figure it out
and he hates his identity to be revealed.

and yes, bobby shen is dannyhamtx, mewto55555, and sttpo123 at the same time

[hide="I know his identity. It is"]available via PM.[/hide]

[quote="dannyhamtx"]
and yes, bobby shen is dannyhamtx, mewto55555, and sttpo123 at the same time[/quote]
lolll. he prolly could be xP jkjk.

[quote="dannyhamtx"]bobby shen is dannyhamtx[/quote]

BOY HE'S TALL!

AMSP seems to have gotten worse from year 2006 like there appears to be less lectures and problem solving sessions and no time to ask the instructor questions. Someone that attended is this true?

[quote="leoxnlin"]Bobby Shen is the eighth-grader best known as the two-time first place Texas MATHCOUNTS individual winner who placed thirteenth and first in the written competition in sixth and seventh grade respectively, but lost to Darryl Wu in seventh grade in an oral match (placing him second). An AoPSer.[/quote]
Bobby Shen will go to AMSP this summar, in Dallas location, I believe.


AMC
AIME
probability
number theory
relatively prime
AIME II

Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $ 50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets $ 0$ points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $ A$ beats team $ B$. The probability that team $ A$ finishes with more points than team $ B$ is $ m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.

http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_II_Problems/Problem_10

In the solution presented in the above link,

$ \frac{{1 \choose 5}}{2^5} \frac{{0 \choose 5}\plus{}{1 \choose 5}}{2^5} \equal{} \frac{5}{1024}.$

What is zero choose five? Is it the same as five choose zero? If it is the same, doesn't the numerator have $ 5\times(1\plus{}5)\equal{}30$?

I think all the combinations on that page are written upside-down.

Yeah... but then you have $ 5\times(1\plus{}5)\equal{}30$ on the top that makes the entire numerator when you add up the cases odd which produces a number greater than 1000 as your answer.

Well its given that A beat B, so the probability that A finishes with more points than B can't possibly be that low, because then B has a 1019/1024 chance of having more points than A in the end, which isn't possible.

$ {5 \over 1024}$ is the probability that A beats B IF A wins exactly one more game (so that makes sense).  The final probability is $ {319 \over 512}$, which is not a problem.

Right. I'm not being very clear.

The following line in the solution doesn't make sense to me.
$ \frac{{1 \choose 5}}{2^5} \frac{{0 \choose 5}\plus{}{1 \choose 5}}{2^5} \equal{} \frac{5}{1024}$

Five chose 1 is five. Five chose zero is one. Five chose one is five. How do you get five from this simple multiplication?

if you actually add up all those fractions, you get $ {613 \over 1024}$, not $ {638 \over 1024}$.  Making the 5 a 30 would yield the sum at the bottom of the page.


geometry
analytic geometry
similar triangles

We got an 8. We missed 10 (yay for not reading unit square) and some other number I don't know.

we did 100 million stupid things. we spent literally half our time on that problem. then, with 1 minute left, i FINALLY realized to use similar triangles. then in the last 30 seconds, we did  something stupid on #6 or 7, and changed #9 from 325 to 324.(my mistake) so yeah...you get the point

The idiots at our chapter competition only gave us 15 min. instead of 20 and we only got a 5.  :mad: But, it's ok 'cause our team was #1.

We thought we got like an 8 on the test, but somehow we got a 10. The Chapter Team Round was pretty hard actually.

9.

I think we missed #10...and I was the one that screwed it up. I knew I should have uses analytic geometry w/ calc...the most sure and mindless way to do this stuff.

I was "supposed" to be the best on my team (all of my teammates had less than 40 points individually, I think), but when I saw #10, I remembered I did a problem like that before, but I blanked out and couldn't figure it out, even though I could solve the other 9 problems. Then, my 7th grade teammate did it with a really complicated way, without using similar triangles but the pythagorean theorem. In short, we got a 10. :D

That really sucks abacadaea. :(

[quote="MathAndKnowledge"]Hey xpmath, the fact that the square in #10 is a unit square is irrelevant.  In fact, I solved the ENTIRE problem without knowing that it's a unit square, and I still got the answer...
It asks for a common fraction, so the common fraction is true for all squares.  I did the rigorous method by calling the side of the square s, and I solved everything in terms of s and got 2/5.
However, you are right that it was hard to see.  One of my team members questioned me, asking me how I did it without knowing the square's side length.  Then she saw that it said "unit square" and wanted me to do the entire problem again, but I told her that the fraction stays the same for any side length.[/quote]
We had a few minutes left when one of my team members told me it was a unit square and that I should start over. I was being stupid, so I actually started again, but time ran up before I finished.

lotsofmath, what you said about your team was not very nice. You compete on this as a team. Maybe if it was not for them ( for those 2 questions) , you guys would not get 4th or 5th  place as a team in Houston. Think about how one of your teammates feel when they read this.

Well, on #10, it would matter, I think, since it asked for the area of the shaded region.
We got that one wrong due to my extreme skills, I got 3/5 for some reason.  :cursing:

This is so funny.
The alternates got the same score as the real people.
Except if the alternates were the regulars, I don't think our average would have changed much at all.

lol, there was one kid in our chapter, if that kid had switched places with the last place person on their team(who moved from 7th to 3rd in cd), their entire team PLUS 1 person would havegone, except, of course, then i wouldnt have gone....

[quote="illusiat"]Our team got 10
I basically did nothing except check the number of triangles in one problem!  :D
Our best person did everything :)[/quote]

That is bad.

Do work.

Now.


algebra
polynomial
number theory proposed
number theory

Let $f(z) \in \mathbb{Z}[z]$ be an integer polynomial whose coefficients do not have a common prime divisor, and with $f(0)$ square-free. Assuming the $abc$-conjecture, prove that there are infinitely many $n \in \mathbb{N}$ for which $f(n)$ is square-free. In fact, there exists a constant $c_{f}> 0$ such that $\# \{n \leq x \mid f(n) \text{ square-free}\}\sim c_{f}x$ as $x \to \infty$.

Prove this result unconditionally in the case $\deg{f}\leq 2$. (In fact, the result has been proven unconditionally for $\deg{f}\leq 3$, but the proof is quite ingenious and rather nontrivial).

I now remember that the following case was already posted [by Fedor Petrov?] and discussed on the forum: there are infinitely many $n \in \mathbb{N}$ such that $n(n+1)(n+2)$ is square-free. There was a one-line solution to this problem (this also hints the general, but not so easy, solution in the case $\deg{f}\leq 3$; nothing is known unconditionally, as far as I am concerned, for the case $\deg{f}\geq 4$). In particular, $n^{2}-1 = (n-1)(n+1)$ and $n^{2}\pm n = n(n \pm 1)$ are square-free for infinitely many $n$. By a similar (elementary) argument (and I remember that this, too, was discussed), show that $n^{2}+1$, too, is square-free for infinitely many $n$. More generally, prove that if $c$ is a square-free integer and $a,b$ are any integers, there are infinitely many $n$ for which $an^{2}+bn+c$ is square-free. All this is elementary.

here are two related links:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=squarefree&t=5949
and
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=squarefree&t=27502

by the way in your question I guess  $c_{f}=\prod_{p\in P}\left(1-\frac{n_{p}}{p^{2}}\right)$ where $n_{p}$ represents the number of values in $0,1,2,...,p^{2}-1$ for which $f(x)$ is divisible by $p^{2}$.

Of course, your guess is correct. One easily sees that (with the assumption that $f(0)$ is square-free, so that all summands are nonzero) this product is normally convergent and $>0$. But it's an open problem to prove in general that the number of square-free terms of $f(1),\ldots,f(n)$ is asymptotic to $c_{f}n$. In fact, so far as I'm concerned, it's even an open problem to prove in general that there are infinitely many square-free values of $f(n)$!

The problem I posted is to show that the $abc$-conjecture implies the claim that the number of square-frees among $f(1),\ldots,f(n)$ is asymptotic to $c_{f}n$ for the $c_{f}$ you give. Or, at least, to show that infinitely many among the $f(n)$ are square-free.

Here is a related problem that I just thought of:

1. For every $d$, give an $f$ of degree $d$ such that there are infinitely many square-free terms in the sequence $f(n)$. (not difficult at all)

3. (open to me) How about an irreducible such $f$? (of every degree)


[quote="chess64"]$\lneq$? is that the same as $<$?[/quote]
[quote="seamusoboyle"]Id say its $\leq$, as if you set them all to $\frac{1}{\sqrt{3}}$ you get an equality[/quote]
I don't think so. $\lneq$ means less and not equal, so $\mathbf{chess64}$ is right: $\lneq$ is the same as $<$. But, unfortunately, inequality with $<$ is wrong since for $a=b=c=\frac{1}{\sqrt{3}}$ we have equality.  :maybe:
[b]EDIT:[/b] I solved it.
Homogenise: \[\frac{\sum a^{2}b}{\sum a^{2}}\leqslant \sqrt{\frac{\sum a^{2}}{3}}.\] From weighted $\mathrm{QM-AM}$: \[\mathrm{LHS}\leqslant \sqrt{\frac{\sum a^{2}b^{2}}{\sum a^{2}}}.\] We have to prove: \[\sqrt{\frac{\sum a^{2}b^{2}}{\sum a^{2}}}\leqslant \sqrt{\frac{\sum a^{2}}{3}}.\] Square: \[\frac{\sum a^{2}b^{2}}{\sum a^{2}}\leqslant \frac{\sum a^{2}}{3}.\] Now substitute $x=a^{2}$ etc. It is equivalent to: \[3\sum xy\leqslant\left(\sum x\right)^{2}\] which is obviously true. :D

I tried but failed. Can you post your solution? Maybe it is easier than mine.

[quote=arkhammedos]try with Cauchy-chwartz[/quote]

No problem $\sum_{cyc}ab\cdot a\le \sqrt{a^2b^2+b^2c^2+c^2a^2}\cdot \sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{a^2b^2+b^2c^2+c^2a^2}\le \frac{1}{\sqrt{3}}$
as for all $x,y,z\ge 0$ holds $\frac{x+y+z}{3}\ge \sqrt{\frac{xy+yz+zx}{3}}$

[quote=tchebytchev]let $a,b$and $c$ a positives reals numbers such that :$a^{2}+b^{2}+c^{2}=1$
prove that $a^{2}b+b^{2}c+c^{2}a\lneq \frac{1}{\sqrt{3}}$[/quote]

Because:
$$a^2b+b^2c+c^2a \leq \frac{1}{3}(a+b+c)(a^2+b^2+c^2) \leq \frac{\sqrt{3(a^2+b^2+c^2)}(a^2+b^2+c^2)}{3}=\frac{1}{\sqrt{3}}$$
Equality holds for $a=b=c=\frac{\sqrt{3}}{3}$.


(From Mock AMC 1, #11-15)

Here are the third five problems from the first AMC Mock Contest.  Please post your solutions here so that we can hear different methods of doing each problem.

I have also included the source for each problem.

11. MathFiend is making cookies.  She cuts circular cookies of radius 1 out from a 10 by 14 rectangular sheet of cookie dough so that each cookie occupies its own 2 by 2 square, and each cookie is tangent to its neighboring cookies or the edge of the sheet of cookie dough.  Find, to the nearest square unit, the area of the region that is between the outermost circles and the edges of the rectangle.
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Source: Alabama State Math Contest, Geometry 2002 (with slight rewording)

12. THE BOB is throwing turtles at a dartboard with two regions.  If he hits the smaller region, he gets 15 points; if he hits the larger region, he gets 7 points.  What is the largest score that Bob cannot get?
(A) 67
(B) 79
(C) 83
(D) 89
(E) 97

Source: K. Hon, a guy at my school

13. For how many positive, integral values of n is n^3 - 8n:^2: + 20n  13 a prime number?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 or more

Source: AMC 12p 2002

14. How many zeroes are at the end of (2003!)/(1003!)?
(A) 100
(B) 203
(C) 250
(D) 272
(E) none of these

Source: K. Hon, a guy at my school

15. Solve 2^(1-3x)  4 = a for x.  Then x =
(A) (5-log_2 a)/3
(B) (log_2 (a+2))/3
(C) (1 + log_2 (a+4))/3
(D) (1  log_2 (a+4))/3
(E) 1-log_2 (a+4):^3:

Source: Alabama State Math Contest, Algebra II/Trig 1999

11) Consider a unit circle inside a square with side length 2. Each corner outside the circle has area = (4-:pi:)/4 = 1 - :pi:/4.
In the dough there are 44 such pieces. Total area = 44 - 11:pi:. [b]C[/b].

12) (15-1)(7-1) - 1 = 83. [b]C[/b].

13) This can be factorised as (n-1)(n^2 - 7n + 13). n=1 gives 0, no. n=2 gives 3, yes. n=3 gives 2, yes. n=4 gives 3, yes. Any bigger n has both brackets bigger than 1. [b]C[/b].

14) 2003! ends in [2003/5] + [2003/25] ... = 499 zeroes. 1003! ends in 249 zeroes. [b]C[/b].

15) 2^(1-3x) = a+4, 1-3x = log_2(a+4), x = (1-log_2(a+4))/3. [b]D[/b].

for 14, pleast at least mention there will be enough factors of 2 to go around

for number 14, there are enough factors of 2 to go around
:)

Actually, the answer to number 11 is D (10) because while there are 22 such pieces, four of them have three corners cut off in the square by the circle (the squares on the corners are the ones I'm talking about) and the other 18 have two corners, giving a total of 12+36=48 corners, so you have 12(4-pi)=about 10 as the answer.

There are only 16 with two corners (2(10/2-2)+2(14/2-2)=16), so I think TripleM's answer was correct.

Thanks for posting the problems.

Well, TripleM was right, I just went too fast and made a stupid mistake counting the squares on number 11. . .I concede defeat   :oops:

can anyone explain #12 to me..?
i dont understand how you get (15-1)(7-1)-1

Umm.. for Number 11,  the question says "the area of the region that is between the [b]outermost[/b] circles and the [b]edges[/b] of the rectangle."

doesnt that mean that you  are subtracting

( 3 ( 4-pi) from the four corners + 3 (4-pi) from the sides with length 10 and 5(4-pi) from the sides with length 14? ) from 44??

that leaves you with 11 (4-pi) which is D...

I mean to say that I inferred from the question the area that the unused portion from the middle should be discarded.....

sorry if I am reading things wrongly


calculus
integration
induction
real analysis
real analysis unsolved

Here is a very very nice problem: we all know (do we?) that if $f$ is continuous on a compact interval and all integrals of $x^n f(x)$ are 0 for all $n$, then $f$ is the zero function. Does the conclusion remain true if $f$ is defined on $(0,\infty)$ and all integrals of $x^n f(x)$ (from 0 to $\infty$ ) are 0?  ;) .
   For Moubinool: please, don't post the answer.

We will construct $f$ by induction and such that $f$ vanish at each positive integer. It suffices to impose that for each $n \geq 1$, $\int_{[0;n]} f(t) t^k dt = 0$ for any $1 \leq k \leq n$ and $\sup_{[n: n+1]} |f(t) t^k| < \frac{1}{n}$ for any $1 \leq k \leq n$ (for the existence of the integrale).
Suppose that $f$ is already constructed on $[0;n]$. Then to construct $f$ on $[n;n+1]$ it suffices that:
1/$f(n)=f(n+1)=0$
2/ $\int_{n}^{n+1} f(t) t^k dt = 0$ for every $1 \leq k \leq n$
3/ $\int_{n}^{n+1} f(t) t^{n+1} dt = - \int_{0}^{n} f(t) t^{n+1} dt$
4/ $\sup_{[n: n+1]} |f(t)| < C$

Hence $f$ must be not too big and in an intersection of a finite number of closed hyperplanes of $C([n;n+1],\mathbb{R})$, which is possible.


A 45.0kg boy walks along a 3.00m long wooden plank of mass 20.0kg that overhangs partially completed deck by 0.75m. How Far from the deck edge can the boy walk before the plank tips?

Don't Know the Right equation to use

I used r1*F1=r2*F2

Why this is wrong?
Thank you, please respond ASAP-for test

Yes this is the right equation, if it is used in the right way.

You can suppose the plank will rotate about the edge of the deck, then the maximal distance of the boy from the edge is such that the torques are in equilibrium, that is, r1*F1=r2*F2. With the correct r1,r2,F1,F2 of course!

i use -9.8m/s^2 for g.  i make it positive for the weight of the board thats opposite the folcrum to the overhang.

$\sum F = 0 = (45kg)(-9.8m/s^2)(d) + (\frac{0.75}{3.00})(20kg)(\frac{.75}{2}m)(-9.8m/s^2) + (\frac{3.00-0.75}{3.00})(20kg)(\frac{3.00 - 0.75}{2}m)(9.8m/s^2)$

$d = 0.2916666666666... = 0.292$

the fractions w/o units are to use just a fraction of the boards weight (i think the technical thing to do would be write m/m)

the fractions when i divide by 2 are to get the distance of the centers of mass of the board section


So, how did the other USAMO-taking Ohioans do?

I think/hope that I got 1, 2, and 5, though my solution to 4 was wrong.

I think my number 5 is pretty rigorous, and 4 being around a 4... but number 1 has a major flaw and will probably only get a 2 or 1.

no one made mop

i must make mop next year

you didn't
tenniskid didn't
carlos didn't

the blue mop list has been released v1a h0n0r4bl3 mentions.

black mop also released

so...

tenniskid = only freshman

didn't make mop
no oh mopper


algebra
polynomial
induction
Recursive Sequences

Let $P(x)$ be a nonzero polynomial with integer coefficients. Let $a_{0}=0$ and for $i \ge 0$ define $a_{i+1}=P(a_{i})$. Show that $\gcd ( a_{m}, a_{n})=a_{ \gcd (m, n)}$ for all $m, n \in \mathbb{N}$.

Induction after $ \max(m,n)$. In fact we can WLOG $ m>n$, and we will do induction after $ m$. Suppose $ m\equal{}1$. Then the result is obvious $ \gcd(a_1,a_0) \equal{} a_1 \equal{} a_{\gcd(1,0)}$.

Suppose that we have proved for all pairs $ (m',n')$ such that $ m>m'>n'$. Then
\[ \gcd(a_m,a_n) \equal{} \gcd(P^{(m\minus{}n)}(a_n),a_n) \equal{} \gcd(P^{(m\minus{}n)}(0),a_n)\equal{} \gcd(a_{m\minus{}n},a_n),\] which by induction is $ a_{\gcd(m\minus{}n,n)}$, which obviously equals $ a_{\gcd(m,n)}$.

sorry, but who can explain for me why $ gcd(P^{(m \minus{} n)}(a_n);a_n) \equal{} gcd(P^{(m \minus{} n)}(0);a_n)$ (and I think that because $ a_0 \equal{} 0$ so $ P^{(m \minus{} n)}(0) \equal{} 0$) is it wrong  :blush:  thanks

$ P^{m\minus{}n}(0)$ certainly need not equal 0. The reason is that $ a_n\equal{}a_n\minus{}0|P(a_n)\minus{}P(0)|P^{(2)}(a_n)\minus{}P^{(2)}(0)|\cdots|P^{(m\minus{}n)}(a_n)\minus{}P^{m\minus{}n}(0)$, so that if $ d|a_n$ and $ d|P^{(m\minus{}n)}(a_n)$ then $ d|P^{(m\minus{}n)}(0)$, and similarly, if $ d|a_n$ and $ d|P^{(m\minus{}n)}(0)$ then $ d|P^{(m\minus{}n)}(a_n)$.


Let's play an anagram game!
I'll start.  The topic for right now is [color=green][size=200]Animals![/size][/color]

Rules:
1) Scramble a word that is topic related.
2) No inappropriate words.
3) No sentences in the first round.  (So that it will be easier for more people to play)
4) Person who gets it right first gets to make up next word.
5) Make it so that the word is in the color [color=red]RED![/color] :D
[size=150][color=red]6) Have fun![/color][/size]

[color=red]1st Word:
thpenlae[/color]

Oh yeah, good luck!

[quote="probability1.01"]elephant

next: thibula[/quote]

Good job!  Hmm...that one is kinda tough.  Is it a certain kind of animal?  Like Polar Bear or something like that?

cadillac

delorean

[quote="junggi"]toyota

ladlacic[/quote]

dude I do not think "toyota" is correct

the best car thingy ever.(hyundai)

aki(lol easy)

You spelled it wrong.

"Hummur"

AsAnAnt


Let $C_1,C_2$ be two secant circles in the points $A,B$. Let $M$ be a mobile point which belongs to the exterior common tangent $t$ for which the point $A$ is nearer of line $t$ than the point $B$. Denote the second intersections $X$, $Y$ of the line $MB$ with the circles $C_1$, $C_2$ respectively. Construct (ascertain) the positions of the point $M\in t$ so that there is the relation $MX\cdot MY=MB^2.$

Nice problem.

If R and S are the common points to t and c1 and c2 respectively, it holds that MX*MB = MR^2 and MB*MY = MS^2. Combining those two relation and the requested one we have that we must determione M so that MB^2 = MR*MS. The tanget at B to the circumcircle c(RSB) intersect t at the requested point M.

[color=red]Thanks, Sprmnt21 ![/color] I am gladly you say this problem is nicely.[hide="Variations on the same subject."]
Let $C_1$, $C_2$ be two secant circles in the points $A$, $B$. Let $t$ be a external common tangent in the points $C\in C_1$, $D\in C_2$ so that it is nearer of the point $A$ than the point $B$. Let $M_1\in t$, $M_2\in t$ be two points for which:
$\blacktriangleright$ The line $M_1A$ is tangent to the circumcircle of the triangle $ACD$ and meets for the second time the circles $C_1$, $C_2$ in the points $U$, $V$ respectively;
$\blacktriangleright$ The line $M_2B$ meets for the second time the circles $C_1$, $C_2$ in the points $X$, $Y$ respectively such that $M_2X\cdot M_2Y=M_2B^2$.
[b]Prove that:[/b]
$\blacksquare\ 1^{\circ}.\ M_1U\cdot M_1V=M_1A^2$;
The line $M_2B$ is tangent to the circumcircle of the triangle $BCD$.
$\blacksquare\ 2^{\circ}$. The point $M\in M_1A\cap M_2B$ belongs to the bisector of the segment $[AB]$.[/hide]


A box with a square base of side (x) is three times taller than it is wide. Express the volume (V) of the box as a function of (x)

V = ????

oooh, is that it, how silllllly i am. Thank you so much


algebra
function
domain
polynomial
abstract algebra
calculus
integration

Supposed $ I$ is an ideal of $ Z[x]$ that is generated by $ x^2\plus{}1$ and $ 3$. Prove or disprove, whether $ I$ is a prime ideal or not. Thank you for the solution before.

We can notice that $ \mathbb{Z}[x] / I \approx \mathbb{Z}_3[i]$ ($ i^2 \equal{} \minus{}1$). Since $ \mathbb{Z}_3[i]$ is a domain, $ I$ is a prime ideal.

Why can we notice that $ Z[x]/I$ is isomorphic to $ Z_3[i]$. I really really sorry but I don't understand about this. Please give me a compelete explanation. Thank you for the solution before.
NB : I would like also to ask the definition of $ Z_3[i]$ and how we can form $ Z[x]/I$

$ \mathbb{Z}_3[\imath]$ is the set of polynomials in $ \imath$ with coefficients in $ \mathbb{Z}_3$.  This ring is in fact finite and has six elements $ 0 \plus{} 0i, 0 \plus{} i, 0 \plus{} 2i, 1 \plus{} 0i, 1 \plus{} i, 1 \plus{} 2i, 2 \plus{} 0i, 2 \plus{} i, 2 \plus{} 2i$.  The isomorphism takes $ a_0 \plus{} a_1 x \plus{} a_2 x^2 \plus{} ...$ to $ \overline{a_0} \plus{} \overline{a_1} \imath \plus{} \overline{a_2} \imath^2 \plus{} ... \in \mathbb{Z}_3[\imath]$ where $ \overline{a_n}$ denotes reduction modulo $ 3$; this map has kernel $ I$.

[b]Proposition:[/b]  $ R/I \text{ is an integral domain} \Leftrightarrow I \text{ is a prime ideal}$.

[i]Proof.[/i]  Suppose $ R/I$ has a zero divisor $ a$ such that $ ab \equal{} 0$.  Then $ ab \in I$ but $ a, b \not \in I$, so $ I$ is not prime, and the converse also holds.  In other words, the condition for $ R/I$ to be an integral domain is exactly equivalent to the condition that $ I$ is a prime ideal.

Sorry, but how can we conclude that $ Z_3[i]$ is principal ? I have proved that $ Z_3[i]$ is an integral domain. Please anyone who can help me and thank you for the solution before.

$ I$ isn't principal; what are you asking?  Once you've proven that $ \mathbb{Z}_3[\imath]$ is an integral domain, you are done.

$ Z_3[i]$ has $ 6$ elements, then $ R/I$ has $ 6$ element also so that $ R/I$ is isomorphic to $ Z_3[i]$. May I know what are the elements of $ R/I$ ? Then I can be sure that $ R/I$ is isomorphic to $ Z_3[i]$. Thank you for your answer before  :lol: .

The inverse of the isomorphism takes $ a_0 \plus{} a_1 i$ to $ a_0 \plus{} a_1 x \plus{} I$ where $ a_0, a_1 \in \{ 0, 1, 2 \}$.  Hence the elements of $ R/I$ are $ 0 \plus{} 0x \plus{} I, 0 \plus{} x \plus{} I, 0 \plus{} 2x \plus{} I, 1 \plus{} 0x \plus{} I, 1 \plus{} x \plus{} I, 1 \plus{} 2x \plus{} I, 2 \plus{} 0x \plus{} I, 2 \plus{} x \plus{} I, 2 \plus{} 2x \plus{} I$.

Six $ \neq$ nine  :lol:

But for the proof itself, it would be much more comfortable if instead of effectively computing the field with nine elements, one would just show that $ x^2\plus{}1$ is prime in $ \mathbb Z/3 \mathbb Z [x]$.

Okay, but I would like to ask, how to get these different elements of $ R/I$ as stated in the last two posts ? Thank you in advance.


Hey everyone,

Just a heads up that the 2007 Bellevue BATH will take place on May 19, 2007. It'll be the best competition of the year! Don't forget to come :D. The information sheet is attached.

let me correct that; he means best after crazy Eastlake's contest... :)

Amir is delusional--BATH is by far the best competition ever created on the face of the planet. Heck, I'm flying 3000 miles to come!

I wrote some hard questions for this.   :D

[quote="xxreddevilzxx"]did this happen last yr? how come ive never heard of this (until now)?[/quote]

Dude, where have you been? Lol. TJ even came last year.

BTW, if anyone thinks they will be coming, send me an email/PM to let me know approximately how many teams they will send. It's not a final registration or anything, we just want to get an idea.

[quote="munkyshi22"]Ninja Pirate Zombie Robot...
Is that KOL?
 :ninja:[/quote]

Eh? That was QC's idea.

Remember, BATH registration due in about a week and a half!


$ \phi n$ l $ n\minus{}1$ Show that n has at least 5 distinct prime factors.

does $ \phi n$ mean the number of natural numbers less than $ n$ and coprime to $ n$? :maybe:
if yes then take $ n\equal{}p$ a prime number,then $ \phi n\equal{}\phi p\equal{}p\minus{}1$ so $ \phi n\mid n\minus{}1$ but $ n$ doesn't have 5 distinct prime divisors... :huh:

[quote="BaBaK Ghalebi"]does $ \phi n$ mean the number of natural numbers less than $ n$ and coprime to $ n$? [/quote]

YES
and n is composite.(I missed that)

Hey no one???
At least show that there are 4 distict prime factors.

I will show that there are at least 3 distinct prime factors, and I am sure someone with more talent than myself can punch in the rest.

First note $ n$ must be squarefree, for if $ n\equal{}p^kn_0$ with $ p$ prime, $ k\ge2$ and $ (p,n_0)\equal{}1$, then $ \phi (n)\equal{}p^{k\minus{}1}(p\minus{}1) \phi (n_0)$, so that $ p| \phi (n)$ and therefore $ p|n\minus{}1$ by transitivity, a contradiction since $ p|n$.

Now write $ n\equal{}pq$ for distinct primes $ p,q$. By assumption, $ n\minus{}1 \equiv 0 \bmod \phi (n)$, i.e., $ pq\minus{}1 \equiv 0 \bmod (p\minus{}1)(q\minus{}1)$ so that $ pq\minus{}p\minus{}q\plus{}1 \equiv 2\minus{}p\minus{}q \bmod (p\minus{}1)(q\minus{}1)$ and thus $ (p\minus{}1)\plus{}(q\minus{}1) \equiv 0 \bmod (p\minus{}1)(q\minus{}1)$, so that $ q\minus{}1|p\minus{}1$ and $ p\minus{}1|q\minus{}1$, forcing $ p\equal{}q$, a contradiction.

There is some simple arithmetic/modular reduction to be filled in here, but I wanted to keep it short.

my solution is very long :D
 suppose that $ n \equal{} p_1p_2...p_k , p_i\in P$,STEVESHANK proved that $ p_1,p_2,...,p_k$ are

distinct,and $ k\ge3$

now,I will prove $ k \equal{} 3,4$ is impossible.and so $ n$ has at least $ 5$ distinct prime factors...

CASE(1) , $ k \equal{} 3$:

$ n \equal{} p_1p_2p_3 , p_1 < p_2 < p_3$

$ \phi(n)|n \minus{} 1 \Rightarrow \\
(p_1 \minus{} 1)(p_2 \minus{} 1)(p_3 \minus{} 1)|p_1p_2p_3 \minus{} 1 (I)$

let $ p_1 \minus{} 1 \equal{} a_1 , p_2 \minus{} 1 \equal{} a_2 , p_3 \minus{} 1 \equal{} a_3$ then:

$ (p_1 \minus{} 1)(p_2 \minus{} 1)(p_3 \minus{} 1)|p_1p_2p_3 \minus{} 1 \Rightarrow a_1a_2a_3|a_1 \plus{} a_2 \plus{} a_3 \plus{} a_1a_2 \plus{} a_1a_3 \plus{} a_2a_3 \\
 \\
\Rightarrow a_1a_2a_3\le a_1 \plus{} a_2 \plus{} a_3 \plus{} a_1a_2 \plus{} a_1a_3\plus{}a_2a_3\Rightarrow \\
 \\
\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \frac {1}{a_3} \plus{} \frac {1}{a_1a_2} \plus{} \frac {1}{a_2a_3} \plus{} \frac {1}{a_1a_3}\ge1 (II)$

obviously $ (p_1 \minus{} 1)(p_2 \minus{} 1)(p_3 \minus{} 1)$ is even,so from $ (I)$ we get that $ p_1p_2p_3$ is odd,and
therefore $ p_1\ge3$

if $ p_1 > 3$ then $ p_1\ge5 , p_2\ge7 ,p_3\ge11$

so $ \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \frac {1}{a_3} \plus{} \frac {1}{a_1a_2} \plus{} \frac {1}{a_2a_3} \plus{} \frac {1}{a_1a_3} \le\frac {1}{4} \plus{} \frac {1}{6} \plus{} \frac {1}{10} \plus{} \frac {1}{4\cdot6} \plus{} \frac {1}{4\cdot10} \plus{} \frac {1}{6\cdot10} \equal{} \frac {144}{240} < 1$ which contradict $ (II)$

therefore $ p_1 \equal{} 3$

so $ p_1p_2p_3 \minus{} 1$ is not divisible by 3,and form $ (I)$ we get that $ (p_2 \minus{} 1)(p_3 \minus{} 1)$ is not divisible

by 3,therefore $ p_2\equiv p_3\equiv 2 (\mod 3)$ and $ p_2\ge5$

if $ p_2 > 5$ then $ p_2\ge11$ and $ p_3\ge17$ then:

$ \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \frac {1}{a_3} \plus{} \frac {1}{a_1a_2} \plus{} \frac {1}{a_2a_3} \plus{} \frac {1}{a_1a_3} \le \frac {1}{2} \plus{} \frac {1}{10} \plus{} \frac {1}{16} \plus{} \frac {1}{2\cdot10} \plus{} \frac {1}{2\cdot16} \plus{} \frac {1}{10\cdot16} \equal{} \frac {240}{360} < 1$
which contradict $ (II)$

therefore $ p_2 \equal{} 5$ and from $ (I)$ we have:

$ 8(p_3 \minus{} 1)|15p_3 \minus{} 1 \Rightarrow 8(p_3 \minus{} 1)|8(15p_3 \minus{} 1) \minus{} 15(8p_3 \minus{} 8) \Rightarrow 8(p_3 \minus{} 1)|14\times 8\Rightarrow p_3 \minus{} 1|14$

which is impossible,since $ p_3\ge11$

therefore $ k \equal{} 3$ is impossible.
case $ k \equal{} 4$ is extermly similar with this case.if you want,I will post it later :D

Your method could be improved by simple observation, that if $ \varphi(n) | n \minus{} 1$, and $ n$ is composite number, then $ \varphi(n)$ is a proper divisor of $ n \minus{} 1$  :)
[color=blue]/Edited: It's wrong, I see it now.. :| Thanks for explanation, and sorry for bothering  :oops:  [/color]
For elimination of small numbers which satisfies estimation one can use Euler's Theorem to show, that $ n$ is a Carmichael Number.

By the way the existence of composite $ n$ such that $ \varphi(n) | n \minus{} 1$ is an open problem.

[quote="ifk"]Your method could be improved by simple observation, that if $ \varphi(n) | n \minus{} 1$, and $ n$ is composite number, then $ \varphi(n)$ is a proper divisor of $ n \minus{} 1$  :)
[/quote]
I dont think so...actually in my solution,$ (II)$ is equivalent with $ 2\phi(n)\le n \minus{} 1$ , not $ \phi(n)\le n \minus{} 1$
but if one prove that $ \phi(n) \equal{} \frac {n \minus{} 1}{2}$ is impossible then my method can be improved :)


For positive $ a,$ $ b$ and $ c$ such that $ abc\equal{}1$ prove that:
1) \[ \frac {a^2\plus{}b^2}{c^2\plus{}a\plus{}b} \plus{}\frac {b^2\plus{}c^2}{a^2\plus{}b\plus{}c} \plus{} \frac{a^2\plus{}c^2}{b^2\plus{}a\plus{}c} \geq 2\]
2) \[ \frac {a^2\plus{}b^2}{c^2\plus{}a\plus{}b} \plus{}\frac {b^2\plus{}c^2}{a^2\plus{}b\plus{}c} \plus{} \frac{a^2\plus{}c^2}{b^2\plus{}a\plus{}c} \geq 2 \plus{} \frac {2(a^2\plus{}b^2\plus{}c^2\minus{}a\minus{}b\minus{}c)}{(a^2\plus{}b^2\plus{}c^2)^2} \]

1) $ \sum_{\text{cyc}}\frac{a^2\plus{}b^2}{c^2\plus{}a\plus{}b}\geq \frac{\left(\sum_{\text{cyc}}(a\plus{}b)\right)^2}{2(a^2\plus{}b^2\plus{}c^2\plus{}2(a\plus{}b\plus{}c))}\geq 2\iff ab\plus{}bc\plus{}ca\geq a\plus{}b\plus{}c$

[quote="boxedexe"]1) $ \sum_{\text{cyc}}\frac {a^2 \plus{} b^2}{c^2 \plus{} a \plus{} b}\geq \frac {\left(\sum_{\text{cyc}}(a \plus{} b)\right)^2}{2(a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(a \plus{} b \plus{} c))}\geq 2\iff ab \plus{} bc \plus{} ca\geq a \plus{} b \plus{} c$[/quote]
But $ ab \plus{} bc \plus{} ca\geq a \plus{} b \plus{} c$ is wrong. :wink:

I am [i]really[/i] stupid when I am sick, that is, I'll go rest.  :blush: Conterexample is $ (a,b,c)\equal{}(\frac{1}{3},6,\frac{1}{2})$.


According to what people have said on the ARML forum, congratulations to Eastern MA for 6th overall and Vlad Firoiu for 2nd individually.

yay vlad! i only got a 4 :( but I wasn't on A so its okay lol. vlad is scary.


function
real analysis
real analysis unsolved

Let $ f$ be a function from $ R^\plus{}$ to $ R$, with the property that for every $ x>0$, the function $ f(nx),n\equal{}1,2,\cdots$ is monotonic increasing.
a. If $ I\equal{}[0,1]$ and the function is continuous on $ I$, is the function is monotonic increasing function on the interval $ R^\plus{}$ ?
b. If $ I\equal{}Q^\plus{}$ and the function is continuous on $ I$, is the function is monotonic increasing function on the interval $ R^\plus{}$ ?


Hi friends,

Some of my friends appeared in the interview. They were asked this problem...

Suppose ur C-compiler provides signed integers of size 32-bits. The max possible number using 32-bits is [b]2^32 - 1[/b]. Note that this number exceeds 4 - million.

Well, we provide u with exactly 4 - million numbers in a file or suppose in an array. Obviously some numbers r missing. Find any one of them with an efficient algorithm.
The 4 - million numbers range from [b]0 to 2^32 - 1[/b]. They r selected at random. No number is repeated. Kindly note, find any one of the missing numbers.

Okay then bye...

If you can afford to use 4 million + 1 bits, check all of the numbers from 0 to 4000000:

[code]for i := 1 to 4000000 do
	if input[i] <= 4000000 then
		bits[input[i]] := false;
for i := 0 to 4000000 do
	if bits[i] = true then
		{the result is i}
[/code]

Thanx 4 ur solution bus and good to see u back... But ur solution is probably using a lot of extra space.

Further I came to know about this dangerous thing, [b]which I have never heard of[/b], that the most optimal solution is thru ---- MULTI LEVEL HASH BUCKETING...(What on earth is this).

Okay then, have a good day...

Well, it only uses 500 kB memory so it isn't really that much.... of course, things would be different if you said there are 4 [b]billion[/b] numbers ;) (which is still less tahn 2^32 but much closer).

Yes, multi level hash bucketing sounds like something nasty... it probably has something to do with reading the numbers one by one and hashing them somehow, taking into account the fact that we only need one number that's not in the input array.... interesting, I'll think about it.


summer program
MathPath
geometry
AMC
USA(J)MO
USAMO
percent

What did you think of the 2005 AMC 10 B?

YAY! I had a half day today, after checking my answers, I think I have a 138! I got 25 and 21 wrong, but 25 was a complete "gusto" guess anyway, since I had 1 second left. Hopefully this will help Mathpath to accept me :) .

140.5
number 21 omit and number 12 (i think) wrong

Do y'all think that its harder or easier than AMC 10 A?

harder, even though i scored about 26.5 points higher

I'm very happy, even though I only got a 113. I had a really good day and, after all, I'm only in 8th grade.

I got a 126.5.  I answered 20 and missed 1 (#15, the bill question).  I wish I had tried #25 during the actual test.  After the test I solved it in about two minutes, though I thought my answer was the "trap" answer since it was kind of obvious.

144

The two tests were about the same...both were roughly the same level of difficulty, and both had an annoying number of counting problems but very little geometry or algebra.

...except I missed the cards numbered 1-10 problem.

140.5... i missed the same one as nightflarer and i skipped the one mathfanatic missed... i still dont get how that one works, but yeah. i agree there were too many dumb counting problems! >=( what happened to good ol' geometry and algebra?!

144

I missed number 13: Find all numbers from 1-2005 which either multiples of three or four but not twelve.

126--all answered, four wrong.
I feel so ashamed...I'm wayyy below average relative to AoPS. Oh well, I can blame the fact that I had a severe cold and eye infection distracting me and making me make stupid mistakes.

Sorry--that 150 was meant for the 12B. Subtract 1 150.

[quote="goddessxx"]Okay i feel sooooooooo stupid.  I think i got 70 somethin on the AMC 10 and im talking gr. 10 math rite now in gr. 9!  u guys are so friggin smart its scaring me (yes, im new here) i made so many stupid mistakes on the amc and no, i dont suppose ill ever go into USAMO or watever (first time hearing about it)!!!! :([/quote]

Much of the math on the AMC exams is not frequently taught in the classroom.  Stick around the site and work problems in the Basics and Getting Started forums and perhaps you'll see a lot of math that is new and challenging to you.

i got a 126.5. I'm happy cause at least i made the AIme. But i feel stupid alongside all you guys. All the other armlers got higher scores than me

[quote="karch"]ah, jesus.  i didn't even answer 16 problems...  i was sitting at the end of the table, and the test guy took my paper before i got the chance to put in some random marks.  now i KNOW for sure i didn't get 120...

:(:(  as you guys can probably tell, i'm new here too.

i wish i had found this place earlier..[/quote]

Welcome to the forums!

It's good that you didn't put random marks.  The AMC exams are scored in such a way that guessing is heavily penalized.  The only times when it's good to guess are if you've narrow down the answer choices or have answered 24 problems with confidence.

mmk...well my teacher signed me up for this test and I had no idea when it was going to be and then the day of the test she told me to go to a classroom during 7th period where it was being held. So, i only had about 30 minutes to take it because I dont have a car and didnt have the oppurtunity to call my ride and say I would be late so i could have the other 55 minutes to finish. I only got 10 of the questions answered. That really pissed me off. yeah...sorry....I had to get that off my chest.

i don't know my grade yet because my new math teacher did not know that we could keep the test booklet so i haven't been able to check my answers.  However, i answered 19 and know i got 1 wrong so the best i have is 123 so i'm hope i get no more wrong and that i am above the min which i highly doubt.

[quote="demji"]one question.  on prob 2 which i left blank there is an x such that %x=4.  does that mean x^2/100=4.  in other words x percent of x equals 4.  Any ideas?[/quote]

I believe the question was asking what x gives x% of x=4. Solve x^2/100=4 to get x=20.

You should really take advantage of the fact that the test is multiple choice. It takes less than 2 minutes to plug in the choices and see what comes out correct. I'm sure you spent much more time deciphering its meaning. Quite a few problems on this year's 12B could have been answered by elimination. (I took the 12s, not 10s, but I'm sure the strategy applies.)

Perfect 150 :D I found the 10B incredibly easy. Last year (as an 8th grader) I took the 10A and got 100, but this year i aced the 10B. Not sure what accounted for the huge jump :O

Sleep helps.  I'm in tenth grade, have been thinking about AMC kinds of math all my life, am a strong thinker and could take the test in a tranquil location - but I got two and a half hours of sleep the night before, and thus only got 128.5.  So as far as your perfect 150 (which really I should have gotten had I not made three stupid, stupid, stupid mistakes, similar to, for your context, forgetting the area formula for a triangle) you probably were awake during the test.

I, myself, was peeling my face off the table to answer a question.
[/quote]

[quote="ziggman7"]I got a 130. Left 24 22 21 and 12 blank mostly because I just ran out of time at the end. Also missed 16, made a stupid mistake and misread the problem. If you qualify on the 10B do you take the second AIME or is that just if you have a conflict during the first one?[/quote]

Only if you have a confict during the first one. AIME I is really the default AIME you take.

is there only one amc 10? I'n confused. I was just wondering what people got and I found this site. I think i took the amc 10b and I got 118.5, answering 16 right and leaving 9 blank! if only I had answered one more correctly... :mad: at least I got all the ones i answered right. and there's always next year, since this is amc [i]10[/i] after all i should do better in [i]10[/i]th grade.

I forget exactly what I got (I don't feel like redoing the checking work, and the high school has the official scores while I'm in middle school) but I got high 120s to low 130s. I made a TERRIBLE mistake with the painter problem, I mistook what was left for what he used...I'm happy though that I remembered the sum of roots/product of roots rule (my old math tutor called it the Viette theorem, is that right?)  correctly and got the roots question right. Now, it's time to get a 0 on the AIME... :D


[quote="MCrawford"]Find the units digit of the sum:
\[ (1!)^2 + (2!)^2 + (3!)^2 + \cdots + (100!)^2.  \][/quote]
[hide]
Notice that the pattern of factorials:

$1!=1$

$2!=2$

$3!=6$

$4!=24$

$5!=120$

$6!=720$

Every factorial higher than $6!$ will end in zeroes. And if you square them, it will only double the zeroes, so only $1!$ through $4!$ will count.

$1!$ squared has a units digit of 1.

$2!$ squared has a units digit of 4.

$3!$ squared has a units digit of 6.

$4!$ squared has a units digit of 6.

Adding them together will yield with the units digit of 7[/hide]


Se dau ecuatiiile :$\frac{1}{xy}+z=\frac{4}{z+1},\frac{1}{yz}+x=\frac{4}{x+1}$ si
$\frac{1}{zx}+y=\frac{4}{y+1}$.Aflati $\ x,y,z$.

OK, dar in final se ajunge la $\ x=y=z=1$.