number theory
prime numbers
number theory theorems

so,what does this silence mean?isnt there any explanation of the existence of the primes?

They obviously exist ;)
Please state the question better or tell us why the existence of $ 2$ doesn't satisfy you.

i am not telling that i am not satisfied with the existence of primes e.g. 2.....i was just wondering why they exist necessarily?
the question just came into my mind.i know it sounds foolish....still i asked.
thank you..... :|

[quote="hzbrl"]when i was learning prime numbers, this question came into my mind but i was unable to find any satisfactory answer.....so plz help me to clear my doubt.

the question is:

can we explain the reason of the existence of prime numbers in the number system?i mean can we give any logic against their existence or just they do?

please reply.......thank you.[/quote]

If you have a set $ S$ and an operator "*" : $ S X S \rightarrow S$ which is associative, commutative and with an identity element e,  you can call "prime element" an element $ p\neq e$ such that the only couples $ (x,y)\in S^{2}$ such that $ p=x*y$ are $ (e,p)$ and $ (p,e)$

With such a definition :
Prime numbers exist in $ (%Error. "MathBB" is a bad command.
{N},multiplication)$. For example : $ 2,3,5,7,11,\ldots$
Prime numbers don't exist in $ (%Error. "MathBB" is a bad command.
{Z},addition)$.
Prime numbers exist in $ (\{0\}\cup%Error. "MathBB" is a bad command.
{N},addition)$. For example : $ 1$ (and it is the only prime number here).
Prime numbers exist in $ (\{1\}\cup\{EvenPositiveIntegers\},multiplication)$. For example : $ 6,10,14,\ldots$
Prime numbers don't exist in $ (%Error. "MathBB" is a bad command.
{Q},multiplication)$.
...
So there is no need for existtence of prime numbers. It just depend on the set and the operator  (and on ... the definition  :wink:) .

A good question. Prime numbers exist in all bases and I don't think the question can be answered easily. Primality/compositeness are not properties of numbers, rather they are properties of [i]relation[/i] by way of multiplication and division. Numbers by themselves do not constitute mathematics and as primitive objects they have no properties. They simply exist. These properties only come into existence with the idea of relation and from relation comes mathematics. Music is an analogy. Notes in themselves are not music. Music only comes into being when the notes are played in relation to each other. Relation is the essence of music. Notes are its 'atoms'. You might read Plotinus on Number and quantity.

Just a note: pco's definition of primes coincedes with the known ones on $ \mathbb N$, but not for a lot of other sets (especially the ring theoretic defintion).

There is no logical reason why one number is prime and another is not. Numbers are primitive. They [i]precede[/i] logic. They simply are what they are.


geometry
geometric transformation
reflection
symmetry
circumcircle
trapezoid
perpendicular bisector

Let $ABC$ be an acute triangle. Points $P,U$ are on segment $BC$, $Q,S$ on segment $CA$, $R,T$ on segment $AB$, such that

$PR,US$ is perpendicular to $BC$
$PQ,ST$ is perpendicular to $AC$
$QR,TU$ is perpendicular to $BA$

Prove that triangles $PQR$,$STU$ are congruent.

you can prove that $QU//AB , PT//AC$ and$RS//BC$
I think it easy

You can easly prove that these 6 points lie on one circle and then the thesis is obvious. :D

[quote="TomciO"]I don't think that's so easy...[/quote]
Really?
 :idea:
I think that when you know that these points are concyclic, it isn't difficult, just prove that $S$ is the reflection of point $P$ in the symmetry through the circumcenter of $\triangle PQR$, then use some inequalities...

Of course. But I don't think that's easy to prove that these points are concyclic...

Oh, I solved this problem (because I take part in the Polish MO) and my proof is very simillar to what nxxx said. Especially, it's really easy to prove that these points are concyclic... :lol:

Let $O$ be the circumcenter of $\triangle PQR$. Denote $P', Q', R'$ the points symmetric by $O$ to $P, Q, R$, respectively. Since $O$ lies on the perpendicular bisector of segment $PQ$, using Thales Theorem we obtain that $P'\in AC$. Analogously, $Q'\in AB$ and $R'\in BC$. Triangle $P'Q'R'$ is congruent to triangle $PQR$ and has it's sides parallel to respective sides of $\triangle PQR$ so $P'Q'$ is perpendicular to $AC$, simillarly for other sides.

Suppose that $Q'$ is different from $T$, for example $AQ'<AT$. Hence $AP'<AS$ (as $P'$ and $S$ are perpendicular projections of points $Q'$ and $T$ on $AC$). In the same way we get $CP'>CS$, then $CR'>CU$, $BU>BR'$, $BT>BQ'$, $AQ'>AT$. Contradiction.

Hence $Q'=T$, $P'=S$, $R'=U$ which means that triangles $PQR$ and $STU$ are congruent so we are done.  :)

Well, this is the proof which I know, but actually I had something different on mind... When nxxx said that you can easily prove that these points are concyclic I thought that he means that it can be showed directly (without reflecting the points at first). And I think that's not that easy!
By the way, it's a nice proof.

It is clear that TRSQ, SQUP, and UPTR are cyclic quadrilaterals.  If the circles around any two of these are equal than the 6 points are concyclic.  Suppose then that all three circles are distinct.  It is easy to see that the radical axes of pairs of these circles are the sides of the triangle ABC.  But the sides of the triangle do not concur.

My proof uses a pretty strange lemma:

[color=blue][b]Third theorem.[/b] Let ABC be a non-degenerate triangle. Let $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ be points on the lines BC, BC, CA, CA, AB, AB, respectively. Assume that

- the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle,
- the points $C_{1}$, $C_{2}$, $A_{1}$, $A_{2}$ lie on one circle,
- the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$ lie on one circle.

Then, all six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle.[/color]

Here, if the points $B_{1}$ and $B_{2}$ coincide, then an assertion like "the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle" has to be understood in the following way: "the circle through the points $B_{1}$, $C_{1}$, $C_{2}$ touches the line CA at the point $B_{1}$". Similar conventions apply if the points $C_{1}$ and $C_{2}$ coincide or the points $A_{1}$ and $A_{2}$ coincide.

[i]Proof of the Third theorem.[/i] Consider the following three circles:
- the circle $k_{a}$ passing through the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$;
- the circle $k_{b}$ passing through the points $C_{1}$, $C_{2}$, $A_{1}$, $A_{2}$;
- the circle $k_{c}$ passing through the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$.

If at least two of these circles $k_{a}$, $k_{b}$, $k_{c}$ coincide, then they form one circle through the six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$, so it follows that the six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle, and the Third theorem is proven.

Thus, it remains to prove that at least two of the circles $k_{a}$, $k_{b}$, $k_{c}$ coincide. In order to prove this, we assume the contrary, i. e. assume that no two of these circles $k_{a}$, $k_{b}$, $k_{c}$ coincide. Then, consider the radical axis of the circles $k_{b}$ and $k_{c}$. This radical axis must pass through the points of intersection of the circles $k_{b}$ and $k_{c}$, i. e. through the points $A_{1}$ and $A_{2}$; thus, this radical axis is the line BC (since the line passing through the points $A_{1}$ and $A_{2}$ is the line BC). So we see that the radical axis of the circles $k_{b}$ and $k_{c}$ is the line BC. Similarly, the radical axis of the circles $k_{c}$ and $k_{a}$ is the line CA, and the radical axis of the circles $k_{a}$ and $k_{b}$ is the line AB. Now, the pairwise radical axes of three circles always concur; applying this fact to the circles $k_{a}$, $k_{b}$, $k_{c}$, we thus obtain that the lines BC, CA, AB concur. But this is obviously wrong, since the triangle ABC is non-degenerate. So we get a contradiction, and thus our assumption (that no two of the circles $k_{a}$, $k_{b}$, $k_{c}$ coincide) was wrong. Hence, we have shown that at least two of the circles $k_{a}$, $k_{b}$, $k_{c}$ coincide, and therefore the Third theorem is proven.

Now, a slightly extended statement of the problem:

[color=blue][b]Problem.[/b] Let ABC be a non-degenerate triangle. Let $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ be points on the lines BC, BC, CA, CA, AB, AB, respectively, such that $A_{1}B_{1}\perp BC$, $B_{1}C_{1}\perp CA$, $C_{1}A_{1}\perp AB$, $C_{2}A_{2}\perp BC$, $A_{2}B_{2}\perp CA$, $B_{2}C_{2}\perp AB$. Then, prove that:

[b](a)[/b] The points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle.
[b](b)[/b] We have $B_{1}C_{2}\parallel BC$, $C_{1}A_{2}\parallel CA$, $A_{1}B_{2}\parallel AB$.
[b](c)[/b] The triangles $C_{1}A_{1}B_{1}$ and $B_{2}C_{2}A_{2}$ are congruent.[/color]

[i]Solution.[/i] Since $B_{1}C_{1}\perp CA$ and $B_{2}C_{2}\perp AB$, we have $\measuredangle B_{2}B_{1}C_{1}=90^{\circ}$ and $\measuredangle B_{2}C_{2}C_{1}=90^{\circ}$. Thus, the points $B_{1}$ and $C_{2}$ lie on the circle with diameter $B_{2}C_{1}$. This shows that the points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle. Similarly, the points $C_{1}$, $C_{2}$, $A_{1}$, $A_{2}$ lie on one circle, and the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$ lie on one circle. Thus, by the Third theorem, all six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$ lie on one circle, and problem [b](a)[/b] is solved.

Since the points $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$ lie on one circle, we have $\measuredangle B_{2}A_{2}C_{1}=\measuredangle B_{2}B_{1}C_{1}$. Together with $\measuredangle B_{2}B_{1}C_{1}=90^{\circ}$, this yields $\measuredangle B_{2}A_{2}C_{1}=90^{\circ}$, so that $C_{1}A_{2}\perp A_{2}B_{2}$. Since $A_{2}B_{2}\perp CA$, this becomes $C_{1}A_{2}\parallel CA$. Similarly, $A_{1}B_{2}\parallel AB$ and $B_{1}C_{2}\parallel BC$. Thus, part [b](b)[/b] is solved.

The fact that $C_{1}A_{2}\parallel CA$ rewrites as $C_{1}A_{2}\parallel B_{1}B_{2}$. Hence, the quadrilateral $C_{1}A_{2}B_{2}B_{1}$ is a trapezoid. Since this trapezoid has a circumcircle (the circle through the points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$), it must be isosceles, so that $B_{1}C_{1}=A_{2}B_{2}$. Similarly, $C_{1}A_{1}=B_{2}C_{2}$ and $A_{1}B_{1}=C_{2}A_{2}$. Thus, the triangles $C_{1}A_{1}B_{1}$ and $B_{2}C_{2}A_{2}$ are congruent, and problem [b](c)[/b] is solved.

  Darij


Let points $A,B,C,D$ be on circle $O$, and a line $l$ pass through $O$. $AB,AC,CD,BD$ cut $l$ at $P,Q,R,S$ respectively. Show that $OP=OR\leftrightarrow OQ=OS$.

If $O$ is unit circle of complex plane and $l$ intersect $O$ at $x$, we easly get that both conditions are equivalent to $x^{2}=\frac{abc+bcd+cda+dab}{a+b+c+d}$.

[quote="malinger"][color=darkred]Let points $A,\,B,\,C,\,D$ be on circle $O$, and a line $l$ pass through $O$. $AB,\,AC,\,CD,\,BD$ cut $l$ at $H,\,J,\,K,\,I$ respectively.
Show that $OH=OK\Longleftrightarrow OI=OJ$[/color].[/quote]
[color=indigo][b]Lemma.[/b] Let be given a quadrilateral $ABCD$ incribed in the circle $(O).$ A mobile point $M$ lies on arc $CD$ ( not contain $A,\,B$).
Defined $X=MA\cap CD,\,Y=MB\cap CD$, then we have: $\frac{XD\cdot YC}{XY}=const$
[b]Proof.[/b] Take point $E$ on the opposite to $CD$ so that $\angle BEC=\angle AMB$

So the quadrilateral $XMEB$ is cyclic
 $\Longrightarrow XY\cdot YE=YM\cdot YB=YC\cdot YD$ $\Longrightarrow XY\cdot CE=XD\cdot YC\Longrightarrow\frac{XD\cdot YC}{XY}=CE (const)$

[b]Proof for the proposed problem.[/b] Denote $E,\,F=l\cap (O)$
$\bullet\quad OH=OK$
Applying this [b]Lemma.[/b] for quadrilateral $BCFE,$ we have: $\frac{EH\cdot JF}{HJ}=\frac{EI\cdot KF}{IK}$

On the otherhand, $OH=OK\Longrightarrow HE=KF\Longrightarrow HF=EK$

Hence: $\frac{JF}{HJ}=\frac{EI}{IK}\Longrightarrow\frac{HF}{HJ}=\frac{EK}{IK}\Longrightarrow HJ=IK\Longrightarrow OI=OJ$

$\bullet\quad OI=OJ$ Similar, we have: $OH=OK$ [/color]

Please show the proof to your lemma, April.  :)

How come $XY\cdot CE=XD\cdot YC$?

Moreover, I do not see which points you applied the lemma to.

[color=indigo]We have: $XY\cdot YE=YC\cdot YD\Longleftrightarrow XY\cdot (YC+CE)=YC\cdot (XY+XD)$ $\Longleftrightarrow XY\cdot CE= XD\cdot YC$

And I applying this [b]Lemma.[/b] for quadrilateral $BCFE$ and two points $A,\,B$ lie on it's circumcircl, then we have $\frac{EH\cdot JF}{HJ}=\frac{EI\cdot KF}{IK}=const$.[/color]

[quote="malinger"]Let points $ A,B,C,D$ be on circle $ O$, and a line $ l$ pass through $ O$. $ AB,AC,CD,BD$ cut $ l$ at $ P,Q,R,S$ respectively. Show that $ OP \equal{} OR\leftrightarrow OQ \equal{} OS$.[/quote]

In other words, you want to prove that O is the midpoint of PR if and only if O is the midpoint of QS. This follows from Theorem 3 in the note "On cyclic quadrilaterals and the butterfly theorem" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url] (in fact, the point O is its own orthogonal projection on the line $ l$, because O lies on $ l$).

  darij


Let m,n be the positive integers.Show that 4mn-m-n can never be a square.

[hide="Solution"]
We prove by contradiction.
Suppose $ 4mn \minus{} m \minus{} n \equal{} x^2$. We factor the left side (a usual trick) like this:
$ 16mn \minus{} 4m \minus{} 4n \equal{} 4x^2$
$ (16mn \minus{} 4m) \minus{} (4n \minus{} 1) \equal{} 4x^2 \plus{} 1$
$ (4m \minus{} 1)(4n \minus{} 1) \equal{} 4x^2 \plus{} 1$
We see that the LHS is divisible by at least one prime in form of $ 4a \minus{} 1$, so the RHS is also divisible by $ 4a \minus{} 1$. But, that is impossible by Fermat's Little Theorem. So, it's never a square. [/hide]


trigonometry
geometry
power of a point
trig identities
Law of Cosines
Pythagorean Theorem

I need to find out the angle of a satellite whos coverage is [u]+[/u]1482 Km. I have included a diagram. The only other information I have is its altitude 833 Km. See if you can solve.
[img]http://img.photobucket.com/albums/v145/sealcock6/sat_swath_angle.gif[/img]

Use law of cosines and law of sines to establish 2 equations with two unknowns and then you can find the angle.

That doesn't work because of the curved side, DPopov.

I think I have a method:
If you know the circumference of the earth, then you can find the fraction of the total circumference 1482 is.  You can make it a degree measurement as a fraction of 360.
Power of a Point is then appliable.

Or something else with Power of a Point - you likely use that 833 somehow.
Hope that helps. :)

Draw $AE$, $AC$, and $EC$ and extend $BE$ to intersect $C$.  Now it works. I wasn't reffering to use the curved side?? Of course make a triangle. You can express $AC$ using the Law of Cosines with $\cos(2x)$ and also $\frac{AC}{2}$ using the Law of Cosines with $\cos(x)$.

How can you use those methods with the curved measurements of AE and EC?

Uhh..they are not curved....EA and AC are segments and NOT arcs.

It seems to me they are.  Look at the problem title, and kilometers would probably only be surface measurement.

Oh, wait, does he mean the arcs....i thought he meant the segments. Then you are correct, I apologize.

so no one knows how to get the angle of ABC?

If the satellite's coverage is $\pm 1482 \text{ km}$ then $A$ and $C$ are points of tangency. Perpendicular radii, etc. solve for r, then use trig to find angle.

$(AB)^2 + r^2 = (BE + r)^2$

$\angle ABC = 2 \angle ABE = 2 \arctan \frac{r}{AB}$

Not nescerraly tangents. Bu if not, then the radius of the earth would probably be needed

Then again, there will always be some error due the earth not being perfectly spherical.

[quote="mathmanman"][quote="seamusoboyle"] Bu if not, then the radius of the earth would probably be needed[/quote]
Indeed, you're thinking about eratosthene method, right?[/quote]
You can help me to solve equation with perfect part, thks

I solved the problem. Connecting A and C to the center of the earth R, then connecting A to C and connecting E to R.
I then used 1482/ 40076 (circum of the earth at the equator). from the internet. Then I multiplied it by 360 and got 0.23 rad. This gave me ARE.
Then I used the radius of the earth 6374 for AR.
I made a new point D, where BR and AC intersected.
Using sin(13.31)x6374 = 1467.71 = AD
and cos(13.31)x6374 =  6202.72 = DR
ER-DR = ED = 6374-6202.72 = 171.28
ED+BE = BD = 833 + 171.21 = 1004.28
Pythagorean Theorem   AD^2 + BD^2 = AB^2
1467.42^2 +1004.21^2 = AB^2, 1778.42 = AB
Using law of cosines
AD^2 = BD^2 + BA^2 – 2(BD*BA)cos(ABC)
cos(ABC) = 0.56, acos(0.56) = ABC
Then multiplied by 180/pi
I got 55.62 degrees


Find the biggest $ k$ such that for all $ x,y,z\ge0$ with $ x \plus{} y \plus{} z \equal{} 3$, we have $ x^ky\plus{}y^kz\plus{}z^kx\le 3$.

[quote="spanferkel"]Find the biggest $ k$ such that for all $ x,y,z\ge0$ with $ x \plus{} y \plus{} z \equal{} 3$, we have $ x^ky \plus{} y^kz \plus{} z^kx\le 3$.[/quote]
I mentioned it here (a hungkhtn's problem),with his result we only need find k to $ 3\ge \frac {3^{k \plus{} 1}k^k}{(k \plus{} 1)^{k \plus{} 1}}$,but how to solve it (hungkhtn 'problem) is very hard for me

http://www.mathlinks.ro/viewtopic.php?t=219968


$ \frac{7x \minus{} 14}{2x^2 \minus{} 9x \plus{}4} \equal{} \frac{A}{x\minus{}4} \plus{} \frac{B}{2x \minus{} 1}$

What is A + B?

Yay partial fractions.
[hide="outline"]Clear denominators, sub in a value that makes a term 0, solve, blah.[/hide]

[hide="Answer"]Do as XPMath said to get $ \boxed{5}$.[/hide]


Find the area of triangle ABC if a=40ft b=25ft and <C=30 degrees

Drop an altitude down to the 25 side. The altitude has length 20 (by 30-60-90 triangle trig) so $\frac{1}{2} \times 20 \times 25 = 250$.


inequalities
blogs
inequalities proposed

If $a,b,x,y,z>0$ then  $ \frac{x}{ay+bz} + \frac{y}{az+bx} + \frac{z}{ax+by} \geq \frac{3}{a+b}$.

The solution goes like so:
using this inequality :

$ \frac{a_{1}^2}{x_{1}}+ \dots + \frac{a_{n}^2}{x_{n}} \geq  \frac{(a_{1}+ \ldots +a_{n})^2}{x_{1}+ \ldots +x_{n}}$ for $n=3 $

we get that:

$ \sum \frac{x^2}{axy+bxz} \geq \frac{(x+y+z)^2}{(a+b)(xy+yz+zx)} \geq \frac{3}{a+b}$.

from the fact that $( \sum(x))^2 \geq 3 \sum(xy)$

Does anyone see another approach? Thank you very much!

The same problem is as an exercise on Titu's Lemma at [url=http://www.mathlinks.ro/weblog.php?w=1396]Potla's blog[/url] as problem 5.


Consider a paper punch that can be centered at any point of the plane and that,
when operated, removes from the plane precisely those points whose distance from the
center is irrational. How many punches are needed to remove every point?


If i want to solve for example $\cos{\frac{x}{2}}=\sqrt{\frac{1+\cos x}{2}}$ does it mean that i have to draw a triangle, choose the angle $x$ and give it the value of $\frac{adjacent}{hypotenuse}$ ?

For example $\frac{a}{c}\cdot \frac{1}{2}= \sqrt{\frac{1+\frac{a}{c}}{2}}$ where $a$ is the adjacent side and $c$ is the hypotenuse. Is it the way to solve this type of problems?

What do you mean by "solve"? If you want to "solve an equation", you can't, because that's an identity. If you want to prove the identity, you're off the wrong start, because $\cos{x\over 2}\neq{a\over c}\cdot{1\over 2}$

:blush:  Indeed i made a terrible mistake and it seems that i lack some basic math vocabulary..

In other words, what are the steps to prove an identity which has some trig functions? What about solving an equation with a trig function?

Hope i made myself clear.

To prove the half-angles you need to have basic mastery of the double angles, which you can derive from the addition identities. For cosine, the double identity is:

$\cos(2a)=\cos^{2}a-\sin^{2}a = 2\cos^{2}a-1$

Expressing $\cos a$ in terms of $\cos 2a$, we get

$\cos(2a)+1=2\cos^{2}a$

$\cos^{2}a=\frac{\cos(2a)+1}{2}$

$\cos a=\sqrt{\frac{\cos(2a)+1}{2}}$

$\cos \left(\frac{a}{2}\right) = \sqrt{\frac{\cos a+1}{2}}$

You can do similar things for all the other identities.


Provethat there are many infinitely n s.t $n!$ is begun by 2005.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=langlois&t=23417

Pierre.


Let positive real numbers $ a,b,c$ satisfying $ abc\equal{}1$. Prove that $ \frac{a}{b^2\plus{}c^2}\plus{}\frac{b}{c^2\plus{}a^2}\plus{}\frac{c}{a^2\plus{}b^2}\ge \frac{3}{2}$

[quote="thaithuan_GC"]Let positive real numbers $ a,b,c$ satisfying $ abc \equal{} 1$. Prove that $ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2}\ge \frac {3}{2}$[/quote]

try a=1/4, b=c=2


This is a favorite problem of mine.

When 44444444 is written in decimal notation, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B. (A and B are written in decimal notation).

If you've seen it before, hold off on responding for a while to give others a chance to work on it.

It's a nice problem. I have seen it before right here on AoPS.

Yes, it's a cute problem, from an IMO I believe.  I showed the problem to my (then) 9-year-old daughter, and she "solved" it -- with my generous hints, of course.  She loved the fact that we could reason about such large numbers.

I have never seen it and I don't know how to solve it... could you provide some hints so that I might get started?

Spoon wrote:some hints so that I might get started?

[hide]mod 9 + rough upper bound[/hide]

I'm curious... WHy did you choose mod 9?

[hide]Working modulo 9 is often a good thing to do for such a problem because it is invariant under sum of digits.[/hide]


limit
real analysis
real analysis unsolved

hello
1/ Prove that :$ : (\forall n\in \mathbb{N}^{*}) (\forall \alpha \in [0,1[) 1\leq (1 \plus{} \frac {\alpha}{n})^{n}\leq \frac {1}{1 \minus{} \alpha}$

2/let $ (a_n)$ a sequence  , and $ (\forall n \in N) a_n > \equal{} 0$
prove : if $ \lim_{n\rightarrow \plus{} \infty}n\alpha_{n} \equal{} 0$
than $ \lim_{n\rightarrow \plus{} \infty}(1 \plus{} \alpha_{n})^{n} \equal{} 1$

thx

1.) Write the problem in the following form:
Show that the graph of $ f$ is below the graph of $ g$ with:
$ f(x) \equal{} 1\plus{}\frac{x}{n}$ and $ g(x) \equal{} \frac{1}{\sqrt[n]{1\minus{}x}}$

2.) Let $ \epsilon > 0$.
Then there is an $ N$ so that $ |n\alpha_n| < \epsilon$ for $ n > N$.
For those n:  $ 1 \le (1 \plus{} \frac {n\alpha_n}{n})^n < (1 \plus{} \frac {\epsilon}{n})^n$
etc.


[quote="math92"]How many lines of symmetry does the symbol $\Omega$ have?[/quote]
[hide]yes absolutely it has only one symmetry line and it is the down middle. :D [/hide]


Let $ X,Y: \Omega \to \mathbb{R}$ be two random variables such that $ Y$ is $ \sigma (X)$ measurable. I'm trying to proove that $ Y\equal{}\Phi (X)$ where $ \Phi: \mathbb{R} \to \mathbb{R}$ is Borel measurable. Our professor told us to take the $ \Phi$ defined by $ \Phi (X (Y^{\minus{}1}(y)))\equal{}y$ and prove that it's a well defined function. I couldnt manage to show that $ \Phi$ is Borel measurable... My attempts to do so seem to depend on whether $ Im( X )$ is a Borel set or not... Any suggestions?

try first with simple functions $ Y_n \equal{} \sum 1_{A_i}$ and remember that any measurable function is a pointwise limit of such functions.


geometry
circumcircle
incenter
geometry proposed

[color=darkblue] Let $DEF$ be the orthic triangle of the [u]acute[/u] triangle $ABC\ ,$ where $D\in BC$ , $E\in CA$ , $F\in AB\ .$ For a interior point $P$ denote the intersection $R\in AP\cap BC\ .$ Prove that the circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $CPR$ if and only if $\triangle DEF\sim \triangle CPR\ .$[/color]

Suppose that the circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $CPR$.

We need to prove that the angles of triangle $CPR$ are the equal to $\hat P,\hat Q,\hat R$ which are $180^\circ-2A,180^\circ-2B,180^\circ-2C$ respectively.

Let $X,Y,Z$ be foots of perpendicular from $O$(the circumcenter of $ABC$) to the sides $CP,RP,PC$ respectively. So we get $OX=OY=OZ$. And because $OA=OC$, $\triangle OXC\cong \triangle OYA$. Because $\widehat{BOC}=2A$, so $\widehat{XOC}=A$. Therefore, $\widehat{YOC}=A$. From $\widehat{AOC}=2B$, we get $\widehat{XOY}=2C$. Because $\Box OXAY$ is cyclic, $\widehat{PRC}=180^\circ-2C$. $\widehat{OCX}=90^\circ-\widehat{COX}=90^\circ-A$. So $\widehat{PCR}=180^\circ-2C$.

Should it be[color=darkblue] the circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $CPR$ if and only if $D=C,E=P,F=R$[/color]?

Yes ! Two similar triangles (in the mentioned notation $\rightarrow$) : $XYZ\sim MNP\Longleftrightarrow X=M\ ,\ Y=N\ ,\ Z=P .$

O.K. then

Suppose that $\hat C,\hat P,\hat R$ are $180^\circ-2A,180^\circ-2B,180^\circ-2C$

Let $X,Y,Z$ be foots of perpendicular from $I$(the incenter of $CPR$) to the sides $CP,RP,PC$ respectively.  Because $\widehat{PRC}=180^\circ-2C$, $\widehat{RAC}=C$. So $RC=RA$. From $RX=RY$, $CX=AY$. And because $IX=IY$, so $\triangle IXC\cong \triangle IYA$. Because $\widehat{PCR}=180^\circ-2A$, $\widehat{XIC}=A$ also $\widehat{YIC}=A$. Then, because $\widehat{XIY}=2C$, we get $\widehat{CIA}=2B$. From $\widehat{CIA}=2B$ and because $IA=IC$, $I$ is the circumcenter of $ABC$.


A lucky number is a positive integer that can be written as the sum of positive integers whose reciprocals sum to $ 1$. A number is unlucky if not lucky. How many unlucky numbers are there? Prove it.

Thank you for the related post, but I still have proof for my answer..

A USAMO problem asked you to prove that all integers greater than 32 are lucky.

Cool! But I'm not a great prover, so could you please give a hint of how to prove this? Thanks a bunch

I've been working on a formula for lucky numbers composed of $ x$ different numbers (so far only successful with $ x\equal{}2$) and finding a periodic number of lucky occurrences when the lucky number is made up of multiples of a certain number (e.g. lucky numbers made up of multiples of $ 2$ occur every $ 6$ numbers ($ 4\equal{}2*2 ; 10\equal{}2\plus{}2*4 ; 16\equal{}4*4 ; 22\equal{}2\plus{}4\plus{}2*8$).  Numbers composed of multiples of $ 3$ occur every $ 9$  ($ 9\equal{}3*3 ; 18\equal{}2*3\plus{}2*6 ; 27\equal{}3\plus{}4*6$).  Numbers composed of multiples of $ 5$ occur every $ 15$ ($ 25\equal{}5*5 ; 40\equal{}4*5\plus{}2*10 ; 55\equal{}3*5\plus{}4*10$). Numbers composed of multiples of $ 7$ occur every $ 21$  ($ 49\equal{}7*7 ; 70\equal{}7*6\plus{}2*14$; Etc.). From here it is easy to see (I still need to show algebraically) that when based solely on multiples of one number $ n$, there is a periodic occurrence of lucky numbers every $ 3n$. Now I feel I oughta find the relation of period of occurrences when composed of multiples of more than one number (e.g. $ 11\equal{}2\plus{}3\plus{}6$) From there, I feel that induction wouldn't be bad. Anyone got an algebraic relation for the $ 3n$ of an idea for numbers composed of multiple multiples? Hopefully my rambling thoughts on this problem are of use..

$ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{a\plus{}b}{ab}$, so $ a\plus{}b\equal{}ab$. Solve for $ a$: $ a\equal{}ab\minus{}b\equal{}b(a\minus{}1)$. Solve for $ b$: $ b\equal{}ab\minus{}a\equal{}a(b\minus{}1)$. Substituting $ a(b\minus{}1)$ for $ b$, we have $ a\equal{}a(a\minus{}1)(b\minus{}1)$. Dividing by $ a$, we get $ (a\minus{}1)(b\minus{}1)\equal{}1$. Since $ (\minus{}1)(\minus{}1)$ does not yeild positive integers for $ a$ and $ b$, we have to use $ (1)(1)$. Now we have $ a\minus{}1\equal{}b\minus{}1\equal{}1$, so $ a\equal{}b\equal{}2$. Thus, $ a\plus{}b\equal{}4$, so $ 4$ is the only lucky number. There are infinite unlucky numbers.

You do not only have to use two positive integers.  For example, $ 11$ is lucky because $ 11 \equal{} 2 \plus{} 3 \plus{} 6$ and $ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{6} \equal{} 1$.

aahmadi:  that seems like an unnecessarily complicated approach.  I do not, however, seem to be able to find this problem in Engel.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1958492651&t=215882]this thread[/url].

@mathelete: There can be more than two variables.

Check these links:
[url]http://www.research.att.com/~njas/sequences/A028229[/url]

[url]http://www.research.att.com/~njas/sequences/A052428[/url]

[quote="1=2"]It said, and I quote: "Given that the integers between 33 and 73 inclusive are lucky,"
I don't know how to prove it either, but I'm working on it![/quote]

Using the ideas from [url]http://www.math.ucsd.edu/~fan/ron/papers/63_02_partitions.pdf[/url]

the transformations:
(a) $ 1 \equal{} \frac {1}{d_1} \plus{} \cdots \plus{} \frac {1}{d_n} \equal{} \frac {1}{2} \plus{} \frac {1}{2d_1} \plus{} \cdots \plus{} \frac {1}{2d_n}$ transforms $ d_1 \plus{} \cdots \plus{} d_n\equal{}m$ to $ 2m \plus{} 2$

(b) $ 1 \equal{} \frac {1}{d_1} \plus{} \cdots \plus{} \frac {1}{d_n} \equal{} \frac {1}{3} \plus{} \frac {1}{6} \plus{} \frac {1}{2d_1} \plus{} \cdots \plus{} \frac {1}{2d_n}$ transforms $ m$ to $ 2m \plus{} 9$


For a concert, 100 more balcony tickets than main-floor tickets were sold. The balcony tickets cost $4 and the main-floor tickets cost$12. If the total sales for BOTH types of tickets was $3056, find the number of balcony tickets that were sold. My work: Let x = number of balcony tickets$4(x + 100) + $12 =$3056

Is this the correct equation to use?

Or do I use this one:

4X + 100 + 12X = 3056?

[quote="anyone"][hide]let $x$=balcony tickets and $y$=floortickets

$x=y+100$
$4x+12y=3056$

Now just solve the system and find x+y
[/hide][/quote]

Edited
[hide="Finishing up the solution"]Substitute equation 1 into equation 2:
\[4(y+100)+12y = 3056 \]

\[16y=2656 \]

\[y=166 \]

\[x=y+100=266 \][/hide]

The book's answer is 266 balcony tickets.

How do I get 266 tickets?

[quote="vishalarul"]
\[4(y+100)+12y=3056\]
\[16y=2056\]
[/quote]
There's your mistake, that should be a $2656$, thus making $y=166$.  Thus, we find that $x=166+100=\boxed{266}$.


function
calculus
derivative
real analysis
real analysis solved

Let f:R->R a function wich has a primitive F:R->(R+)*.
Show that for every  \varepsilon >0 there exists a real x such that
If(x)I< \varepsilon

Let's assume there is a c>0 s.t. |f(x)|>=c for all x \in R. Since f is a derivative, it has Darboux's property, so it can't change signs, so we may assume f(x)>=c for all x \in R.

The property stated above means that the function G(x)=F(x)-cx is increasing. G(x)->00 as x->-00 (because F(x)>0 for all x \in R), so we have a contradiction (we can find, for example, a small enough x<0 s.t. G(x)>G(0), but G(x)<=G(0) because G is supposd to be increasing).

The samt thing happens if we assume f(x)<-c  for all x\in R.

It's obvious that all of this works for all f which have a bounded primitive F.


So I finally got around to buying this book recently. I hadn't seen it since I was in high school, when it was in it's second edition. I mostly liked it because it had proofs of some things that I hadn't seen before. Like the proof that $\pi$ is irrational and $e$ is transcendental. Upon seeing it again though, I have a couple of comments.

First of all, its a really well written and thought out textbook. I like Spivak's writing style and the problems lead you to some interesting results (as is typical of all of his textboks). But the book seems to fall into an awkward place in terms of curriculum.

Its too technical in my opinion to be used as into a "Calculus" textbook, and at the same time not quite modern or thorough enough to be used for an "Introduction to Analysis" textbook. So I'm not sure that it has much of an audience aside from being a cool book to have on the shelves of people who've already studied single variable calculus.

The reason why I say its too technical for a "Calculus" book, is that I believe students should get aquinted with the techniques and applications of Calculus before they get into the proofs of its theorems. Of course the two pedagogies aren't mutually exclusive, but this book lies heavily on the side of the latter. Nearly all of the exercises are of an advanced nature and very few of them deal with applications or purely conceptual understanding.

For intro to Calculus, I like the books [u]Thomas' Calculus[/u] by Thomas, Weir, Hass, and Giorgano or [u]Calculus[/u] by Stewart. They are for the most part what people call standard "plug and chug" textbooks, but they have some good problems in them (especially Thomas' applications and Stewarts Concept Check and Problem Corner problems).

The reason why I say its not modern or thorough enough to be used for an "Analysis" book, is that well.. it isn't. Just compare it to one of the standard Intro to Analysis books, Rudin's [u]Principles of Mathematical Analysis[/u], and you will see that the coverage of topics is vastly different. I don't think this book alone would prepare someone for upperclass work in Mathematics.

One way that I could see this book being used is for motivated students who finish the typical BC/IB Calculus curriculum early in high school and want to go back and learn more technical details of WHY the theorems are true. This is something that can be done in place of, or in addition to, the more typical Multivariable/Diff. Eqn's/Lin. Algebra courses that high schoolers do after BC Calculus. But I doubt there are many high school teachers who could help you with the problems in this book.

[quote="gauss202"]
One way that I could see this book being used is for motivated students who finish the typical BC/IB Calculus curriculum early in high school and want to go back and learn more technical details of WHY the theorems are true. This is something that can be done in place of, or in addition to, the more typical Multivariable/Diff. Eqn's/Lin. Algebra courses that high schoolers do after BC Calculus. But I doubt there are many high school teachers who could help you with the problems in this book.[/quote]

That sounds like something I might do.  But what do you think would be the best route for someone just taking a BC class?  I am going to try to change my schedule next year so I can take Linear Algebra and/or the next level of Calculus at a local college.

From your last paragraph I get the impression that Spivak's book is a [i]good[/i] book for high school students under those circumstances but are there books that are better for the same types of students, i.e. do you know of any books that are better than Spivak's?

[i][edit: I just read the thread "Sequence of Learning Advanced Math" and it answered a bunch of my questons][/i]

Well, it is a great source of calculus problems, regardless of whether or not its used in a curricular setting.

[quote="blahblahblah"]Rudin has a proof of irrationality of $e$ (it's a standard exercise in all analysis texts; e.g, it's in all four introductory analysis texts I own).  As I recall, the one for $\pi$ isn't much different, expand taylor series, blah blah blah blah.[/quote]

The proof for $e$ is an exercise, the proof for $\pi$ is incomparably harder.

Re: the use of Spivak

When I was a senior in high school, I had the standard high school calculus course, that is, all mechanics, no theory.  I can't even remember which book we used.

In freshman year of college, the math department grouped the smart math majors (or were we self-selected?) into a two-year honors sequence.  The first year was Calculus, using Spivak's book.  I thought it fit the bill nicely -- I loved the book.

[quote="blahblahblah"]The proof of irrationality of $e$ is on p.65 of the third edition.

As for the proof of $\pi$, yes, it is harder - I was mistaken in what I said.  They both end up with proving that if $\pi$ or $e$ were rational, there would be an integer between $0$ and $1$, which is probably why I remembered them as similar.[/quote]

Actually when I wrote the comments above I was thinking of the proofs that $\pi$ is transcendental, not irrational.
Still, proving $e$ is irrational is at the level of an olympiad problem (there is only one move you can make, so make it...) whereas it would be astonishing if anybody were to come up with the irrationality proof for $\pi$ within a few hours of learning that the result is true.

re: Spivak's book, I made several possibly-relevant postings about it in the thread "Best Calculus Textbook", on this board around Dec 2004.


real analysis
calculus
integration
function
trigonometry
limit
abstract algebra

If $ f: \mathbb R\rightarrow \mathbb R$ ,such that :

\[ f(x)\equal{}\left\{ \begin{array}{crl} x\plus{}1 &,&\mathrm{ x \ is \ rational } \\ 2x\minus{}1 &,& \mathrm{x \ is \ irrational } \end{array}\right.\]

 find the Leb. $ \int_{\mathbb R} f$

It doesn't exist. The "rational" part is irrelevant (sets of measure zero never matter) and the function $ x\mapsto 2x\minus{}1$ is not Lebesgue integrable over $ \mathbb R$.

[quote="fedja"]It doesn't exist. The "rational" part is irrelevant (sets of measure zero never matter) and the function $ x\mapsto 2x \minus{} 1$ is not Lebesgue integrable over $ \mathbb R$.[/quote]

What about to put $ \cos(x)$ instead of $ 2x\minus{}1$.

Thx

Indeed , I'm warping about to find the value of it .

[quote="mlok"]The integral of $ \cos x$ over $ \mathbb R$ does not exist either in Lebesgue or improper-Riemann sense. One could argue for it being equal to $ 0$ in some Pickwickian sense. (Like in summability methods for series)[/quote]

what is that ?Pickwickian sense? Sorry  :blush:

For example, I can define a generalization of Lebesgue integral of $ f$ over $ \mathbb R$ as
$ \lim_{\epsilon\to 0}\int_{\mathbb R}\exp(\minus{}\epsilon^2 x^2) f(x)\,dx$ when the limit exists. Note that the factor $ \exp(\minus{}\epsilon^2 x^2)$ converges to $ 1$ uniformly on compact subsets as $ \epsilon\to 0$. It is also uniformly bounded. One can show that if $ f$ is Lebesgue integrable on $ \mathbb R$, the above limit is equal to the integral.

Now take $ f(x)\equal{}\cos x$:
$ \int_{\mathbb R}\exp(\minus{}\epsilon^2 x^2) \cos (x)\,dx\equal{}\frac{\sqrt{\pi}}{\epsilon} \exp(\minus{}1/(4\epsilon^2))$. This converges to $ 0$ as $ \epsilon\to 0$. Hence the Pickwickian integral of $ \cos x$ over $ \mathbb R$ is $ 0$.

(It may have a real name, unrelated to Dickens, but I don't know it.)

That's a summability method, of course. It's not exactly Abel summability, but it's related.

Two-sided Abel summability would be $ \lim_{t\to 0^\plus{}}\int_{\minus{}\infty}^{\infty}e^{\minus{}t|x|}f(x)\,dx.$

The Abel sum of this particular integral would be $ \lim\frac{2t}{1\plus{}t^2}\equal{}0,$ also.

Or you could try Cesaro means: $ \lim_{N\to\infty}\frac1N\int_0^N\int_{\minus{}y}^yf(x)\,dx\,dy$

In the case of this particular integral, that would also be $ \lim_{N\to\infty}\frac2N(1\minus{}\cos N)\equal{}0.$

For all of these methods, including your no-name method, it shouldn't be too difficult to prove the "abelian" theorem: if the integral exists as an improper integral, then it is summable to the same number. There should also be an array of "tauberian" theorems: if $ f$ satisfies such-and-such a condition (perhaps a size condition at $ \infty$), and the integral is summable, then it converges as an improper integral. Of course, $ f(x)\equal{}\cos x$ isn't going to satisfy any of those tauberian conditions.


Prove that for all nonnegative real numbers $ a,b,c$ we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

[hide="Solution 1"] By Minkowski Inequality we have

$ \quad a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \\
\equal{} \sqrt {(a^2)^2 \plus{} \left(a\sqrt {bc}\right)^2} \plus{} \sqrt {(b^2)^2 \plus{} \left(b\sqrt {ca}\right)^2} \plus{} \sqrt {(c^2)^2 \plus{} \left(c\sqrt {ab}\right)^2} \\
\ge \sqrt {\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2}.$

Hence, it suffices to show

$ \left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2 \ge 2(ab \plus{} bc \plus{} ca)^2.$

Which is equivalent to

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \ge 3abc(a \plus{} b \plus{} c).$

By Schur Inequality

$ a^4 \plus{} b^4 \plus{} c^4 \ge a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \minus{} abc(a \plus{} b \plus{} c).$

Therefore we just need to prove

$ \quad a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \\
\ge 4abc(a \plus{} b \plus{} c).$

Using AM-GM Inequality we have

$ a^3 b \plus{} a^3 c \plus{} abc\sqrt {bc} \plus{} abc\sqrt {bc} \ge 4a^2 bc$
$ b^3 c \plus{} b^3 a \plus{} abc\sqrt {ca} \plus{} abc\sqrt {ca} \ge 4b^2 ca$
$ c^3 a \plus{} c^3 b \plus{} abc\sqrt {ab} \plus{} abc\sqrt {ab} \ge 4c^2 ab.$

Adding them up we get the result.[/hide]

[hide="Solution 2"]Applying Schur Inequality we have

$ a\sqrt{a^2\plus{}bc}\plus{} b\sqrt{b^2\plus{}ca}\plus{} c\sqrt{c^2\plus{}ab} \ge \frac{1}{\sqrt{2}}( a(a\plus{}\sqrt{bc})\plus{} b(b\plus{}\sqrt{ca})\plus{} c(c\plus{}\sqrt{ab}))\\
\ge \sqrt{2}(\sqrt{ab}(a\plus{}b)\plus{}\sqrt{bc}(b\plus{}c)\plus{}\sqrt{ca}(c\plus{}a)) \ge \sqrt{2}(ab\plus{}bc\plus{}ca)$[/hide]

Solution 2 is from Zheng Fan from Shanghai High School.

[quote="shalex"]Prove that for all nonnegative real numbers $ a,b,c$ we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

[hide="Solution 1"] By Minkowski Inequality we have

$ \quad a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \\
\equal{} \sqrt {(a^2)^2 \plus{} \left(a\sqrt {bc}\right)^2} \plus{} \sqrt {(b^2)^2 \plus{} \left(b\sqrt {ca}\right)^2} \plus{} \sqrt {(c^2)^2 \plus{} \left(c\sqrt {ab}\right)^2} \\
\ge \sqrt {\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2}.$

Hence, it suffices to show

$ \left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2 \ge 2(ab \plus{} bc \plus{} ca)^2.$

Which is equivalent to

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \ge 3abc(a \plus{} b \plus{} c).$

By Schur Inequality

$ a^4 \plus{} b^4 \plus{} c^4 \ge a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \minus{} abc(a \plus{} b \plus{} c).$

Therefore we just need to prove

$ \quad a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \\
\ge 4abc(a \plus{} b \plus{} c).$

Using AM-GM Inequality we have

$ a^3 b \plus{} a^3 c \plus{} abc\sqrt {bc} \plus{} abc\sqrt {bc} \ge 4a^2 bc$
$ b^3 c \plus{} b^3 a \plus{} abc\sqrt {ca} \plus{} abc\sqrt {ca} \ge 4b^2 ca$
$ c^3 a \plus{} c^3 b \plus{} abc\sqrt {ab} \plus{} abc\sqrt {ab} \ge 4c^2 ab.$

Adding them up we get the result.[/hide]

[hide="Solution 2"]Applying Schur Inequality we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \frac {1}{\sqrt {2}}( a(a \plus{} \sqrt {bc}) \plus{} b(b \plus{} \sqrt {ca}) \plus{} c(c \plus{} \sqrt {ab})) \\
\ge \sqrt {2}(\sqrt {ab}(a \plus{} b) \plus{} \sqrt {bc}(b \plus{} c) \plus{} \sqrt {ca}(c \plus{} a)) \ge \sqrt {2}(ab \plus{} bc \plus{} ca)$[/hide]

Solution 2 is from Zheng Fan from Shanghai High School.[/quote]
Nice solution, shalex.

My solution:
$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

$ \Leftrightarrow \sum \sqrt{\left(\frac{a}{bc}\right)^2\plus{}\frac{1}{bc}} \ge 2\sum \frac{1}{a}$

By Minkowski inequality:
$ \sum \sqrt{\left(\frac{a}{bc}\right)^2\plus{}\frac{1}{bc}} \ge \sqrt{\left(\sum \frac{a}{bc}\right)^2\plus{}\left(\sum \frac{1}{\sqrt{bc}}\right)^2}$

We have:
$ A\equal{}\left(\sum \frac{a}{bc}\right)^2\plus{}\left(\sum \frac{1}{\sqrt{bc}}\right)^2\equal{}\sum \frac{a^2}{b^2c^2}\plus{}2\sum \frac{1}{a\sqrt{bc}}\plus{}2\sum \frac{1}{a^2}\plus{}\sum \frac{1}{ab}$

By AM-GM inequality:
$ \frac{a^2}{b^2c^2}\plus{}\frac{2}{a\sqrt{bc}} \ge \frac{3}{bc}$
$ \Rightarrow \sum \frac{a^2}{b^2c^2}\plus{}2\sum \frac{1}{a\sqrt{bc}} \ge 3\sum \frac{1}{bc}$

So $ \sqrt{A} \ge \sqrt{2\left(\sum \frac{1}{ab}\right)^2}\equal{}\sqrt{2}(\sum \frac{1}{a})$

[quote="shalex"]Prove that for all nonnegative real numbers $ a,b,c$ we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

[hide="Solution 1"] By Minkowski Inequality we have

$ \quad a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \\
\equal{} \sqrt {(a^2)^2 \plus{} \left(a\sqrt {bc}\right)^2} \plus{} \sqrt {(b^2)^2 \plus{} \left(b\sqrt {ca}\right)^2} \plus{} \sqrt {(c^2)^2 \plus{} \left(c\sqrt {ab}\right)^2} \\
\ge \sqrt {\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2}.$

Hence, it suffices to show

$ \left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2 \ge 2(ab \plus{} bc \plus{} ca)^2.$

Which is equivalent to

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \ge 3abc(a \plus{} b \plus{} c).$

By Schur Inequality

$ a^4 \plus{} b^4 \plus{} c^4 \ge a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \minus{} abc(a \plus{} b \plus{} c).$

Therefore we just need to prove

$ \quad a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \\
\ge 4abc(a \plus{} b \plus{} c).$

Using AM-GM Inequality we have

$ a^3 b \plus{} a^3 c \plus{} abc\sqrt {bc} \plus{} abc\sqrt {bc} \ge 4a^2 bc$
$ b^3 c \plus{} b^3 a \plus{} abc\sqrt {ca} \plus{} abc\sqrt {ca} \ge 4b^2 ca$
$ c^3 a \plus{} c^3 b \plus{} abc\sqrt {ab} \plus{} abc\sqrt {ab} \ge 4c^2 ab.$

Adding them up we get the result.[/hide]

[hide="Solution 2"]Applying Schur Inequality we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \frac {1}{\sqrt {2}}( a(a \plus{} \sqrt {bc}) \plus{} b(b \plus{} \sqrt {ca}) \plus{} c(c \plus{} \sqrt {ab})) \\
\ge \sqrt {2}(\sqrt {ab}(a \plus{} b) \plus{} \sqrt {bc}(b \plus{} c) \plus{} \sqrt {ca}(c \plus{} a)) \ge \sqrt {2}(ab \plus{} bc \plus{} ca)$[/hide]

Solution 2 is from Zheng Fan from Shanghai High School.[/quote]
One more solution is
Squaring both sides, we can rewrite the inequality as
\[ \sum a^4\plus{}2\sum ab\sqrt{(a^2\plus{}bc)(b^2\plus{}ca)} \ge 2\sum a^2b^2\plus{}3\sum a^2bc\]
By the Cauchy Schwarz, we have
\[ \sum ab\sqrt{(a^2\plus{}bc)(b^2\plus{}ca)}  \ge \sum ab(ab\plus{}c\sqrt{ab})\]
It suffices to show that
\[ \sum a^4\plus{} 2\sum ab(ab\plus{}c\sqrt{ab}) \ge 2\sum a^2b^2\plus{}3\sum a^2bc\]
\[ \Leftrightarrow \sum (a^4\plus{}2abc\sqrt{bc} \minus{}3a^2bc) \ge 0\]
By the AM-GM Inequality, we get
\[ a^4\plus{}2abc\sqrt{bc} \ge 3\sqrt[3]{a^6b^3c^3}\equal{}3a^2bc\]
The inequality is proved. :)

Actually, the stronger result is also true
\[ \sum a\sqrt{a^2\plus{}bc} \ge (2\minus{}\sqrt{2})(\sum a^2\plus{}\sqrt{2}\sum ab)\]
:)

[quote="shalex"][quote="NguyenDungTN"][quote="shalex"]Prove that for all nonnegative real numbers $ a,b,c$ we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

[hide="Solution 1"] By Minkowski Inequality we have

$ \quad a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \\
\equal{} \sqrt {(a^2)^2 \plus{} \left(a\sqrt {bc}\right)^2} \plus{} \sqrt {(b^2)^2 \plus{} \left(b\sqrt {ca}\right)^2} \plus{} \sqrt {(c^2)^2 \plus{} \left(c\sqrt {ab}\right)^2} \\
\ge \sqrt {\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2}.$

Hence, it suffices to show

$ \left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2 \ge 2(ab \plus{} bc \plus{} ca)^2.$

Which is equivalent to

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \ge 3abc(a \plus{} b \plus{} c).$

By Schur Inequality

$ a^4 \plus{} b^4 \plus{} c^4 \ge a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \minus{} abc(a \plus{} b \plus{} c).$

Therefore we just need to prove

$ \quad a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \\
\ge 4abc(a \plus{} b \plus{} c).$

Using AM-GM Inequality we have

$ a^3 b \plus{} a^3 c \plus{} abc\sqrt {bc} \plus{} abc\sqrt {bc} \ge 4a^2 bc$
$ b^3 c \plus{} b^3 a \plus{} abc\sqrt {ca} \plus{} abc\sqrt {ca} \ge 4b^2 ca$
$ c^3 a \plus{} c^3 b \plus{} abc\sqrt {ab} \plus{} abc\sqrt {ab} \ge 4c^2 ab.$

Adding them up we get the result.[/hide]

[hide="Solution 2"]Applying Schur Inequality we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \frac {1}{\sqrt {2}}( a(a \plus{} \sqrt {bc}) \plus{} b(b \plus{} \sqrt {ca}) \plus{} c(c \plus{} \sqrt {ab})) \\
\ge \sqrt {2}(\sqrt {ab}(a \plus{} b) \plus{} \sqrt {bc}(b \plus{} c) \plus{} \sqrt {ca}(c \plus{} a)) \ge \sqrt {2}(ab \plus{} bc \plus{} ca)$[/hide]

Solution 2 is from Zheng Fan from Shanghai High School.[/quote]
Nice solution, shalex.

My solution:
$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

$ \Leftrightarrow \sum \sqrt {\left(\frac {a}{bc}\right)^2 \plus{} \frac {1}{bc}} \ge 2\sum \frac {1}{a}$

By Minkowski inequality:
$ \sum \sqrt {\left(\frac {a}{bc}\right)^2 \plus{} \frac {1}{bc}} \ge \sqrt {\left(\sum \frac {a}{bc}\right)^2 \plus{} \left(\sum \frac {1}{\sqrt {bc}}\right)^2}$

We have:
$ A \equal{} \left(\sum \frac {a}{bc}\right)^2 \plus{} \left(\sum \frac {1}{\sqrt {bc}}\right)^2 \equal{} \sum \frac {a^2}{b^2c^2} \plus{} 2\sum \frac {1}{a\sqrt {bc}} \plus{} 2\sum \frac {1}{a^2} \plus{} \sum \frac {1}{ab}$

By AM-GM inequality:
$ \frac {a^2}{b^2c^2} \plus{} \frac {2}{a\sqrt {bc}} \ge \frac {3}{bc}$
$ \Rightarrow \sum \frac {a^2}{b^2c^2} \plus{} 2\sum \frac {1}{a\sqrt {bc}} \ge 3\sum \frac {1}{bc}$

So $ \sqrt {A} \ge \sqrt {2\left(\sum \frac {1}{ab}\right)^2} \equal{} \sqrt {2}(\sum \frac {1}{a})$[/quote]

A typo at last; it should be

$ \sqrt {A} \ge \sqrt {2\left(\sum \frac {1}{a}\right)^2}$[/quote]
 :oops: You're right.

[quote="can_hang2007"]Actually, the stronger result is also true
\[ \sum a\sqrt {a^2 \plus{} bc} \ge (2 \minus{} \sqrt {2})(\sum a^2 \plus{} \sqrt {2}\sum ab)
\]
:)[/quote]
Here is my solution for the weaker one :)

[quote="can_hang2007"][quote="shalex"]Prove that for all nonnegative real numbers $ a,b,c$ we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

[hide="Solution 1"] By Minkowski Inequality we have

$ \quad a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \\
\equal{} \sqrt {(a^2)^2 \plus{} \left(a\sqrt {bc}\right)^2} \plus{} \sqrt {(b^2)^2 \plus{} \left(b\sqrt {ca}\right)^2} \plus{} \sqrt {(c^2)^2 \plus{} \left(c\sqrt {ab}\right)^2} \\
\ge \sqrt {\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2}.$

Hence, it suffices to show

$ \left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \plus{} \left(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}\right)^2 \ge 2(ab \plus{} bc \plus{} ca)^2.$

Which is equivalent to

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \ge 3abc(a \plus{} b \plus{} c).$

By Schur Inequality

$ a^4 \plus{} b^4 \plus{} c^4 \ge a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \minus{} abc(a \plus{} b \plus{} c).$

Therefore we just need to prove

$ \quad a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \plus{} 2abc\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \right) \\
\ge 4abc(a \plus{} b \plus{} c).$

Using AM-GM Inequality we have

$ a^3 b \plus{} a^3 c \plus{} abc\sqrt {bc} \plus{} abc\sqrt {bc} \ge 4a^2 bc$
$ b^3 c \plus{} b^3 a \plus{} abc\sqrt {ca} \plus{} abc\sqrt {ca} \ge 4b^2 ca$
$ c^3 a \plus{} c^3 b \plus{} abc\sqrt {ab} \plus{} abc\sqrt {ab} \ge 4c^2 ab.$

Adding them up we get the result.[/hide]

[hide="Solution 2"]Applying Schur Inequality we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \frac {1}{\sqrt {2}}( a(a \plus{} \sqrt {bc}) \plus{} b(b \plus{} \sqrt {ca}) \plus{} c(c \plus{} \sqrt {ab})) \\
\ge \sqrt {2}(\sqrt {ab}(a \plus{} b) \plus{} \sqrt {bc}(b \plus{} c) \plus{} \sqrt {ca}(c \plus{} a)) \ge \sqrt {2}(ab \plus{} bc \plus{} ca)$[/hide]

Solution 2 is from Zheng Fan from Shanghai High School.[/quote]
One more solution is
Squaring both sides, we can rewrite the inequality as
\[ \sum a^4 \plus{} 2\sum ab\sqrt {(a^2 \plus{} bc)(b^2 \plus{} ca)} \ge 2\sum a^2b^2 \plus{} 3\sum a^2bc
\]
By the Cauchy Schwarz, we have
\[ \sum ab\sqrt {(a^2 \plus{} bc)(b^2 \plus{} ca)} \ge \sum ab(ab \plus{} c\sqrt {ab})
\]
It suffices to show that
\[ \sum a^4 \plus{} 2\sum ab(ab \plus{} c\sqrt {ab}) \ge 2\sum a^2b^2 \plus{} 3\sum a^2bc
\]

\[ \Leftrightarrow \sum (a^4 \plus{} 2abc\sqrt {bc} \minus{} 3a^2bc) \ge 0
\]
By the AM-GM Inequality, we get
\[ a^4 \plus{} 2abc\sqrt {bc} \ge 3\sqrt [3]{a^6b^3c^3} \equal{} 3a^2bc
\]
The inequality is proved. :)

Actually, the stronger result is also true
\[ \sum a\sqrt {a^2 \plus{} bc} \ge (2 \minus{} \sqrt {2})(\sum a^2 \plus{} \sqrt {2}\sum ab)
\]
:)[/quote]

Do you have the solution for this one?  :)

[quote="shalex"]
Do you have the solution for this one?  :)[/quote]
Yes, I have. :)

[quote="shalex"]
Applying Schur Inequality we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \frac {1}{\sqrt {2}}( a(a \plus{} \sqrt {bc}) \plus{} b(b \plus{} \sqrt {ca}) \plus{} c(c \plus{} \sqrt {ab})) \\
\ge \sqrt {2}(\sqrt {ab}(a \plus{} b) \plus{} \sqrt {bc}(b \plus{} c) \plus{} \sqrt {ca}(c \plus{} a)) \ge \sqrt {2}(ab \plus{} bc \plus{} ca)$

Solution 2 is from Zheng Fan from Shanghai High School.[/quote]

Anyway, a typo here,

$ \sqrt {2}(\sqrt {ab}(a \plus{} b) \plus{} \sqrt {bc}(b \plus{} c) \plus{} \sqrt {ca}(c \plus{} a))$

should be

$ \frac {1}{\sqrt {2}}(\sqrt {ab}(a \plus{} b) \plus{} \sqrt {bc}(b \plus{} c) \plus{} \sqrt {ca}(c \plus{} a))$

[quote="shalex"]Prove that for all nonnegative real numbers $ a,b,c$ we have

$ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ca} \plus{} c\sqrt {c^2 \plus{} ab} \ge \sqrt {2} (ab \plus{} bc \plus{} ca).$

[/quote]

This one is from me, http://hi.baidu.com/shfdfzhjj/blog/item/991a61d9b8cbf9ed38012fd2.html :)


modular arithmetic
search
number theory unsolved
number theory

Prove that there exist such number $x$ that $x$ is not divisable with any $n^2,n\in N$ and numbers \[x+1, x+2,..., x+2004\] are all divisable with squares of some numbers.

CRT gives you the divisibility of $x+1$ etc by any squares you please --- but you need to ensure that $x$ is square-free: that's not a congruence condition.

Well, I don't know $C.R.T$ but now i am going to learn it, thanx a lot (can you post some nice link, from where to learn it  :lol:  :lol:)...

[quote="grobber"]You can even find prime $x$'s with the required property, but you need to employ Dirichlet's Theorem as well:

Pick any $2004$ primes $p_1,\ldots,p_{2004}$, none of them not dividing any of the numbers $1,2,\ldots,2004$, and determine (by the Chinese Remainder Theorem) $k$ s.t. $x\equiv k\pmod{p_1^2\cdot\ldots\cdot p_{2004}^2}\Rightarrow x\equiv -i\pmod{p_i^2},\ \forall i\in\overline{1,2004}$, and then take $x$ to be a prime with the property $x\equiv k\pmod{p_1^2\cdot\ldots\cdot p_{2004}^2}$ (such a prime number exists, by Dirichlet's Theorem, since our choices ensure that $(k,p_1^2\cdot\ldots\cdot p_{2004}^2)=1$).[/quote]

What do you mean by Dirichlet's theorem: In infinit aritmetic sequence there are infinit prime numbers...?
I never saw some problem where that can be used, and I am not sure that you can use that theorem in competition....
Nice solution, but I am looking for some more elementary, CRM OK but dirichlet theorem ;)

Yes, that's the theorem I was referring to. Where did you find the problem? I'd like to know that before I even attempt to solve it in an elementary manner. :)

Let $S$ be the set of positive integers $x$ such that $x + 1$, $x + 2$, ..., $x + 2004$ are not square-free.  By the Chinese Remainder Theorem, we know that $S$ is not empty.  Let $x_0$ be the smallest element of $S$.  Because $x_0 - 1$ is not in $S$ but $x_0$ is, we know that $x_0$ must be square-free.  Thus $x_0$ is the required value of $x$.

Thanx Ravi B nice sol.

Grobber, I found this problem in some papers from some trening, but they are for 9th grade....

As a follow-up, is there an elementary proof that every AP contains infinitely many square-free numbers if it contains two?

(bump)
[b]A[/b]rithmetic [b]P[/b]rogression. :)


calculus
integration
trigonometry
logarithms
algebra
partial fractions
calculus computations

[quote="mofidy"]Compute this integral:
\[\int{\sqrt{\tan (x)}dx}\]
[/quote]
It was already here. there was many many methods.

Yep.

Just set $u^{2}=\tan x$, and let the imagination blow up your mind.

Start by making the substitution $\tan x = \frac{1}{u^{2}}, x' = \frac{-2u}{1+u^{4}}$, and we get

[hide="Click to reveal hidden content"]$\int \sqrt{\tan x}dx =-2\int \frac{1}{1+u^{4}}du$.[/hide]

Finding this last primitive is a nice exercise, and you should get

[hide="Click to reveal hidden content"]$\int \frac{1}{1+u^{4}}du = \frac{1}{4\sqrt{2}}\ln\left(\frac{u^{2}+\sqrt{2}u+1}{u^{2}-\sqrt{2}u+1}\right)+\frac{1}{2\sqrt{2}}\arctan\left(\frac{\sqrt{2}u}{1-u^{2}}\right)+C$.[/hide]

Now you just have to multiply this by -2 and make the reverse substitution:

[hide="Final Answer"]$\int \sqrt{\tan x}dx =-\frac{1}{2\sqrt{2}}\ln\left(\frac{1+\sqrt{2 \tan x}+\tan x}{1-\sqrt{2 \tan x}+\tan x}\right)-\frac{1}{\sqrt{2}}\arctan\left(\frac{\sqrt{2 \tan x}}{\tan x-1}\right)+C$[/hide].

[quote="mofidy"]Compute this integral:
\[ I=\int{\sqrt{\tan (x)}\,dx}\]
[/quote]
First set a change of variables according to $ u =\sqrt{\tan x}$, the integral becomes to

$ \int\frac{2u^{2}}{u^{4}+1}\,du$

Nasty expression, but applyin' partial fractions yields

$ I =\int\frac{1}{2\sqrt2}\left(\frac{2u}{u^{2}-\sqrt 2u+1}-\frac{2u}{u^{2}+\sqrt 2u+1}\right)\,du$

Finally

\begin{eqnarray*}I & = &\frac{1}{{2\sqrt 2 }}\int{\left({\frac{{2u-\sqrt 2+\sqrt 2 }}{{u^{2}-\sqrt 2 u+1}}-\frac{{2u+\sqrt 2-\sqrt 2 }}{{u^{2}+\sqrt 2 u+1}}}\right)\,du}\\ & = &\frac{1}{{2\sqrt 2 }}\left({\ln\left|{\frac{{u^{2}-\sqrt 2 u+1}}{{u^{2}+\sqrt 2 u+1}}}\right|+\int{\frac{{\sqrt 2 }}{{\left({u-\frac{1}{{\sqrt 2 }}}\right)^{2}+\left({\frac{1}{{\sqrt 2 }}}\right)^{2}}}\,du}+\int{\frac{{\sqrt 2 }}{{\left({u+\frac{1}{{\sqrt 2 }}}\right)^{2}+\left({\frac{1}{{\sqrt 2 }}}\right)^{2}}}\,du}}\right)\\ & = &\frac{1}{{2\sqrt 2 }}\left({\ln\left|{\frac{{u^{2}-\sqrt 2 u+1}}{{u^{2}+\sqrt 2 u+1}}}\right|+2\left[{\arctan\left({\sqrt 2 u-1}\right)+\arctan\left({\sqrt 2 u+1}\right)}\right]}\right)+k,\,k\in\mathbb{R}\\ \end{eqnarray*}

Back substitute and we have the desired $ \blacksquare$


percent
geometry
rectangle
congruent triangles

1. A store charges $28 for a certain type of sweater. This price is 40 percent more than the amount it costs the store to buy one of these sweaters/ At an end of season sale, store employees can urchase any remaining sweaters at 30 percent off the store's cost. How much would it cost an employee to purchase a sweater of this type at this sale?

2. In rectangle ABCD, point E is the midpoint of line BC. If the area of quadrilateral ABED is 2/3, what is the area of rectangle ABCD?

[hide="1"]Isn't it just 28*0.7=19.60? [/hide]

[hide="2"]Draw lines ED, AE, and EF, where F is the midpoint of AD. There are four congruent triangles created. ABED is three of them. The are fo the rectangle is 4/3*2/3=8/9. [/hide]

I'm not sure I understand the wording on #1. When it says the store's cost, I assume it's referring to the 28 dollar price? If so, why does it bother to include the 40% increase in what the store pays for it? And if it's referring to the price that the store bought the sweater for, it isn't very clear...

#2. (Note: $[ABC...]$ denotes the area of polygon $ABC...$.) $E$ is the midpoint of rectangle $ABCD$, and $E$ is the midpoint of $BC$. If we take the midpoint of $AD$ and label it $F$, then $[ABEF]=[FCDE]=\frac{1}{2}\cdot [ABCD]$. $ED$ divides rectangle $ECDF$ in half, since it is a diagonal, which means that $[EDF]=[ECD]=\frac{1}{2}\cdot [ECDF]$. Therefore, we have $\frac{1}{2}\cdot [ABCD]+\frac{1}{2}\cdot \frac{1}{2}\cdot [ABCD]=\frac{3}{4}\cdot [ABCD]=\frac{2}{3}$. Therefore, $[ABCD]=\frac{8}{9}$.

I hope you understood that. If not, please tell me what needs clarification.

I think the 40% was just a trick? They do that a lot on tests, but not usually on SAT's.

Q1: let x be the cost of a sweater for the store.

then (28 - x)/x =0.4

at the end, you will find out x = 20

the answer: x * 70% = 14

Cheerful Coffin, welcome to AoPS! And your solution is correct.

[quote="woohooxd"]but the answer is 14.00[/quote]

Okay, so they mean that the employees can buy a sweater for 30% off the price that the store pays to buy the sweaters. You can find the price that the store pays, although it isn't necessary, but I'll explain it that way first. The store sells sweaters for 28 dollars, which is 40% more than they buy it for. Therefore, $1.4x=28$, where $x$ is the price the store pays. Solving that gives $x=20$, so the store pays 20 dollars. The employees get 30% off of that price, meaning that $y=.7\cdot x \Rightarrow y=14$, so the employees pay 14 dollars.

The other, quicker way of solving it is the following: $1.4x=28$, $.7x=y \Rightarrow 1.4x=2y \Rightarrow 2y=28 \Rightarrow y=14$. :)


In the accompanying figure $ \overline{CE}$ and $ \overline{DE}$ are equal chords of a circle with center $ O$. Arc $ AB$ is a quarter-circle. Then the ratio of the area of triangle $ CED$ to the area of triangle $ AOB$ is:
[asy]defaultpen(linewidth(.8pt));
unitsize(2cm);

pair O = origin;
pair C = (-1,0);
pair D = (1,0);
pair E = (0,1);
pair A = dir(-135);
pair B = dir(-60);

draw(Circle(O,1));
draw(C--E--D--cycle);
draw(A--O--B--cycle);

label("$A$",A,SW);
label("$C$",C,W);
label("$E$",E,N);
label("$D$",D,NE);
label("$B$",B,SE);
label("$O$",O,N);[/asy]
$ \textbf{(A)}\ \sqrt {2} : 1\qquad \textbf{(B)}\ \sqrt {3} : 1\qquad \textbf{(C)}\ 4 : 1\qquad \textbf{(D)}\ 3 : 1\qquad \textbf{(E)}\ 2 : 1$

Since arc $ AB$ is a quarter circle, arc $ AB$ measures $ 90$ degrees. Since $ \overline {CD}$ passes through the center, $ \overline {CD}$ is a diameter, making the measure of arc $ CD$ equal to $ 180$ degrees. Thus, the ratio is $ 2: 1$.

the triangle CED is clearly equivalent to two quarters of circle, thus the answer is (E)

We know that angle $ AOB\equal{}90^o$ because the arc $ AB$ is a quarter circle. Say the radius of the circle is $ r$. The area of $ AOB$ is $ \frac{r^2}{2}$. We know that $ CD$ is $ 2r$ because it is the diameter. We also know that angle $ CED\equal{}90^o$

because we have a triangel inscribed in a circle with one of the sides being the diameter. The height of triangle $ CED$ is $ r$ and the base is $ 2r$. The area of triangle $ \frac{2r^2}{2}\equal{}r^2$. Our answer is $ r^2: \frac{r^2}{2}\equal{}2: 1$. That is answer choice $ \mathbf{(E)}$.


Can anyone tell me,how many questions are there  in an one hour SAT physics test? :maybe:
Thanks beforehand :) !

There are 75 questions on the SAT Subject Test in Physics.

All of them are concept questions.  I don't think formulas helped me, except to give me an idea of what happens to one variable when another goes through [insert change].  Relativity is asked.  I took the subject test last May right before the Physics B exam, and relativity was really the only difference between what was covered, c'est à dire, if you've only had physics C, there's going to be more than E&M covered, and definitely no calculus is required.  There are some practice questions on collegeboard's website to give you an idea of how the subject test is.

How many questions did you  manage to answer during 60 minutes? :maybe:
I think the time is too short to answer 75 questions :maybe: .

I think I did all of them when I took it. But I know I didn't get all of them right. I think you can miss a significant number of questions and still get an 800. (I got 800.)

Part of the reason it's fast is that (as quantum leap said) a lot of the questions are concept questions, you don't have to do any calculations. A good strategy for a test where there is not enough time is to skip the problems that you know will take a lot of time and do the quicker ones first, so that you are sure to get all of the easy ones. Then go back and work on the ones that will take more time.

Because the designers of the SAT II tests can't be sure what high school curriculum test-takers have had, the tests are designed to have a lot of questions about very many topics. And then the scoring is such that most test-takers can miss several questions and still get the highest standard score of 800.

I took an old SAT II United States history test recently, for my own practice. (My son is taking an AP United States history course this year.) I missed about ten questions out of ninety, but still got the peak score of 800. This just shows that you don't have to be perfect to obtain the highest possible score.

Definitely most SAT I or SAT II tests are easier for persons who read English rapidly. I think you are a second-language reader of English, and that will be taken into account by the colleges to which you apply. If your specific test scores of English proficiency are high enough, and your other test scores are high, even if not the highest possible, that will be quite impressive. After all, it's difficult to take a test in a second language. I think you know that it is very competitive to get into the most famous United States universities if you are an international student, because so many international students apply to those universities, but you have some prospect of being admitted, so don't worry too much.

Good luck in your applications.

With what Tokenadult said, I took the SAT 2 Physics in October, and walking out of that test (never taken a physics course, just studied from a prep book) I was sure I had got at least 12 wrong, not counting miscalculations and careless mistakes. I ended up with a 780, so my best guess at the curve is around 12 raw score.

Sorry for reviving this very old thread. But since I intend to take a Physics Test on October 4th, I'd like to ask some questions. First, I just learn all of the background knowledge by myself throughout three main books: KAPLAN, BARRON and Schaum's College Physics Outline. I see that all of them have the same number of chapters which cover similar topics. But I don't know whether I just need to learn from those stuffs or need other materials? I really want to get the high score on this test but I seem to can't remember all of the abstract knowledge in such topics: light and sound motion, interference and diffraction, color and light,electric potential and potential difference. Wow! I know that's a big hole but I'm curious how many questions are asked about those topics?(just approximation). Does anybody take the test tell me how to fill up those holes?I can't remember all of the phenomenons.Tell me what should I do now is the best for preparation outside as well as inside the room test.

I guess it is not the best place,but I just don't want to create another topic,so I will ask right here:
Is there any electronic material,which can help to prepare for the SAT Physics Subject test?I mean e-book,some flash application or anything else.Also what is more preferable for you: to take a test in math 1 or math 2?I have never taken this test before,that's why I have no representation which of them is better and what are advantages of each of them.
Thanks in advance...

[quote="mechanicalpencil"]Math level 2 has precalc/trig, math level 1 just has algebra 2 and below. If you've taken precalc, take level 2, or if not, wait until you've taken precalc, because I've heard it's actually easier to get a very high score on level 2 if you've taken a high enough level of math. For the University of California (my state, so I know those requirements best), math level 1 doesn't count as one of your required subject tests, but I think it does for most other colleges.

The sparknotes website linked in the previous post also has stuff for the math subject tests.[/quote]
Thanks,this web-site is really awesome.
Could you explain a bit clearer what math level 1 contains besides algebra?

What is the University of California,you have mentioned in the text?Los Angeles,Berkeley,Davis or something else...

Maybe you or someone else know about requirements of other universities,like CalTech,Berkeley,Stanford.
I know that MIT requires either math 1 or math 2,but what's more preferable for their admission?

[url=http://en.wikipedia.org/wiki/University_of_California]University of California[/url] is a collection of the [url=http://www.universityofcalifornia.edu/campuses/welcome.html]following state schools in California.[/url] Stanford is a private school in Palo Alto, California (part of Silicon Valley). Mostly they recquire the basic tests as SAT/ACT, subject specific tests, essays, and then references of people who know you academically/socially for admission.

There will also be a [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]MIT Math Jam[/url] on 23rd of Octobre if you consider this school.

Caltech is a private university close to Los Angeles. You may want to consult the mathlinker [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=9911]Gjergji Zaimi[/url] for that matter. I think Caltech and MIT are at the same level, and actual choice is rather due to subject preference. I think this forum is a bit MIT-biased for some reason!  :D Of course, it depends on how much fin aid you will get, type of area you prefer etc. I mean discussing weather matters gives a hint where you should study as you may not want to study the whole day.  :wink:

[quote="orl"][url=http://en.wikipedia.org/wiki/University_of_California]University of California[/url] is a collection of the [url=http://www.universityofcalifornia.edu/campuses/welcome.html]following state schools in California.[/url] Stanford is a private school in Palo Alto, California (part of Silicon Valley). Mostly they recquire the basic tests as SAT/ACT, subject specific tests, essays, and then references of people who know you academically/socially for admission.

There will also be a [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]MIT Math Jam[/url] on 23rd of Octobre if you consider this school.[/quote]
Thanks,it was really helpful info.
What about CalTech?Do they have some specific requirements or preferences in SAT?What is the best choice among the mentioned universities,if I am going to have math major.Also please have in mind that I am international applicant.
It seems strange that I haven't heard anything about these classes yet?But it sounds really helpful in any case.
We are straying away from the subject,so I think,it makes sense to change title,what do you think mods?
How do you feel about "several advices conerning passing SAT and some info about American Universities"?

Well...probably nobody saw my last reply,so I will repeat:
Could you suggest me some top US universities,if I am international applicant,I would like to  have math major and it is preferable for me to pass TOEFL,not SAT...
Thanks in advance.
[i]Added after reading [b]orl's[/b] reply above:[/i]
Actually,I mainly consider MIT,but as an alternative option I will apply for other universities as well.As I said I would better choose math major and it is quite preferable for me to pass TOEFL,and MIT is the best place due to all these factors.
But I do really wonder if there exists any other such universities?

Regarding the test: don't worry about answering all of the questions. Before I took the test (last year), I took one practice test for the Physics subject test and found a comfortable number of questions I knew I could answer in the time without sacrificing accuracy -- I left about 6 questions blank and still got the score.


linear algebra
matrix
IMC
college contests

The 16th IMC 2009 will be in Budapest, Hungary on 25th - 30th July, 2009.
It is being co-organized by University College London and Eotvos Lorand University.

http://www.imc-math.org.uk/

Schedule:

First Day - 25th July - Arriving and Accommodation, Registration
Second Day - 26th July - Opening Ceremony, Additional Registration, Meeting of the Jury
Third Day - 27th July - First Exam Day
Fourth Day - 28th July - Second Exam Day
Fifth Day - 29th July - Closing Ceremony, Final Dinner
Sixth Day - 30th July - Departure

So it seems there will be no excursion day and therefore there will be a time for one coordination session only (instead of two in 2008, i.e. the first in the evening on excursion day and the second in the morning on closing day). :maybe:

do u know anything about the accommodation?

[quote="hana"]In our invitation letter is written this:

The participants will be accommodated in the student
residences BME Kármán Tódor Kollégium (Budapest, Irinyi J. u. 911, H1111)
and BME Schönherz Zoltán Kollégium (Budapest, Irinyi J. u. 42, H1111).

 :)[/quote]

Well, you seem to be better informed than me. :)
Anyway, here are some photos I found of the rooms:
Kármán dormitory: http://ghk.ktk.bme.hu/demoszoba.php
Schönherz dormitory: http://picasaweb.google.com/SaTi85/SchSzoba# (this one's a student's photos about his room)

Further information is available here: http://www.imc-math.org.uk/index.php?year=2009&item=travel

[b]kdano,[/b] would you be stayin in the dormitory too?  :)

Btw, I've found clear ncie and big  map of Budapest ~ 6mb, 4880x3710
Direct link: http://www.orangesmile.com/destinations/img/budapest-map-big.jpg
edit: it's was just a part of the city... noway you can find the XI district there =)
http://www.fsz.bme.hu/hungary/budapest/cgi-bin/search_tkp is bigger map and the Irinyi str is in the upper right corner of dictrict XI  :wink:

[quote="rustam"][b]kdano,[/b] would you be stayin in the dormitory too?  :)

Btw, I've found clear ncie and big  map of Budapest ~ 6mb, 4880x3710
Direct link: http://www.orangesmile.com/destinations/img/budapest-map-big.jpg[/quote]

No, I'm gonna sleep at home :)
If you want to use this map for sightseeing, it's quite alright, though it has some inconveniences:
First of all, none of the IMC-related buildings (the dormitories, the campus) are on this map :) they're simply more to the south.
Another one is that there's been a major reorganization in the public transport system. Couple of routes got a new number (like bus 78 is now 178 and 178A) , some got extended (like bus 5 now goes all the way through Rákóczi út), but eg. bus 7A no more exists, while others, like trams 4, 6 didn't change at all (neither did undergrounds :P). So don't expect the routes to be exactly like this :D

can someone attach the problems of IMC 2009 ?

can someone tell me this year cut off? any mathlinkers there?.. so we only have 10 problems this year.

i think, 2nd day problem 3 in the solution paper was already purposed before > :maybe:

thanks

[quote="ma_go"]it's funny how problem 3 from the second day changes from the competition paper to the solution paper ;)
which one is the real problem 3?[/quote]

That [i]is[/i] funny :P
The actual second day problems are that of the "solutions" file.

Another funny (?) thing noted by my friend:

The 2nd day Problem No 3:

The condition can be written in this way:

$ A(AB\minus{}BA)\equal{}(AB\minus{}BA)A.$

Then the statement of the problem is the special case ($ n\equal{}1$) of the Linear Algebra No 2
problem of Iran 2009 olympiad (15-17 July), see

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=291882[/url]

I also noticed that.
A very similar problem appears as an exercise in Matrix Analysis Book by Horn and Johnson.
Also the ISOM problem is a problem in V.V Prasolov, Problems and Theorems in Linear Algebra.

Even the idea has already discussed here in mathlinks a few times, one thread is (posted not long before IMC and ISOM )
http://www.mathlinks.ro/viewtopic.php?t=280822

and also the percentage of participants who solved this problem in IMC is relatively low.. :maybe:

During the competition I tried to solve first of all this problem because I have a feeling that I've seen it somewhere before. After 2 hours of struggling, I got 10 points worth solution  :lol:

Thanx a lot for mathlinks again!

Please: Mathematical Olympiad Challenges, 2.9, problem 11.

[quote="EUCLA"]Out of curiosity: anybody knows where will next IMC be held?  :D[/quote]

Probably Bulgaria (but this time not Blagoevgrad, somewhere also not too far from Sofia). At least in Budapest J.J. said that he thought about.

By the way in 2010 three olympiads for university students are supposed to be in Bulgaria - [url=http://www.seemous.eu/]SEEMOUS[/url] in Plovdiv (March 2010), Open National Olympiad [url=http://translate.google.com/translate?prev=hp&hl=en&js=y&u=http%3A%2F%2Fnsom.ru.acad.bg%2F&sl=bg&tl=en&history_state0=]"NSOM'[/url] in Veliko Tarnovo (May 2010) and probably IMC (July 2010?).


[hide="kind of messy but it is the solution"]If you notice, we have a cubic measurment of 12.5, and we are introducing a an element of 2.  So we need to cube the 2 and multiply it by 12.5 giving us 100.

If you dont see that, consider the following:

Let x be the side of the cube.

$x^3=volume=12.5$

If side is doubled we know have side = 2x

So:

$2^3 x^3=8x^3=100$[/hide]

[quote="MCrawford"]The volume of a cube is 12.5 cubic centimeters.  If each side is doubled, what is the number of cubic centimeters in the new volume?[/quote]

[hide="Quick and easy"]
Since a cube has three dimensions, doubling the side length multiplies the volums by 2:^3:, or 8.

12.5 x 8 = [b]100[/b]
[/hide]

[hide][color=indigo]
If the side is x, then the volume is x*x*x.
That means if the side is doubled the volume is 2x*2x*2x.
So the volume is increased by 8.
8*12.5=[/color][color=green]100[/color][/hide]


Posted before. To prove thi is equivalent to proving without the fourth powers I.E.

(ab+1)/(a+1)+.......>=4, in which case a triple aplication of harmonic mean finishes it off.

YOu first homogenixe by letting a=x/y b=y/z c=z/w d=w/x and the apply AMHM.

Bomb


anyone think the two most recent olymon (june july august)  are really hard?

i beasted the 1998 aime and the 1984 ahsme with near perfect scores just now (timed)... then i tried olymon and got owned on the hard questions

Good, then its not just me.  Yes, it seems harder than I remembered it being before.

The questions aren't in increasing diffuculty, right?


modular arithmetic
algorithm
number theory
algebra unsolved
algebra

7a ≡ 1 (mod 30)  I have the resolution of the problem but i dont understand the first step.

7a ≡ 1 (mod 30) <--> [color=red]7a ≡ 91 (mod 30)[/color] <--> 7a ≡ 7.13 (mod 30)

And using that (7:30)=1 its like saying a ≡ 13 (mod 30), then

a=30q + 13

So i dont understand where the 91 comes from, i dont see it.

[color=red]7a ≡ 91 (mod 30)[/color]

Thanks N.T.TUAN, so have then to look at the properties of cngruence and because:

a ≡ b (m)  --> b ≡ a (m)

and trnasitivity

a ≡ b (m) and b ≡ c (m) --> a ≡ c (m)

That would be the theory to explain right?

What can i use to find such a number as this 91 but for other exercises, is there any algorithm or just by "eye" ?

For $ax\equiv b\pmod{m}$, where $\gcd{(a,m)}=1$, we can slove same as before. Find $c$ such that $ca\equiv 1\pmod{m}$.

[quote="sanchez45"]Thank you very much! :D[/quote]
No problem, I think you need a text book on Number Theory ( rank Olympiad, if you want to *MO ) . Example I.Niven or  D.Burton.

Actually i have a book in spanish from a profesor of my university but i am studying this by myself.


number theory
greatest common divisor
least common multiple

I suddenly realise that the scope of problems based on number system is much larger than i thought initially... now what should i know to strengthen my fundamentals....

What are number systems? Could you give an example? Is it like number theory? :maybe:

EDIT: Or are you refering to bases?

Number Theorey and all... lol.... Number system just sounds weird... I just realised it! GCD, LCM,Digits, and so on...


geometry
geometric transformation
rotation
Pythagorean Theorem

Let A, B fixed in the Euclidean plane.  find the geometric locus of  all points M in the plane satisfying the equality MA^2 - MB^2 = k, where k is some real constant

Please Help me i don't know where to start

Put a line through the points $A, B$, pick an arbitrary 3rd point $P$ on this line and erect a normal at this point. Now, pick an arbitrary point $M$ on the normal and imagine 2 squares constructed on the segments $MA$ and $MB$. By Pythagorean theorem, the 1st square has the side $MA = \sqrt{PA^2 + MP^2}$ and the area $MA^2 = PA^2 + MP^2$. Likewise, the 2nd square has the side $MB = \sqrt{PB^2 + MP^2}$ and the area $MB^2 = PB^2 + MP^2$. The difference of the areas of these 2 squares is

$MB^2 - MA^2 = (PB^2 + MP^2) - (PA^2 + MP^2) = PB^2 - PA^2$

Does this difference depend on the position of the point $M$ on the normal at $P$ ? So, what is the locus ? What remains to be done is to find the position of the point $P$ (the foot of the normal), given the segment $AB$ and some real number $k$. Obviously,

$k = PB^2 - PA^2 = (PA + AB)^2 - PA^2 =$

$= PA^2 + 2 \PA \cdot AB + AB^2 - PA^2 = AB \cdot (2PA + AB)$

$PA = \frac 1 2 \left(\frac{k}{AB} - AB\right)$

Start by assuming that A and B are both points on the x-axis. Let $A = (-a,0)$ and $B = (a,0)$ for some number $a$. So you are looking for the set of points $(x,y)$ such that $[(x+a)^2 + y^2] - [(x-a)^2 + y^2] = k$. Try and solve this for x and y.

Once you have that, you pretty much have your solution - because every other placement of A and B will just be translations and rotations of that solution.


ceiling function
number theory unsolved
number theory

Find the number of elements that a set $B$ can have, contained in $(1, 2, ... , n)$, according to the following property: For any elements $a$ and $b$ on $B$ ($a \ne b$), $(a-b) \not| (a+b)$.

Obviously we cannot choose any numbers whose difference is 1 or 2, so an upper bound is $\lceil{\frac{n}{3}}\rceil$. This is also attainable; just choose all numbers that are 1 mod 3.


induction
modular arithmetic
number theory proposed
number theory

Show that there are an infinity of positive integers $n$ such that $2^{n}+3^{n}$ is divisible by $n^{2}$.

i) $a^{2}|2^{a}+3^{a}$ and $b^{2}|2^{b}+3^{b}\Longrightarrow (ab)^{2}|2^{ab}+3^{ab}$

ii)5 is 'good' and 55 is 'good'

using this 2 facts, we have one solution, because we easily see that the numbers $5^{k}55^{l}$ are 'good's'.

[quote="TomciO"]I don't get why i) is true for $a, b$ not being coprime. If it would be true for all $a, b$ you could use just $5$ and you would have $5^{2k}|2^{5^{k}}+3^{5^{k}}$ but that's false. Am I wrong?[/quote]
You are right $5^{2k}\not |2^{5^{k}}+3^{5^{k}}$ for $k>1$, but $55^{2k}|2^{55^{k}}+3^{55^{k}}$.

So it probably follows from the induction easily (I'm too lazy to check) and we are done, but it doesn't follow from i), I think.

[quote="Rust"]but $55^{2k}|2^{55^{k}}+3^{55^{k}}$.[/quote]

It seems to me that this is false too. Look,

we know that if $\parallel a-b\parallel _{p}=k \geq1$ then $\parallel a^{n}-b^{n}\parallel _{p}=k+\parallel n\parallel _{p}$; but assuming that your clame is correct we should have that $2k \leq \parallel 2^{55^{k}}-(-3)^{55^{k}}\parallel _{5}=\parallel 2+3\parallel _{5}+\parallel 55^{k}\parallel _{5}=1+k$ so $k=1$????But you clamed that $55^{2k}|2^{55^{k}}+3^{55^{k}}$ for all positive integer $k$s,don't you?

[hide]
We know there exists a positive integer solution, since 1 clearly works. Suppose there are finitely many, then there exists a largest integer $M$ for which this property holds.

For all positive integers $M$, we have $2^{M}+3^{M}>M^{2}$. (In fact $3^{M}>M^{2}$ which is easliy verifiable through induction and $3M^{2}>(M+1)^{2}$ for $M\geq 2$.). Therefore, there exists a prime $p$ such that $2^{M}+3^{M}$ is divisible by $pM^{2}$. We will now show that $2^{pM}+3^{pM}$ is divisible by $(pM)^{2}$, which will give us the contradiction we need.

To prove the above assertion, it's clear that $p,M$ are both odd and not divisible by 3 in order to have $2^{M}+3^{M}$, an odd number not divisible by 3, divisible by $pM^{2}$. We now have
\[2^{pM}+3^{pM}=(2^{M}+3^{M})\left(\sum_{k=0}^{p-1}(-1)^{k}2^{kM}3^{(p-1-k)M}\right) \]
It remains to prove that the second factor in the right hand side is divisible by $p$. Observe that $2^{M}\equiv-3^{M}\pmod{p}$, so that
\[(-1)^{k}2^{kM}3^{(p-1-k)M}\equiv 3^{(p-1)M}\equiv 1\pmod{p}\]
Then the right hand sum, consisting of $p$ terms all congruent to 1 modulo $p$, is divisible by $p$, as desired.
[/hide]

We will construct the infinite sequence $ {n_{k}}$ such that for any member $ n_{k}$ of this sequence satisfies the condition:
$ 2^{n_{k}} + 3^{n_{k}}$ is divisible by $ n_{k}^{2}$. Easy to verify that $ n_{1} = 1$ and $ n_{2} = 5$. Assume we constructed it until $ n_{k}$. We will easily find $ n_{k + 1}$. Let $ 2^{n_{k}} + 3^{n_{k}} = an_{k}^{2}$,where $ a$ is odd positive integer(as $ n_{k} > = n_{2} = 5$, easy to show that $ 2^{n_{k}} + 3^{n_{k}} > n_{k}^{2}$) and $ a > 1$. Take $ p$ the odd prime divisor of $ a$. Substitute $ 2^{n_{K}}$ by $ x$ and $ 3^{n_{k}}$ by $ y$. Then, obviously ${ x^{p} + y^p} = (x + y)(x^{p - 1} - yx^{p - 2} + ... + y^{p - 1})$. As $ x + y$ is divisible by $ pn_{k}^{2}$,
if we define $ r$ to be the reminder of $ x mod p$, than the reminder $ y mod p$ will be $ - r$. So, $ x^{p - 1}, - yx^{p - 2},...,y^{p - 1}$ numbers have the same reminder if we divide them by $ p$, but the number of them is $ p$. So, $ x^{p - 1} - x^{p - 2} + ... + y^{p - 1}$ is divisible by $ p$. It follows
$ x^{p} + y^{p}$ is divisible by $ (pn_{k})^{2}$. Now we take $ n_{k + 1} = pn_{k}$. As $ p > 1$, $ n_{k + 1} > n_{k}$ and
$ 2^{n_{k + 1}} + 3^{n_{k + 1}}$ is divisible by $ n_{k + 1}^{2}$, which completes our proof.
P.S. That's not my own solution. I have read this solution before.

The official solution was(i'm sorry,if it was posted before):
If $ n_k$ is the number,such that $ n_k^2|3^{n_k}\plus{}2^{n_k}$ then $ n_{k\plus{}1}\equal{}\frac{3^{n_k}\plus{}2^{n_k}}{n_k}$ also satisfies to the given condition.
So by taken $ n_1\equal{}1$ it follows that $ n_2\equal{}5,n_3\equal{}55\dots$ and we are done.

Obviously $5$ satisfy $n^2 | 2^n + 3^n$.
Now, we'll prove that for each odd number $a$ satisfy, we could construct another larger odd number satisfy.
Take an odd prime $q | 2^a + 3^a$ that hasn't been used before. This must exists due to [b]Zsigmondy Theorem[/b].
We'll now prove $aq$ satisfy.
Now, notice that our hypothesis gives us
\[ a^2 | 2^a + 3^a | (2^a)^q + (3^a)^q \]
Moreover, $v_q (2^{aq} + 3^{aq}) = v_q (2^a + 3^a) + v_q (q) \ge 2 $.

Using this method, one could construct arbitrarily long sequence $a_n$ starting with $a_1 = 5$, and applying the strategy above to find $a_n$ by using $a_{n - 1}$ we're then finished.


function
inequalities
algebra unsolved
algebra

let function $f$:[0,1]-R such that:

$f(1/n)$=$(-1)^{n}$, :forall  $n$=1,2..

Prove that: no exists 2 functions creasing $h(x)$,$g(x)$:[0,1]-$R$  such that:
$f(x)$=$g(x)$-$h(x)$  :forall $x$ :in [0,1]

$var(f(x))=var(g(x))+var(h(x))=\infty$, therefore or $var(h(x))=\infty$ or $var(h(x))=\infty .$ It  give contradition for increase or decrease function.

Official solution:
Suppose that there exist such functions $g$ and $h$. Then for any $k\in\mathbb{N}$ we have \[g\left(\frac1{2k+1}\right)= f\left(\frac1{2k+1}\right)+h\left(\frac1{2k+1}\right)=-1+h\left(\frac1{2k+1}\right)<-1+h\left(\frac1{2k}\right)= 1+h\left(\frac1{2k}\right)-2= f\left(\frac1{2k}\right)+h\left(\frac1{2k}\right)-2= g\left(\frac1{2k}\right)-2< g\left(\frac1{2k-1}\right)-2.\] Applying this inequality we get \[g(0)< g\left(\frac1{2k+1}\right)< g\left(\frac1{2k-1}\right)-2< g\left(\frac1{2k-3}\right)-4< \ldots< g\left(\frac13\right)-2(k-1)< g(1)-2k.\] Thus $g(1)-g(0)>2k$ for any $k$, which is impossible.
@[b]Rust[/b]: maybe it's not a clever question, but what is $var$?


consider this problem,
problem:there are unknown number of doors for a secret place.Each door has 4 different locks.There are 8 people posessing keys to locks in such a way that any combination of 4 people can open all keys of all doors.
                                                     what is the minimum number of doors? and how many keys?

so for any combination of 3 of them they can't open the door .
so we have $\binom{8}{3}=56$ locks so we have $\frac{56}{4}=14$ doors. ;)

Thank you a_vakilian.The solution is incomplete in two aspects,so let me complete the formalities.
(1)First let me prove that "for every group of three people there exists a unique  lock which cant be opened".
    On contrary,let there be a lock that cant be opened by atleast two groups of three people.This means if a       person from one of the two groups joins the other group , the four cant still open the lock. This is contradiction.
(2)The number of locks is exhaustive.
    Let us say there may exist some more lock(s).If it is not the lock already enumerated then it must be such that three can open it.Again contradiction.