Prove that

$ \frac{3\sqrt{3}}{4}\leq \frac{yz}{x(1\plus{}yz)}\plus{}\frac{zx}{y(1\plus{}zx)}\plus{}\frac{xy}{z(1\plus{}xy)}\leq \frac{x\plus{}y\plus{}z}{4}$,

where $ x,y,z$ are positive reals such that $ x\plus{}y\plus{}z\equal{}xyz$

[quote="outback"]Prove that

$ \frac {3\sqrt {3}}{4}\leq \frac {yz}{x(1 \plus{} yz)} \plus{} \frac {zx}{y(1 \plus{} zx)} \plus{} \frac {xy}{z(1 \plus{} xy)}\leq \frac {x \plus{} y \plus{} z}{4}$,

where $ x,y,z$ are positive reals such that $ x \plus{} y \plus{} z \equal{} xyz$[/quote]
$ \sum_{cyc}\frac {yz}{x(1 \plus{} yz)}\leq\frac {x \plus{} y \plus{} z}{4}\Leftrightarrow\sum_{cyc}\frac {yz}{x\left(\frac {xyz}{x \plus{} y \plus{} z} \plus{} yz\right)}\leq\frac {(x \plus{} y \plus{} z)^2}{4xyz}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {yz}{2x \plus{} y \plus{} z}\leq\frac {x \plus{} y \plus{} z}{4}\Leftrightarrow\sum_{cyc}\left(\frac {x}{4} \minus{} \frac {yz}{2x \plus{} y \plus{} z}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {2x^2 \plus{} xy \plus{} xz \minus{} 4yz}{2x \plus{} y \plus{} z}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(x \minus{} y)(x \plus{} 2z) \minus{} (z \minus{} x)(x \plus{} 2y)}{2x \plus{} y \plus{} z}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x \minus{} y)\left(\frac {x \plus{} 2z}{2x \plus{} y \plus{} z} \minus{} \frac {y \plus{} 2z}{2y \plus{} x \plus{} z}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x \minus{} y)^2(2z \plus{} x \plus{} y)(x \plus{} y \minus{} z)\geq0.$
Let $ x\geq y\geq z.$
Hence, $ S_z \equal{} (2z \plus{} x \plus{} y)(x \plus{} y \minus{} z)\geq0,$ $ S_y\geq0$ and
$ \sum_{cyc}(x \minus{} y)^2S_z\geq(x \minus{} z)^2S_y \plus{} (y \minus{} z)^2S_x\geq\frac {x^2}{y^2}(y \minus{} z)^2S_y \plus{} (y \minus{} z)^2S_x \equal{}$
$ \equal{} \frac {(y \minus{} z)^2(x^4 \plus{} x^3y \minus{} 4x^2y^2 \plus{} xy^3 \plus{} y^4 \plus{} z(z \plus{} 2x \plus{} 2y)(x^2 \plus{} y^2))}{y^2}\geq0.$

For  $ LP$ it remains to prove that $ 3\sum_{cyc}\frac{z^2xy}{(2x\plus{}y\plus{}z)(2y\plus{}x\plus{}z)}\geq\frac{27xyz}{16(x\plus{}y\plus{}z)},$
which is equivalent to $ \sum_{cyc}(14x^3\plus{}x^2y\plus{}x^2z\minus{}16xyz)\geq0.$
Done!


induction
combinatorics proposed
combinatorics

Some $ n > 2$ lamps are cyclicaly connected: lamp $ 1$ with lamp $ 2$, ... , lamp $ k$ with lamp $ k \plus{} 1$, ... , lamp $ n \minus{} 1$ with lamp $ n$, lamp $ n$ with lamp $ 1.$ At the beginning all lamps are off. When one pushes the switch of a lamp, that lamp and the two ones connected to it change status (from off to on, or vice-versa). Determine the number of the configurations of lamps reachable from the initial one, through some set of switches being pushed.

[i]Dan Schwarz[/i]

There are $ 2^n$ ways of switching the lamps. We denote every such way by a bit sequence $ x_1,\ldots,x_n$ with $ x_i\equal{}1$ meaning "lamp $ i$ is switched" and $ x_i\equal{}0$ meaning "lamp $ i$ is not switched".
Note that lamp $ i$ is switched on, iff $ x_{i\minus{}1}\plus{}x_i\plus{}x_{i\plus{}1}$ is odd.

Now let us investigate to distinct sequences $ x_1,\ldots,x_n$ and $ y_1,\ldots,y_n$ that lead to the same configuration.
If $ x_i\ne y_i$ holds for all $ i$, we get a contradiction (since then all lamps are switched differently).
Hence $ x_i\equal{}y_i$ must hold for some $ i$, and since the sequences are distinct also $ x_j\ne y_j$ must hold for some $ j$. Hence there exist consecutive indices $ i$ and $ i\plus{}1$ such that  $ x_i\equal{}y_i$ and $ x_{i\plus{}1}\ne y_{i\plus{}1}$. Assume wlog that $ i\equal{}1$.

Then $ x_1\equal{}y_1$ and $ x_2\ne y_2$. Since $ x_1\plus{}x_2\plus{}x_3$ and $ y_1\plus{}y_2\plus{}y_3$ have the same parity, this implies $ x_3\ne y_3$.
An easy induction shows that $ x_{3k\plus{}1}\equal{}y_{3k\plus{}1}$ and $ x_{3k\plus{}2}\ne y_{3k\plus{}2}$ and $ x_{3k}\ne y_{3k}$ for all $ k$.
If $ n$ is not a multiple of 3, this yields a contradiction.

If $ n$ is a multiple of 3, then the pattern closes and yields two equivalent configurations.
There are only two other ways of getting sequences that yield identical configurations:
$ x_1\ne y_1$, $ x_2\equal{}y_2$, and $ x_3\ne y_3$ (and repeat in blocks of three).
$ x_1\ne y_1$, $ x_2\ne y_2$, and $ x_3\equal{}y_3$ (and repeat in blocks of three).
In fact that the third way corresponds follows by putting the first and second way in sequence.
Hence, every sequence allows 2x2=4 equivalent sequences.

The answer is:
$ 2^{n\minus{}2}$ if $ n$ is a multiple of 3.
$ 2^n$ if $ n$ is not a multiple of 3.


function
trigonometry
limit
induction
algebra unsolved
algebra

Find all functions $f: Q_+ \to  Q_+$  satisfy: $f(x+1)=f(x)+1$ and $f(x^k)=f^k(x)$ where $k \geq 2$ be natural number.<$Q_+$ be set of positive rational numbers>

I hope it works:

$f(x+1)=f(x)+1$ give us :

$\frac{f(x+1)-f(x)}{(x+1)-x}=1$. And this implies that $f(x)$ is a affine function. So, $f(x)=x+b$.

But we have $(x+b)^k=f(x)^k=f(x^k)=x^k+b$
So, $b=0$

So, $f(x)=x$.

[quote="Arne"][quote="Labiliau"]
$\frac{f(x+1)-f(x)}{(x+1)-x}=1$. And this implies that $f(x)$ is a affine function[/quote]

Why is this "immediate"? :)[/quote]

He has a point - for example, $f(x) = x + \sin \frac{x}{2\pi}$ also satisfies this property.   :)

Edit:

In the second condition, set $x = 1$ to get $f(1) = f(1)^k$, so $f(1) = 0, 1$, which means $f(n) = n, n-1, n \in \mathbb{N}$.  Now, $f(n^k) = f(n)^k = n^k, n^k - 1 = n^k, (n-1)^k$ so it is clear $f(n) = n$ for naturals.

It should be equally clear that $f(x)$ must be of the form $f(x) = x + g(x)$ where $g(0) = 0$ and $g(x+1) = g(x)$.  Moreover, we have $f(x^k) = x^k + g(x^k) = (x + g(x))^k \implies g(x^k) = (x + g(x))^k - x^k$.

Suppose $|x| < 1$.  Then $\lim_{k \to \infty} g(x^k) = \lim_{k \to \infty} g(x)^k + k g(x)^{k-1} x + ... = g(0) = 0$, so $|g(x)| < 1$ for $|x| < 1$, and because $g(x)$ has period $1$ we know $|g(x)| < 1$ for all $x$.

Assume $g(x) \neq 0$.  Then for $x \gg 1$ we have $g(x^k) = g(x)^k + k g(x)^{k-1} x + ... < 1$, however, because $x$ is very large, we can choose a $k$ such that the RHS is $> 1$ - contradiction - hence, $g(x) = 0$ for all $x$ and our function is $f(x) = x$.


[quote="13375P34K43V312"]1337 has 1337 posts at the beginning of 11/26, and grobber has 7500 posts. Grobber has 5 posts per day and 1337 has 6 posts per day. At this rate, when will 1337 have as many posts as grobber?[/quote]
[hide]$7500-1337=6163$[/hide]

[hide]So October 9th, 2023? I think that is 6163 days afterwards but it might be a little off. Or a lot off. I dunno  :maybe:[/hide]

oop

[hide]
a gain on average of 1 post
so...7500-1337=6163
6163 days
6163/365=16 323/365
and subtract 4 days because of 4 leap years
so 16 years and 319 days
we know it'll be in 2023 + 283 days

So, October 10th

Final Answer:
Tuesday, October 10th, 2023
[/hide][hide]
I might be off by a day or two, depending on your point of view on how to count the 26th of november :) :P [/hide]

-jorian

[quote="jhredsox"]Tomorrow :P
-jorian[/quote]

And when is tommorow? Can you guys please be a bit more precise.

Also, how are you supposed to answer the problem? With a date?

I don't know...however YOU interperate the problem?

You could show no effort and say its in x amount of days if u wanted to :P

-jorian


find all pairs $ (x,y)$ in $ R^2$ such that :

$ (x^2\plus{}y^2\minus{}x)(xy\minus{}1)^2\plus{} \sqrt{y^2\minus{}x^2} \equal{}0$  :(

Look at the equation, we see $ \sqrt{y^2\minus{}x^2} >\equal{}0$ and $ (xy\minus{}1)^2 >\equal{}0$ for every pair of $ x,y$.
So the problem is we are not sure about $ (x^2\plus{}y^2\minus{}x)>\equal{}0$ or not.
If $ x<\equal{}0$. Then that thing is true.
If $ x>\equal{}1$. Then $ x^2>x$ so that thing true.
Hence we have $ (x^2\plus{}y^2\minus{}x)(xy\minus{}1)^2 >\equal{}0$  and $ \sqrt{y^2\minus{}x^2} >\equal{}0$ But the equation need to equal to zero, Hence you have $ (x^2\plus{}y^2\minus{}x)(xy\minus{}1)^2\equal{}0$ and $ \sqrt{y^2\minus{}x^2}\equal{}0$
$ (\frac{1}{2},\frac{1}{2}) ; (1, 1)$

If $ 0<x<1$. I don't know how to do  :D

in fact , in my partial proof , it remains the case : $ (x,y) \leq \frac {1}{2}$  , and I proved that : $ y^2\geq x^2 \geq \frac{y^2}{2}$  :(


calculus
function
Asymptote
derivative
calculus computations

hi, everyone.  I was reading my calculus textbook. On my text book, it says a function is not differentiable if it has a vertical tangent.  I am not quiet understand what vertical tangent means.

 If anyone here can explain to me what vertical tangent is, I would really appreciate it.   Thank you very much.

Consider, for example, the line tangent to $y^3=x$ at $x=y=0$.  Basically, it's where the derivative goes to $\infty$.

I got it.  Thanks again for all the replies.


algebra
function
domain
factorial
complex numbers
Ring Theory
number theory proposed

Let $ Z[i]$ denote the set of complex numbers of the form $ a\plus{}bi$, where $ a,b$ are integers.

Let $ a,b,c,d\in Z[i]$ be such that $ ab\equal{}cd$. Prove that there exist $ w,x,y,z\in Z[i]$ such that $ a\equal{}wx, b\equal{}yz, c\equal{}wy, d\equal{}xz$.

[quote="nguyentrang"]Let $ Z[i]$ denote the set of complex numbers of the form $ a \plus{} bi$, where $ a,b$ are integers.

Let $ a,b,c,d\in Z[i]$ be such that $ ab \equal{} cd$. Prove that there exist $ w,x,y,z\in Z[i]$ such that $ a \equal{} wx, b \equal{} yz, c \equal{} wy, d \equal{} xz$.[/quote]

Same argument as in http://www.mathlinks.ro/Forum/viewtopic.php?t=150434 for $ \mathbb{Z}\left[i\right]$ instead of $ \mathbb{Z}$ - here you use that $ \mathbb{Z}\left[i\right]$ is an Euclidean domain (though factorial domain is already enough).

  darij

How is $ \gcd$ defined in $ \mathbb{Z}\left[i\right]$? (Sorry i'm not familiar)

A $ \gcd$ of some set $ \{a_1, a_2, ...\}$ of numbers (elements of a ring) is defined to be a number $ d$ such that $ d$ divides all $ a_i$ and for any $ e$ dividing all $ a_i$ we have $ e$ dividing $ d$.


What is your favorite type of pizza? Mine is cheese+olives

Unless meetza is cheese + olives, cheese is my fave. Cheese.

4 choices only? come on...

I dunno...I really like pepperoni + pineapple actually.

I am reminded of a quote from a book, [i]The Last Continent[/i], by Terry Pratchett:

[quote][There are such things as] platonic burgers made of beef instead of cow lips and hooves.  There are fish 'n' chips where the fish is more than just a white goo lurking at the bottom of a batter casing and you can't use the chips to shave with.  There are hot dog fillings which have more in common with meat than mere pinkness, whose lucky consumers don't apply mustard [i]because that would spoil the taste[/i].  It's just that people can be trained to [i]prefer[/i] the other sort, and seek it out.  It's as if Machiavelli had written a cookery book.

Even so, there is no excuse for putting pineapple on pizza[/quote]

 :D

I don't understand why nobody thinks of the combination pepperoni+mushrooms.

Nevertheless, I like CPK's pizzas the best.

how about spinach...or...canadian bacon+pineapple

Mmmm...Peperonni, double sausage, green pepper, mushrooms, ham, and sweet onions. But NOT ANCHOVIES. Especially ones eaten in AMC problems and that feel like tricking test-takers.

meetza is like a pizza with tons of meat toppings.   :blush: I forgot hawaiian, whoops

My favorite is green pepper, usually in a combination of ham or pepperoni.

cheese, cheese, and cheese

I have nothing against, pepperoni or beef

pinapples! pineapplew! pineapples!

yum! :P

veggie with lots of cheese....yummmmmm....

 :lol:

Cheese, Pepperoni, Chicken, they are all good but it's the pizza that matters most :D

elephant pizza is really delishes.

i m not kidding. ;)

we have elephant pizza in iran.


inequalities
number theory unsolved
number theory

Prove that for any integer $ n>1$ the sum $ S$ of all divisors of $ n$ (including $ 1$ and $ n$)satisfies the inequalities:
$ k\sqrt{n}<S<\sqrt{2k}n$, where $ k$ is the number of divisors of $ n$.

from the formula
$ \sum_{d|n} \frac{1}{d}\equal{}\frac{\sigma(n)}{n} \implies n.\sum_{d|n} \frac{1}{d}\equal{}\sigma(n), n.\frac{k^2}{\sigma(n)} \le n.\sum_{d|n} \frac{1}{d}\equal{}\sigma(n) \implies \text{the left side of the inequality!}$
note that since $ n>1$ the inequality becomes strict.
for the right side we have :
we must prove that :
$ \frac{\sigma(n)}{n}\equal{}\sum_{d|n} \frac{1}{d}<\sqrt{2k}$.
but we obviously have
$ \sum_{d|n} \frac{1}{d} < \sum_{i\equal{}1}^{k} \frac{1}{i}$
but we have
$ \frac{\sum_{i\equal{}1}^{k} \frac{1}{i}}{k}< \sqrt{\frac{\sum_{i\equal{}1}^{k} \frac{1}{i^2}}{k}}<\sqrt{\frac{2}{k}} \implies \sum_{i\equal{}1}^{k} \frac{1}{i}<\sqrt{2k}$.
we r done!


MATHCOUNTS Gauss FTW ARML AMC USA(J)MO USAMO

I know!!!! MD is gonna be pretty tough to beat this year, mostly because of Hyun and Zbarsky who both made nats cd last year. :o

Sigh gauss your statement is somewhat funny, for a specific reason.

Also, if you want to get enough posts for ftw, just post in Missouri. Nobody cares.

Too bad Linus Hamilton's leaving.  Otherwise, we'd really pwn

May has National Matchounts, ARML, Science Bowl, ACSL, and (technically end of April) USAMO   :(

Who is xpmath? Also, bob, you did very well. 6th out of ~150 is a good place.

[quote="bobma99"]NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
i didnt make like most of those -.-
sniff sniff really hoping to make nationals this year -.-
SOOOO CLOSE just had to make so many retarded mistakes and get 6th.
and i think i coulda made usamo this year too -.-
emo corner time[/quote]

I know how you feel. I got 7th. Blech.
Emo corner for me too. Well I won't exactly cut myself, but sad time.

[quote="cpast"]What grade are you in? You can only be emo if you're in 8th or if there will be 4 amazing people next year, leaving you no chance.[/quote]
well i'm in 8th so i'm gonna have to go emo...
of course without cutting myself like jonathan chou

[b]jonathanchou711 wrote:[/b][quote]jonathanchou711
[quote]bobma99 wrote:
NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
i didnt make like most of those -.-
sniff sniff really hoping to make nationals this year -.-
SOOOO CLOSE just had to make so many retarded mistakes and get 6th.
and i think i coulda made usamo this year too -.-
emo corner time[/quote]

I know how you feel. I got 7th. Blech.
Emo corner for me too. Well I won't exactly cut myself, but sad time.[/quote]
I have 3 words for you guys. Let. It. Go. Yes, I feel sorry for you, especially since screwed up at state and got 5th, but at some point you just have to [b]let it go and move on[/b]. Especially if this isn't your last year.


Dear Pohoatza and Mathlinkers,
Khoa Lu Nguyen has proved in Message 216910, that II' is orthogonal to the orthotransversale of I in the general case when I is a general point P.
With my response to http://www.mathlinks.ro/Forum/viewtopic.php?t=216910, the conclusion follows.
Sincerely
Jean-Louis

[quote="Leonhard Euler"]Let $ I^'$ be the isogonal conjugate $ I$ with respect to triangle $ A^'B^'C^'$ and $ X,Y,Z$ be the projection of $ I^'$ on $ B^'C^',C^'A^',A^'B^'$. We have $ YZ\perp A^'I$,or, equivalently, $ YZ\perp AI$. Denote $ I_aI_b,I_c$ as excenter of angle $ A,B,C$. From the fact that $ I_bI_c\perp AI$ and together with $ YZ\perp A^'I$, we get $ I_bI_c\parallel YZ$ and similarly $ I_cI_a\parallel ZX,I_aI_b\parallel XY$. So triangle $ I_aI_bI_c$ and $ XYZ$ are homothety.  By applying fact that $ I_aO\perp B^'C^'$([url]http://www.mathlinks.ro/viewtopic.php?t=168196[/url]), we get $ I_aO\parallel XI^',I_bO\parallel YI^',I_cO\parallel ZI^'$ implying $ O,I^'$ are corresponding points of $ \triangle I_aI_bI_c\sim \triangle XYZ$. Let $ P,Q$ be the circumcenter of triangle $ I_aI_bI_c$ and $ XYZ$. By homothety of triangle $ I_aI_bI_c$ and $ XYZ$, $ OP\parallel I^'Q$. Since $ Q$ is midpoint of $ I^',I$, $ I^',Q,I$ are collinear and $ OP\parallel I^'Q$ become $ OP\parallel I^'I$. Now it remain to proving $ O,P,I$ are collinear but it is clear since $ O,P,I$ are Nine-Point center,circumcenter,orthocenter of triangle $ I_aI_bI_c$, respectively that lying on the Euler line of triangle $ I_aI_bI_c$.[/quote]

$ I_aO\perp B^'C^'$ can be proved by lemma: Let $ D,E$ be the points on ray $ AC$ beyond $ C$ and ray $ AB$ beyond $ B$. Prove that $ I_aO\perp DE$. This lemma is extraversion of http://www.mathlinks.ro/Forum/viewtopic.php?t=147540. By this lemma, it is suffice to show $ DE\parallel B^'C^'$ which is directly from Pappus theorem.

[quote="pohoatza"]Let $ ABC$ be a triangle and let $ I$ be its incenter. Denote by $ A'$, $ B'$, $ C'$ the intersections of $ AI$, $ BI$, $ CI$ with the sides $ BC$, $ CA$, and $ AB$, respectively. Prove (synthetically) that the isogonal conjugate of $ I$ with respect to triangle $ A'B'C'$ lies on the line $ OI$, where $ O$ is the circumcenter of triangle $ ABC$.[/quote]

Let $X,Y,Z$ be the points of tangency of the incircle  with $BC,AC,AB$ respectively.Does the isogonal conjugate  of $I$ w.r.t. triangle $XYZ$ also lies on $OI$?

My solution:

Let $ I' $ be the isogonal conjugate of $ I $ with respect to triangle $ A'B'C' $ .

From [url=http://www.artofproblemsolving.com/community/c6h216910]On mixtilinear incircles[/url] we get $ IO $ is perpendicular to the orthotransversal of $ I $ WRT $ \triangle ABC $ .

From [url=http://www.artofproblemsolving.com/community/c6h105611]Isogonal conjugate and perpendicularity[/url] we get $ I'I $ is perpendicular to the orthotransversal of $ I $ WRT $ \triangle ABC $ .

From the discussion above we get $ I, I', O $ are collinear.

Q.E.D

Another solution:

Rewrite $ A', B', C' $ as $ A_1, B_1, C_1 $ .
Let $ A_2, B_2, C_2 $ be the projection of $ I $ on $ BC, CA, AB $ .
Let $ A_3, B_3, C_3 $ be the projection of $ I $ on $ B_1C_1, C_1A_1, A_1B_1 $ .

Invert with respect to $ (I) $ and denote $ P' $ as the image of $ P $ under this inversion.

Easy to see $ A_2=A_2', B_2=B_2', C_2=C_2' $ and $ A', B', C' $ is the midpoint of $ B_2'C_2', C_2'A_2', A_2'B_2' $ .

Since $ B_3, C_3, A_1, A_2, I $ all lie on $ (A_1I) $ ,
so $ B_3', C_3', A_1', A_2' $ are collinear and $ A_1' $ is the projection of $ I $ on $ B_3'C_3' $ .
Similarly, we can get $ B_1', C_1' $ is the projection of $ I $ on $ C_3'A_3', A_3'B_3' $ .
Since $ (A_1I,A_1A_2;A_1C_3,A_1B_3)=-1 $ ,
so $ A_2' $ is the midpoint of $ B_3'C_3' $ .
Similarly, we can get $ B_2', C_2' $ is the midpoint of $ C_3'A_3', A_3'B_3' $ .

Since $ (I) $ is the nine point circle of triangle $ A_3'B_3'C_3' $ ,
so $ I $ is the nine point center of triangle $ A_3'B_3'C_3' $ .
Since the center of $ (A'B'C'),(A_3'B_3'C_3') $ all lie on the Euler line of triangle $ A_3'B_3'C_3' $ ,
so the center of $ (ABC),(A_3B_3C_3) $ and $ I $ are collinear.
ie. the isogonal conjugate of $ I $ with respect to triangle $ A_1B_1C_1 $ lie on $ OI $

Q.E.D

[quote="IDMasterz"]Since the reflections of $I'$ on $A'B'C'$ and $I$ and $I'$ form homothetic configuration to $Be$ and $I_aI_bI_c$ and $O$, so $BeO \parallel II' \implies I' \in IO$.[/quote]

Nice solution !
Let me add more detail for this nice proof :) :

Let $ I' $ be the isogonal conjugate of $ I $ WRT $ \triangle A'B'C' $
Let $ I_a, I_b, I_c $ be three excenters of $ \triangle ABC $ .
Let $ X, Y, Z $ be the reflection of $ I' $ in $ B'C', C'A', A'B' $, repsectively .
Let $ B_e $ be the Bevan point of $ \triangle ABC $ .

Easy to see $ I $ is the circumcenter of $ \triangle XYZ $ .

Since $ YZ \perp AI, ZX \perp BI, XY \perp CI $ ,
so we get $ \triangle XYZ $ and $ \triangle I_aI_bI_c $ are homothetic .
Since it's well known that $ I_aO \perp B'C', I_bO \perp C'A', I_cO \perp A'B' $ ,
so we get $ \triangle XYZ \cap I \cap I' \sim \triangle I_aI_bI_c \cap O \cap B_e $ ,
hence we get $ II' \parallel OB_e \equiv OI $ . ie. $ I' \in OI $

Q.E.D

Let $\mathcal{H}$ be the feuerbach hyperbola of $\triangle ABC$ and $I'$ be the isogonal conjugate of $I$ wrt $\triangle A'B'C'$ ,clearly $OI$ is tangent to $\mathcal{H}$ and from [url=http://artofproblemsolving.com/community/c6h1576460p9705843]Orthogonal hyperbola through a point [/url] we have $II'$ is tangent to $\mathcal{H}$ so we conclude that $O,I,I'$ are collinear   $\blacksquare$

Observe that orthotransversal of $I$ w.r.t. $ABC$ is radical axis of $\odot(ABC)$ and degenerate circle $\odot(I)$, which clearly perpendicular to $AI$. Hence from [url=http://artofproblemsolving.com/community/c28312h1198358p5879552]here (post #3)[/url] we can conclude that $I'\in OI$.


calculus
limit
integration
geometry
calculus computations

If you already have credit for Calculus II and/or AP Calculus BC, you're supposed to already know this stuff. This is an in-class exam, not a set of puzzles. You have 50 minutes.

1. Find thise limits. Show or explain your reasoning.

a) $\lim_{x\to\infty}\left[\frac12\ln(x^{2}+10)-\ln(3x+1)\right]$

b) $\lim_{n\to\infty}\frac{n^{4}+3^{n}}{n^{3}+4^{n}}$

2. Find the center of mass of the region in the first quadrant of the plane that lies inside the circle $x^{2}+y^{2}=4.$

3. Determine whether or not the following improper integrals converge or diverge, and compute exact values for the ones that converge:

a) $\int_{0}^{\infty}\frac{dx}{x^{2}+4x+3}$

b) $\int_{0}^{2}\frac{x^{2}\,dx}{x-1}$

4. Find $\int\frac{e^{2x}\,dx}{(e^{4x}+4)^{3/2}}$

5. Let a curve be given by the parametric equations

$x=t^{3},\ y=3t^{2},\ 0\le t\le 2.$

a) Find the straight-line distance between the endpoints of the curve.

b) Find the arclength of this curve.

c) Find the area of the surface of revolution obtained by rotating this curve about the $x$-axis.

Let me just add that an actual classroom test should not be posted here without the explicit permission of the instructor of the course. In this case, we do have explicit permission, since I am the instructor.

Seems appropriate for a Calc II final. No leaky buckets lifted by ropes with variable density.

I covered topic #5 today.  :lol:

Well, not a final exam - there's still a lot of material to be covered, especially on on sequences, series and power series.

[quote]No leaky buckets lifted by ropes with variable density.[/quote]
That was in the chapter covered by this test, of course. If I liked such problems, I could have included one - apparently I don't like them all that much.


In 1996, there were 500 cases of measles reported in the U.S. In 1960 there were 450,000 cases reported. What is the ratio of the number of cases in 1996 to the number of cases in 1960? Express your answer as a common fraction.


Anyone has a proof for sequences that

suppose the sequence (an) tends to lambda as n tends to infinity
prove using the definition of convergence that constant C x sequence = C x limit of sequence.

PLS HLP!!!!

By the definition of a sequence limit:

$(\forall\varepsilon>0)(\exists n_0)(\forall n\geqslant n_0)(|a_n-\lambda|<\varepsilon)$

If this holds, then it means that if you pick some $\varepsilon$ and then take $\varepsilon_1={\varepsilon\over |C|}$, you will also find some $n_0'$ (generally DIFFERENT from $n_0$) such that

$(\forall n\geqslant n_0')(|a_n-\lambda|<{\varepsilon\over |C|})$

But then just get rid of the fraction:

$(\forall n\geqslant n_0')(|Ca_n-C\lambda|<\varepsilon)$

and you get what you need.


calculus derivative geometry trapezoid integration rectangle IMC

for problem 4:

Expand $f(1)$ and $f(-1)$ at $x=0$, respectively:
$f(1) = f(0) + f'(0) + \frac{f''(0)}{2}  + \frac{f'''(\theta_1)}{6}$
$f(-1) = f(0) - f'(0) + \frac{f''(0)}{2}  - \frac{f'''(\theta_2)}{6}$
Subtract them
$f(1)-f(-1) = 2f'(0) + \frac{f'''(\theta_1)+f'''(\theta_2)}{6}$
or
$\frac{f(1)}{2}-\frac{f(-1)}{2}-f'(0) = \frac{f'''(\theta_1)+f'''(\theta_2)}{2}\cdot \frac{1}{6}$

Notice that a derivative (it is $f'''$ here) has Darboux property, hence there exists $\xi \in [\theta_2,\theta_1]$ such that $f'''(\xi) = \frac{f'''(\theta_1)+f'''(\theta_2)}{2}$. The conclusion follows.

Oki, here is solution using quadrature formulae. Here we have the folowing trpezoid formulae.

 [b] Trapezoid Rule. [/b]

    Let $g\in C^{2}[a, b]$. Then there is $c\in (a,b)$ such that the following formula is valid:

                     \[ \int_{a}^{b} g(x)dx=\frac{b-a}{2}[g(a)+g(b)]-\frac{(b-a)^{3}}{12}\cdot g''(c).  \]

   Now, let's use this formula. By taking $g=f'$ and $a=-1, b=1$ we get

                     \[ f(1)-f(-1)=\frac{1}{2}[f'(-1)+f'(1)]-\frac{1}{12}f'''(c).  \]

   I'm sorry, we do not obtain the problem given in contest. :blush:  Now, let's use the following formula.

   [b] Rectangle Rule. [/b]

      Let $g\in C^{2}[a,b]$. Then there exists $c\in (a,b)$ such that the folowing formula is valid

                                    \[ \int_{a}^{b} g(x)dx=(b-a)g(\frac{a+b}{2})+\frac{(b-a)^{3}}{24}\cdot g''(c).  \]

   Now, let's use this formula. By taking $g=f'$ and $a=-1, b=1$ we get

                                    \[ f(1)-f(-1)=2f'(0)+\frac{1}{3}f'''(c).  \]
   Now, we clearly obtain the problem given in contest. :)

[quote="Moubinool"]Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a three times differentiable function. Prove that there exists $w \in [-1,1]$ such that \[ \frac{f'''(w)}{6} = \frac{f(1)}{2}-\frac{f(-1)}{2}-f'(0).  \][/quote]
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a three times differentiable function. Let $c := \frac{f(1)}{2}-\frac{f(-1)}{2}-f'(0)$, and let $g(x) = f(x)-cx^3$. Then $\frac{g(1)}{2}-\frac{g(-1)}{2}-g'(0) = 0$.

Let $h(x) = g(x)-g(-x)$, then $h'(x) = g'(x)+g'(-x)$. MVT gives $\xi \in (0,1)$ s.t. $h(1) = h(1)-h(0) = h'(\xi)$. So:
\[ (g'(\xi)-g'(0)) - (g'(0)-g(-\xi)) = h'(\xi)-2g'(0) = g(1)-g(-1)-2g'(0) = 0 \]
Using MVT again gives $\xi_1 \in (0,\xi), \xi_2 \in (-\xi,0)$, s.t. $g'(\xi)-g'(0) = \xi g''(\xi_1), g'(0)-g(-\xi) = \xi g''(\xi_2)$. So:
\[ g''(\xi_1) - g''(\xi_2) = 0 \]
Using the MVT again gives $w \in (\xi_2,\xi_1) \subseteq [0,1]$ s.t. $g'''(w) = 0$, i.e. $f'''(w) - 6c = 0$, which was what we wanted.


The students in Mrs. Reed's English class are reading the same
760-page novel. Three friends, Alice, Bob and Chandra, are in the
class. Alice reads a page in 20 seconds, Bob reads a page in 45
seconds and Chandra reads a  page in 30 seconds. \\ \\
If Bob and Chandra both read the whole book, Bob will spend how many
more seconds reading than Chandra?

Since Chandra "pulls away" by 15 seconds per page, we just need to compute $ 760\times15$ which is $ \boxed{11400}$


limit
logarithms
algebra
polynomial
function
algebra proposed

[i]Let $ l ; m$ be fixed positive integers , $ \mathbb{P}(x) \ \in \ \mathbb{R} [x]$ such that :

         $ \mathbb{P}(x) \equal{} a_0 x^{m \plus{} 1} \plus{} a_1 x^m \plus{} ...... \plus{} a_m x$

  Construct the sequence $ (u_n)$ as follows :

    $ u_n \ \equal{} \ \sum_{k \equal{} 0}^{ln} \mathbb{P} \left( \frac {1}{n \plus{} k} \right)$

    Find $ \lim_{n \ \to \ \plus{} \infty} u_n$[/i]

[hide="solution"]Denote $ H(n) \equal{} \sum_{k \equal{} 1}^n \frac {1}{k}$
\[ u_n \equal{} \sum_{k \equal{} n}^{n(l \plus{} 1)}\left(\frac {a_m}{k} \plus{} \mathcal{O}\left(\frac {1}{k^2}\right)\right) \\
\equal{} \mathcal{O}\left(\frac {1}{n}\right) \plus{} a_m \left[H(n(l \plus{} 1)) \minus{} H(n)\right] \\
\Rightarrow \lim_{n\to \infty}u_n \equal{} \lim_{n\to \infty}a_m\left(\log n(l \plus{} 1) \minus{} \log n \right) \equal{} \boxed{a_m \log (l \plus{} 1)}\]
where in the third line we have used the fact $ \gamma \plus{} \log k < H(k) < \gamma \plus{} \log (k \plus{} 1)$ for all $ k\in \mathbb{N}$.
Note: The fact that $ P$ is a polynomial is irrelevant.  Replacing $ P$ by any function $ f$ satisfying $ \lim_{x\to 0}\frac {f(x)}{x} \equal{} a_m$ would yield the same conclusion.[/hide]


vector
geometry
area of a triangle
Heron\u0027s formula

[quote="aufha"]How do you find the equation of a plane given three points $ (2,0,0)$,$ (0,3,0)$, and $ (0,0,5)$?

How would you go about finding the shortest distance from the origin to the plane in this case?

When would it be appropriate to use the generalized formula to find the distance from a point to a plane?[/quote]

The most general equation of a plane:

$ ax \plus{} by \plus{} cz \equal{} 1$
Now the Plane satisfies our 3 points

$ 2a \equal{} 1 \equal{} > a \equal{} 1/2$
$ 3b \equal{} 1 \equal{} > b \equal{} 1/3$
$ c \equal{} 1/5$

Distance of the Plane from the origin:
$ |\frac{1}{\sqrt{a^2 \plus{} b^2 \plus{} c^2}}|$

Tell me if you want the proof. :)

The proof scared me a bit... http://www.physicsforums.com/showthread.php?t=135082
What's a 3d vector? It isn't an arrow like a 2d vector is it?

Do you know a proof an average high schooler might understand? I'm still interested in the proof at the bottom of the above link because of its shortness.

Here's how I learned it:

A unique plane is defined by a point on the plane and a vector normal to it.
Let the point be $ P(x_1,y_1,z_1)$ and the vector be $ \vec{n}\equal{}<a,b,c>$.
Let $ Q(x,y,z)$ be any point for which $ PQ$ is orthogonal to $ \vec{n}$.

Then $ \vec{n} \cdot \overrightarrow{PQ} \equal{} 0$

Then doing the dot product gives

$ a(x\minus{}x_1)\plus{}b(y\minus{}y_1)\plus{}c(z\minus{}z_1)\equal{}0$

and this is the standard form of a plane.

By inspection, $ 15x\plus{}10y\plus{}6z\equal{}30$ seems to work as your plane.

To find the distance from the origin, we recall that $ D^2 \equal{} x^2 \plus{} y^2 \plus{} z^2$, and consider the two lists $ (15,10,6)$ and $ (x,y,z)$.

Operating Cauchy-Schwartz over this pair of lists gives:
$ (15x \plus{} 10y \plus{} 6z)^2 \leq (x^2 \plus{} y^2 \plus{} z^2)(15^2 \plus{} 10^2 \plus{} 6^2) \implies \frac {30^2}{15^2 \plus{} 10^2 \plus{} 6^2} \leq D^2$

Thus, $ D^2_{\textbf{min}} \equal{} \frac {30^2}{15^2 \plus{} 10^2 \plus{} 6^2} \equal{} \frac {900}{361} \implies \boxed{D_{\textbf{min}} \equal{} \frac {30}{19}}$

[quote="aufha"]What's a 3d vector? It isn't an arrow like a 2d vector is it?[/quote]

It is an arrow, but instead of in a plane it's pointing to some direction in space.  In other words, instead of being a directed line segment between two points in a plane, it's a directed line segment between two points in space.  A three-dimensional vector has three components instead of two.

The vector approach to this problem is very nice, but it requires a strong understanding of the dot product.  If you want a proof that "an average high school might understand," it'll be fairly tedious, starting with the proof that the vector $ \left< a, b, c \right>$ is perpendicular to the plane $ ax \plus{} by \plus{} cz \plus{} d \equal{} 0$.

[quote="Cyber-Shreyas"]The most general equation of a plane:

$ ax \plus{} by \plus{} cz \equal{} 1$[/quote]

You'll notice that planes passing through the origin can't be given by such equations.  The [i]most[/i] general equation is simply given by $ ax \plus{} by \plus{} cz \plus{} d \equal{} 0$.

[quote="123456789"][quote="guile"]The first one was a bit easy. What about this? I have no idea on how to solve it.

 (-2,3,5),(1,0,1) and (2, 1, 3)?[/quote]Try to learn how to solve it instead of just getting the answer.[/quote]

Of course its impossible to get the answer unless you know how to solve them. Anyway, I have already learned the process last night and I can share it, should anybody want me to.

[quote="Altheman"]For the problem given in the original post, you can just use volume.[/quote]

Or for any three points, such that their containing plane is not parallel to an axis, if you find the plane and then find its intercepts.

Outline:
- The volume of (a,0,0), (0,b,0), (0,0,c), and (0,0,0) is abc/6.
- We can find the pairwise distances between the first three points, and find the area of their triangle via Heron's Formula
- Then, we can use V=Ah/3, or h=abc/(2A), where h is the distance from the plane to the origin, and A is the area of the first three points.

Another technique, that's also been touched upon, is the normal vector method. Write the plane as:
ax+by+c=d, where a^2+b^2+c^2=1 (if it's not 1, you can scale the coefficients so that it is 1). Then, (a,b,c) is a normal vector to the plane.

Now, find the value of d such that (ad,bd,cd) satisfies the equation of the plane. That value of n is the distance from the plane to the origin.


function
algebra
functional equation
real analysis
real analysis unsolved

$(a_n)_{n \geq 1}$ - sequence of positive integers:
$a_1=1$, $a_2=2$
$a_{n+1}=a_n+a_{[(n+1)/2]}$, $n \geq 3$, $[]$ - denote the integer part
Can we determine $a_n$?

Do you mean finding a closed-form of $a_n$?

We can rewrite the recursive formula as follows.

\[a_{n+1}=a_n +\frac{n+1}{2}\ \ (n=1,3,\cdots)\]

\[a_{n+1}=a_n +\frac{n}{2}\ \ \ \ (n=2,4,\cdots)\]

No we can't, kunny- those are arguments of the sequence.

If we let $f(x)=\sum_{n=1}^\infty a_nx^n$, the recursion gives us the functional equation $f(x)=xf(x)+(1+x)f(x^2)+x$, which can be rearranged to $(1+x)f(x^2)+x=(1-x)f(x)$.

Unfortunately, I don't know what to do with this functional equation.

[quote="jmerry"]$(1+x)f(x^2)+x=(1-x)f(x)$.
Unfortunately, I don't know what to do with this functional equation.[/quote]
The only idea I have is to write it as $f(x)=\frac{x}{1-x}+\frac{1+x}{1-x}f(x^2)$ and iterate. After canceling some factors, we'll get the representation $f(x)=\frac 1{1-x}\sum_{n\geqslant 0}\frac{x^{2^n}}{(1-x)(1-x^2)\dots(1-x^{2^{n-1}})}=\frac1{1-x} \left[x+\frac{x^2}{1-x}+\frac{x^4}{(1-x)(1-x^2)}+\dots\right]$. This is not an elementary function so no "simple closed formula" for $a_n$ is possible but it is still possible that its $n$-th coefficient can be expressed in terms of the binary representation of $n$ or something like that. Anyway, the right question to ask seems to be whether we can compute $a_n$ in time essentially less than $n$.


let a,b,c>0 if a^2+b^2+c^2+2abc=1

then proove 2ab+2bc+2ca<=1+4abc

http://www.mathlinks.ro/Forum/viewtopic.php?t=156012
Hope it can helps :-)


geometry
rectangle
quadratics
special factorizations
algebra
difference of squares
quadratic formula

The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

[hide="Solution"]We will first express everything in terms of variables.

Square: Let $s$ be the number of people in each side of the square. The total number of members would be $s^2 +5$.

Rectangle: If we let $c$ be the number of people in a column of the rectangle, the number of members would be $c(c+7)=c^2+7c$

Setting these equal and rearranging the terms, we get $(s+c)(s-c)=7c-5$. Now we need to find the largest possible value.

Testing $s=c+4$, we get $8c+16 >7c-5$. This is clearly true because we have positive values of c.

Next, testing $s=c+3$, we find that this one works. Setting the two equations equal, $6c+9=7c-5$, we find that $c=14$.

Our answer is therefore $17^2+5=14*21=294$[/hide]

[hide="Answer"]$n=x^2+5=y(y+7)=y^2+7y\Rightarrow x^2-y^2=7y-5=(x+y)(x-y)$
Testing possible values of $x$ in terms of $y$, we see that $x=y+3$ is the largest possible value for positive $x,y$, so we have $6y+9=7y-5\Rightarrow y=14$, so our answer is $\boxed{294}$.[/hide]

[quote="314.math"][quote="4everwise"]Testing $ s \equal{} c \plus{} 4$, we get $ 8c \plus{} 16 > 7c \minus{} 5$. This is clearly true because we have positive values of c.[/quote]

Where did $ s \equal{} c \plus{} 4$ come from?  Thanks![/quote]

Could someone please help and answer my question?

It looks like 4everwise wanted to express $ s$ in terms of $ c$ in order to compare the left side with the right side. You want to find an upper limit on how big $ s$ can be. Perhaps $ s\equal{}c\plus{}4$ was chosen because it resulted in a slightly bigger $ c$-coefficient than the right side's. Or it was obtained just by trying $ s\equal{}c\plus{}1, c\plus{}2, \ldots$ and the others were left out because they didn't provide any insight.

[quote="Bictor717"]It looks like 4everwise wanted to express $ s$ in terms of $ c$ in order to compare the left side with the right side. You want to find an upper limit on how big $ s$ can be. Perhaps $ s \equal{} c \plus{} 4$ was chosen because it resulted in a slightly bigger $ c$-coefficient than the right side's. Or it was obtained just by trying $ s \equal{} c \plus{} 1, c \plus{} 2, \ldots$ and the others were left out because they didn't provide any insight.[/quote]

I still don't completely get.  Could someone please clarify?  Thanks a lot!

[quote="dgreenb801"]I don't get it either, can someone explain it?
Here is what I did:
$ x^2 \plus{} 5 \equal{} y(y \plus{} 7)$
Multiply by 4 and completing the square,
$ 4x^2 \plus{} 20 \equal{} (2y \minus{} 7)^2 \minus{} 49$
Using difference of squares,
$ 69 \equal{} (2y \plus{} 2x \minus{} 7)(2y \minus{} 2x \minus{} 7)$
Letting the first factor equal 69 and the second 1 gives
x=17, y=21, and our answer is 294.[/quote]
I don't really understand 4everwise's solution but i understand this one and they give the same answer. Are they the same thing?
Can someone please explain?

[code]
$\boxed{5}$
[/code]

comes out as

$ \boxed{5}$

Alternatively, you can use \fbox{}.

[quote=GeoffreyS][quote="Average Math Student"]A different solution that may or may not be easier to uderstand:

[hide]Let $ x$ be the number of people on each side of the square and let $ y$ be the number of people on the shorter side of the rectangle.  We then have the relation
$ x^2\plus{}5\equal{}y^2\plus{}7y$.

Solving for $ y$ using the quadratic formula (and ignoring the smaller root) yields

$ y\equal{}\frac{\minus{}7\plus{}\sqrt{4x^2\plus{}69}}{2}$

Since $ y$ has to be an integer, $ 4x^2\plus{}69$ must be a perfect square and we're looking for the largest such $ x$.  This gives

$ 4x^2\plus{}69\equal{}m^2$

for some value of $ m$.  Solving for $ x$ gives

$ x\equal{}\frac{1}{2}\sqrt{m^2\minus{}69}$

and we want the largest possible integer value for $ x$.  This occurs when $ m$ takes on its largest integer value, which occurs when

$ \sqrt{m^2\minus{}69}\equal{}m\minus{}1$

This is simply going to be the value of $ m$ such that

$ m\plus{}\left(m\minus{}1\right)\equal{}69$

so $ m\equal{}35$.  Plugging in gives $ x\equal{}17$ and plugging that into the original gives that the number of members in the marching band is 294.[/hide]

If anyone's reading this, can you tell me how to box my answer?[/quote]

can somebody explain this statement:

This occurs when $ m$ takes on its largest integer value, which occurs when

$ \sqrt{m^2\minus{}69}\equal{}m\minus{}1$[/quote]

[quote=professordad]Yes, I also don't get

[quote="Average Math Student"]
This is simply going to be the value of $m$ such that

$m + (m - 1) = 69$
[/quote][/quote]

It's due to Difference-of-Squares.
[hide=Clarification to #10]To better explain the finish of post #10, we can set up $$4k_1^2+69=k_2^2 \implies (k_2-2k_1)(k_2-2k_1)=69 \implies k_2-2k_1=1,~k_2+2k_1=69 \implies k_1=17,~k_2=35.$$ Squaring $k_1=17$ and adding $5$ gives the desired result.[/hide]


For puzzle solving instructions, please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13586]Puzzle 1[/url].

---

[b]Cryptic Equestrianship[/b]

In a horse race, the following conversations are overheard:

- I heard they're all among the top 26 horses in the country!

- Q, R, T, V and W all lagged behind the rest of the bunch. My guess is that they're ranked at least 20 already.

- I like the odds of C, E, F, H, I, O, R, T, V and X.

- Even A, M, N, Q, S, U, W, Y and Z aren't bad, either.

- C, E, H, T, U and X are primetime favourites. Did you know that C, H and another horse who's not here are triplets? E also has a twin horse who's not present; so does X.

- Oh look, F, U, C, H, S, E and V standing in a line, in that order. How cute. Almost reminds me of that rabbit tree thingy by that crazy Italian dude.

- A, C, E, M, N, O, S, U and Y don't get along well with X. I heard that they won't even be residents in the same square stable with X.

- F, I, R and Z are too fat - squarish, even.

- A, O, R and W are all ranked behind H; there are same numbers of horses between each of them.

- M and N ranked one place behind and one place before E, respectively.

- A lot of horses couldn't make this game. The stables of #4, #6, #7, #11, #17, #22 and #24 were burnt down and they were trapped inside. How tragic.

WSHFMXTHYIXAVNIVXCFHXFYZVQQNHURWOERVH

Ah, this ones much easier. I won't give it away though.  I didn't use the "I heard that they won't even be residents in the same square stable with X. " clue at all though. In fact, there could have been a lot more trimming down if you really wanted to. But fun.


Dear africans , now that you had a look at this topic , don't forget to post your signature and introduce yourself if possible !! (just a way to gather more members and show your belonging and presence) :)
Long life to the african community !

hello everyone well I'm from morocco I hope we'll pass a good time together at this forum .
Thanks everyone :arrow:


Hi All,

I am trying to reconcile two approaches used in counting problems. The first approach uses combinations and the other uses probability. I understand the combinations approach, but not able to comprehend the probability approach. Consider the following example,

[color=red]A carton contains 12 toasters, 3 of which are defective (D). If four toasters are sold at random, find the probability that exactly one will be defective. [/color]

[b]By the combinations approach[/b]: 3C1 * 9C3 / 12C4 = 252/495 = 0.509

[b]Now the probability approach[/b] :

I am purposely want to use here permutations and not combinations.

total possible permutations = 12P4 =11880

Another way of finding the total permutations is to find the chance of getting any permutation = 1/12 * 1/11* 1/10*1/9 = 0.000084. The reciprocal of this number gives the number 11880.

My confusion starts now. Consider the case DGGG, where G = Good, and D= Defective.

P (DGGG) = 3/12 * 9/11 * 8/10 * 7/9 = 0.127

If we divide this number by the probability of each permutation, 0.000084, we get 1512.
Thus, DGGG accounts for 1512 permutations or 1512 / 24 = 63 combinations.
[b]
I am having a hard time understanding how these 1512 permutations or 63 combinations look like. [/b]

Could you please help me with this?

Thanks,

MG.

[quote="musicgold"]total possible permutations = 12P4 =11880 [/quote]

i dont think its permutations because if he is sold toaster 1,2,3,4 that is the same number of defective toasters as 4,3,2,1

Thanks. I know that this problem needs to be solved with combinations. I am trying to understand what permutations/combinations are represented by the DGGG case.


What is the smallest number of 10'' by 15'' rectangular tiles required to form a square with no cutting or gaps?

If they are arranged in three rows width-wise it would give a 30 by 30 square
so isnt the answer 6?
not 15?

yea, but the answer says that its 15. i got 6, and i have no idea why that is wrong.

Lol, I think that you should report it.  I have no idea why they would say $15$; $10 \cdot 15 \cdot 15 = 2250$ which is not even a perfect square :P


calculus
integration
function
complex analysis
complex analysis unsolved

Given a complex power series $ \sum_{n\equal{}0}^{\infty}a_{n}z^{n}$ with finite radius of convergence $ r$, there exists a point on the circle of radius $ r$ such that our series is singular there. It seems likely that this problem has been solved somewhere in mathlinks' history but I failed to find it...

It should be proved in any text on complex analysis.  Perhaps based on the Cauchy integral formula and compactness.

The basic idea without filling in the details is that if there wasn't one, then you could extend the radius of convergence. I'm not convinced Cauchy Integral Formula is the way to go, but use compactness.

Every point on the boundary of the disk ($ \partial(r\mathbb{D})$)of convergence has a positive radius of convergence for the series centered there (since we assumed non-singular). So these cover the boundary, and hence have a finite subcover. Their boundaries form a closed curve $ \gamma$ around the original disk. But the function is analytic inside this new region and the min of $ dist(\partial(r\mathbb{D}), \gamma)>0$. This contradicts the maximality of the radius of convergence.

Note that the details aren't fully filled in.

I saw this problem somewhere and I think it's made me realize I either never knew or have forgotten the definition of "singular"... Is the power series $ \sum_{n\equal{}1}^{\infty}n^{\minus{}2}z^{n}$ not a counterexample? If not some clarification would be nice, thanks.

Ok, I guess that singular means that in this point the analyticity of the function $ f$ breaks. Your example does not imply that the sum is analytic function on the unit disc. :wink:


superior algebra
superior algebra unsolved

Let F be a field (or division ring), n>1 an integer and let R be the set of upper triangular nxn matrices over F for which all of the entries on the diagonal are equal. Show that R is a local ring.

A ring $R$ with idendity is local $\Longleftrightarrow$ $x \in R$ is not invertible,then $1 - x$ is invertible.
So in this case (Field) We know the determinant of matrices under consideration is product of diagonal entries,so If $A$ belongs to our ring and not invertible,then all its diagonal entries are zero, So $I -A$ has all its diagonal entries $-1$ and hence invertible and hence the result follows.

... and the unique maximal ideal consists of the matrices whose diagonal is $0$. The residue field is isomorphic to $F$.


function
ratio
algebra
rational function
algebra proposed

Find all the non-constant rational function of $f(x)$ and real numbers $a$ satisfying $\{f(x)\}^2-a=f(x^2).$
Here a rational function of $x$ is the equation expressed by the ratio of two polynominals of $x.$

take f= P/Q , And without less of generality we can assume Q is monic , now we can see; Q(x^2)= Q^2(x) , then Q= x^n , and we can easily find P .


algebra
polynomial
Rational Root Theorem

How to find the roots of the equation ?

4x³ - 24x² + 47x - 30 = 0

If you can't easily see that, you can do it the long way and use the Integer Root Theorem. This involves A LOT of plugging and chugging + OR - the factors of 30,though.  :(

Hello Andrew,
But how can I see that $ 4x^{3}\minus{}24x^{2}\plus{}47x\minus{}30\equal{}0$  == $ (x\minus{}2)(4x^{2}\minus{}16x\plus{}15)\equal{}0$ ?
Is this a easy method ?

Basically, any time you have to find roots of such polynomials, when they have integer coefficients, you should think about using the rational root theorem or the integer root theorem. The latter states that the roots of the polynomial (with integer coefficients) must be divisors of the constant term, which in this case is -30. So, naturally we start with 1, -1, (clearly 0 doesn't work), 2, -2, etc. In this case 2 happens to be a root, thus you can factor out $ (x \minus{} 2).$

Ah ok. Thanks limac. It's a nice method. Thanks.

rational root theorem, or cardano's theorem. But rationa; root theorem is the easiest way.


1) Let $ x\diamond y$ be defined as $ x^2+y^2-xy$.  How many solutions are there to $ x\diamond y=x$?

[hide="choices"]
a) 0
b) 1
c) 2
d) 3
e) $ \infty$
[/hide]

2) If $ S$ is a finite set, let $ |S|$ be the number of elements of $ S$.  Let $ A$, $ B$ and $ C$ be finite sets.  If $ |A|+|B|=|A\cap B|+|A%Error. "icap" is a bad command.
C|+|B\cap C|$, $ |A\cap B\cap C|=2$, and $ C\subseteq B\subseteq A$, then what is $ |A\cup B\cup C|$?

[hide="choices"]
a) 1
b) 2
c) 3
d) 4
e) Not enough information.
[/hide]

[hide="Question 2"]$ C \subseteq B \subseteq A$ means the following:
$ A \cap B \cap C \equal{} A \cap C \equal{} B \cap C \equal{} C$
$ A \cap B \equal{} B$
$ A \cup B \cup C \equal{} A$
This simplifies the given equations so much that the problem is trivial.[/hide]

[quote="Cyber-Shreyas"]Question 1: (I'm assuming you want integral solutions)

Answer is 3.

Write the given equation as a quadratic in y. Then x can only be 0,1. For which the solns will be
(0,0), (1,0) , (1,1)[/quote]

If you want solution in reals. Answer is infinite.


[quote="orl"][quote="anirudh"]I have a question. Is there any way I can be a moderator?[/quote]

First task is to learn the lessons of [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=2]Valentin's[/url] signature, especially the point with the search function. Once there was a topic about it and then yet another one which was closed...  ;)[/quote]

where do i learn that?

visit this
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28268[/url]

I think they have enough moderators for now.


function
inequalities
algebra proposed
algebra

[color=darkred]Given are four positive numbers $ a$ , $ b$ , $ c$ $ d$ . Ascertain the range of the real function $ f(x,y) \equal{} cx \plus{} dy$

with two real variables, where the real pair $ (x,y)$ verifies the inequality $ x^2 \plus{} \ y^2\ \le\ 2\cdot (\ ax\ \plus{} \ by\ )$ .[/color]

[hide="Answer."]

$ |c(x \minus{} a) \plus{} d(y \minus{} b)|\ \le\ \sqrt {\left(a^2 \plus{} b^2\right)\left(c^2 \plus{} d^2\right)}$ $ \Longleftrightarrow$ $ \underline {\overline {\left|\ (ac \plus{} bd) \minus{} \sqrt {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}\le f(x,y)\le ac \plus{} bd \plus{} \sqrt {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}\ \right|}}$[/hide]
[url=http://mateforum.ro/viewtopic.php?p=5354#5354][size=117][color=darkred][b]See and here ![/b][/color][/size][/url]

If for a point $ P\in{(x \minus{} a)^2 \plus{} (y \minus{} b)^2\leq a^2 \plus{} b^2}$ which is the disk centered at $ C(a,\ b)$ with radius $ \sqrt {a^2 \plus{} b^2}$, $ cx \plus{} dy$ take some value $ k$,
then the disc has more than one intersection point with the line $ cx \plus{} dy \equal{} k$.
Thus the condition gives the distance between the center $ C(a,\ b)$ and  the line $ l : cx \plus{} dy \minus{} k \equal{} 0\leq$ the radius $ \sqrt {a^2 \plus{} b^2}$

$ \Longleftrightarrow \frac {|ac \plus{} bd \minus{} k|}{\sqrt {c^2 \plus{} d^2}}\leq \sqrt {a^2 \plus{} b^2}$

$ \Longleftrightarrow |k \minus{} (ac \plus{} bd)| \leq \sqrt {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}$, yielding $ ac \plus{} bd \minus{} \sqrt {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}\leq f(x,\ y) \leq ac \plus{} bd \plus{} \sqrt {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}$.


For n=1 it is obviously true.
Suppose it holds for n=k ($ (x^k\plus{}y^k)/2>\equal{}((x\plus{}y)/2)^k$
Multiplying both sides by $ (x\plus{}y)/2$ we get that
$ (x\plus{}y)*(x^k\plus{}y^k)/4>\equal{}((x\plus{}y)/2)^(k\plus{}1)$
So we need to prove that
$ (x^(k\plus{}1)\plus{}y^(k\plus{}1))/2>\equal{}(x\plus{}y)*(x^k\plus{}y^k)/4 \equal{}>2x^(k\plus{}1)\plus{}2y^(k\plus{}1)>\equal{}x^(k\plus{}1)\plus{}xy^k\plus{}yx^k\plus{}y^(k\plus{}1)$
which is equivalent to saying that $ (x^k\minus{}y^k)(x\minus{}y)>\equal{}0$ which is true if $ x\minus{}y>\equal{}0$ then $ x^k\minus{}y^k>\equal{}0$ so both factors are positive or zero whereas if $ x\minus{}y<0$ then $ x^k\minus{}y^k<0$ so both factors are negative
So it is proven

Remark:I really don't think that this belongs in Calculus...


If you read the contest request thread, this is the same post as on the thread. Basically, respond if interested, or if you have any questions/comments. Also, I have a good idea of the questions I'm going to ask already, so any comment on a time you'd think is good for me too post the contest is also welcome:

[quote]
I'm considering posting a contest, but how many people are interested?

I haven't actually fully designed the contest yet, but this is the general format:

There will be 5 questions in each of the 3 groups A, B, and C. The group A problems will be on the level of the harder (around National) Mathcounts questions, and could be solved by a relatively gifted middle-schooler. All the group A questions will have numerical answers. You do not have to show work for any of the group A questions.

The group B problems will be slightly harder than the group A questions, and contain both proofs and numerical answer questions. You will have to show your work for the group B questions. These questions will be challenging, but I've seen harder on this forum.

The group C problems will be the hardest, and probably the level of the harder AIME questions, but will include proofs. Again, you'll have to show your work for group C problems.

You can choose one of the 3 groups of questions to complete, and PM the answers/solutions to me.
Part A will be marked out of 5 (you get 1 mark for a right answer, 0 marks for a wrong),

Part B and C will be marked out of 25 (you may get partial marks)

Of course, more clarification will be available when I post the contest.
So if you want to participate/have any suggestions, just reply. I want to have a good amount of how many people are interested before I actually post the contest.
[/quote]

i'm interested too  :D

edit - also i was wondering if you could enter more than one... i am a 10th grader so I was wondering which group(s) i should enter in thanks

Okay, now that I have some people interested in the contest, I'll post the problems soon (probably in 3 days to a week, so it doesn't coincide with the 4th COP).

Also, feel free to enter solutions if you just have some time on your hands and I'll mark them. Note that you don't really need to register on this thread, but it is nice to have an idea of how many people are writing it.)

So anyway, good luck. The problems will be up shortly  :lol:


geometry
3D geometry
sphere
trapezoid
Pythagorean Theorem

A truncated cone has horizontal bases with radii $ 18$ and $ 2$. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

$ \textbf{(A)}\ 6 \qquad
\textbf{(B)}\ 4\sqrt5 \qquad
\textbf{(C)}\ 9 \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ 6\sqrt3$

[hide]I think you can extend the fact that two tangents to a circle from the same point have equal length to a sphere, so that the slant height is $ 2\plus{}18\equal{}20$. Then from the sphere, we have that the height of the truncated cone is $ 2r$, so using the Pythagorean Theorem,
\[ \left(2r\right)^2\plus{}16^2\equal{}20^2 \implies 2r\equal{}12 \implies r\equal{}6 \rightarrow \boxed{\textbf{A}}.\][/hide]

[quote="Truffles"][quote="123456789"]I think you can extend the fact that two tangents to a circle from the same point have equal length to a sphere[/quote]

Hmm I'm not sure I understand this, does anyone know what the name of this fact is?[/quote]Consider the great circle that the tangents are touching at; and I don't think there is a special name for that... perhaps "Two Tangents Sphere Theorem"? :D

The kick-off to this problem is to figure out that the slant height is $20$ as 12345789 did but I, unfortunately, don't see why this is the case. Forgive my ignorance!

The reason is quite simple.  Consider a circle with center $O$ inscribed in an isosceles trapezoid $ABCD$, with $AB \parallel DC$ and $BC = DA$.  Let the points of tangency to the bases $AB$, $CD$ be $M$, $N$, respectively, and the point of tangency to $BC$ be $P$.  Then clearly $OM = OP = ON = r$, the inradius.  Furthermore, $MB = PB$ and $NC = PC$.  Hence $BC = BP + PC = BM + NC$; that is to say, the slant height of the trapezoid must be equal to the average of the base lengths.

There is a very nice solution with similar triangles; you can write $\frac{18}{r}=\frac{r}{2}\implies{\boxed{r=6}}$

[b]Solution:[/b] Firstly, take the cross section exactly down the middle of the cone. So then you're going to have a circle inscribed in a trapezoid. Extend the trapezoid to form a triangle. The area of that triangle will be $36 \cdot \frac{9}{8}r = \frac{81}{2}r$. This means that the perimeter will be $81$ by using $A=rs$. So the side length of the triangle we made are $45/2, 45/2, 36$. Then using similar triangles, we can find the slant length of the cone to be $20$. This means the radius is $\frac{\sqrt{20^2 - 16^2}}{2} = \boxed{6}$.


the sequence of real numbers $a_{1},a_{2},a_{3}\cdots$ is such that $a_{1}=1$ and

$a_{n+1}=\left(a_{n}+\frac{1}{a_{n}}\right)^{t}$

where $t\geq 1$ . Prove that for $n\geq 2$

$\frac{a_{n+1}}{a_{n}}>2^{(t-1)t^{n-1}}$


[quote="epsilon07"]P(x)  Q(x) and R(x) are polynomial such that P(x^5)+x.Q(x^5)+x^2.R(x^5)  is divisible by x^4 +x^3 + x^2 + x + 1. Then find P(1)...[/quote]

Let $ x_k\equal{}e^{\frac{2k\pi}5i}$, for $ k\in\{1,2,3,4\}$.  We get $ P(1)\plus{}x_k.Q(1)\plus{}x_k^2.R(1) \equal{}0$ and so $ P(1)\equal{}Q(1)\equal{}R(1)\equal{}0$


Let $f: N \to N$ be a function satisfying
(a) For every $n \in N, f(n+f(n))=f(n)$.
(b) For some $n_{0}\in N, f(n_{0})=1$.
Show that $f(n)=1$ for all $n \in N$


Find all solutions to the Diophantine equation $ 2^x \equal{} 3^y \plus{} 5$.

$ (x, y) \equal{} (3, 1)$ and $ (x, y) \equal{} (5, 3)$ are solutions.

These solutions even I could find and I guess these are the only solutions. But how to show that there are no more solutions.

$ 2^x \minus{} 5 \equal{} 3^y$
taking mod$ 3$ it proves $ x$ is odd.

similarly, $ 2^x \equal{} 3^y \plus{} 5$
taking mod$ 4$ it proves $ y$ is odd.

so you have both $ x,y$ odd.

now try

I have tried this but unfortunately it didn't work. I could get other condition on $ x$ and $ y$ by reducing the equation modulo 16,17,27 etc. but none of them seem to help.

x,y odd. So try using $ t^2\minus{}1$=$ (t\minus{}1)(t\plus{}1)$.

But this doesn't lead to any where further.

Factorize it, notice that the maximum power of 2 that divides both handsides is 2, getting that n=1. And also 5 divides the LHS, so m is odd. That means, y is divisible by 3 and it should now move.

No, mistake!m isn't odd. But notice that 5 really divides the LHS, and if $ m > 1$, then 5 doesn't divide the RHS. So m=1 and the only solutions are the above mentioned.


geometry
circumcircle
trigonometry
geometry proposed

$D$ is a point in triangle $ABC$ and the real numbers $a,b,c,d$ exist such that :
$AB=ab$, $AC=ac$, $AD=ad$, $BC=bc$, $BD=bd$, $CD=cd$
Prove that:
$\angle ABD+\angle ACD=60^{\circ}$

$\frac{AC}{BC}= \frac{ac}{bc}= \frac{ad}{bd}= \frac{AD}{BD}$

$\frac{BA}{CA}= \frac{ba}{ca}= \frac{bd}{cd}= \frac{BD}{CD}$

$\frac{CB}{AB}= \frac{cb}{ab}= \frac{cd}{ad}= \frac{CD}{AD}$

Thus D is the concurrency point of the 3 Apollonius circles of $\triangle ABC$ inside its circumcircle, i.e., the 1st isodynamic point. For D to be inside of $\triangle ABC,$ the largest internal angle must be less than $120^\circ.$ Let X, Y, Z be the feet of normals from D to BC, CA, AB. From the cyclic quadrilaterals CXDY, AYDZ, BZDX with their circumcircle diameters CD, AD, BD, we get $XY = CD \sin C,\ YZ = AD \sin A,\ ZX = BD \sin B$ and

$\frac{XY}{YZ}= \frac{CD}{AD}\cdot \frac{\sin C}{\sin A}= \frac{CB}{AB}\cdot \frac{\sin C}{\sin A}= 1$

i.e., XY = YZ and similarly, YZ = ZX. Therefore, $\triangle XYZ$ is equilateral. From the cyclic quadrilaterals CXDY, BZDX,

$\angle ABD+\angle ACD \equiv \angle ZBD+\angle YCD = \angle ZXD+\angle YXD = \angle ZXY = 60^\circ$


[quote="GameBot"]The figure in stage 1 was obtained from the figure in stage 0 by connecting midpoints of the opposite sides of the square and shading the lower right square. To obtain the figure in stage 2, the midpoints of the opposite sides of the unshaded squares from stage 1 were connected and the lower right square was shaded. This pattern is continued. What fractional part of the square in stage 3 will be shaded? Express your answer as a common fraction.

[asy]size(5cm,5cm);

pair A,B,C,D,A1,B1,C1,D1;

A=(0,0);
B=(4,0);
C=(4,4);
D=(0,4);

draw(A--B--C--D--A);

A1=(6,0);
B1=(10,0);
C1=(10,4);
D1=(6,4);

draw(A1--B1--C1--D1--A1);

pair a1,a2,a3,a4;

a1=(1/2)*A1+(1/2)*B1;
a2=(1/2)*B1+(1/2)*C1;
a3=(1/2)*C1+(1/2)*D1;
a4=(1/2)*D1+(1/2)*A1;

fill((a1--B1--a2--(8,2)--cycle),gray);
draw (a1--a3);
draw (a2--a4);

pair W,X,Y,Z;

W=(12,0);
X=(16,0);
Y=(16,4);
Z=(12,4);

draw(W--X--Y--Z--W);

pair O,b1,b2,b3,b4;

O=(14,2);

b1=(1/2)*W+(1/2)*X;
b2=(1/2)*X+(1/2)*Y;
b3=(1/2)*Y+(1/2)*Z;
b4=(1/2)*Z+(1/2)*W;

fill((b1--O--b2--X--cycle),gray);

fill(((13,0)--(14,0)--(14,1)--(13,1)--cycle),gray);

fill(((13,2)--(14,2)--(14,3)--(13,3)--cycle),gray);

fill(((15,2)--(16,2)--(16,3)--(15,3)--cycle),gray);

draw(b1--b3);
draw(b2--b4);

pair c1,c2,c3,c4;

c1=(1/2)*W+(1/2)*b1;
c2=(1/2)*Z+(1/2)*b3;

draw(c1--c2);

c3=(1/2)*Z+(1/2)*b4;
c4=(1/2)*Y+(1/2)*b2;

draw(c3--c4);

pair c5,c6,c7,c8;

c5=(1/2)*b1+(1/2)*O;
c6=(1/2)*W+(1/2)*b4;

draw(c5--c6);

c7=(1/2)*b2+(1/2)*O;
c8=(1/2)*Y+(1/2)*b3;

draw(c7--c8);

label("Stage $0$",midpoint(A--B),S);

label("Stage $1$",midpoint(A1--B1),S);

label("Stage $2$",midpoint(W--X),S);[/asy][/quote]

In each of the unshaed squares, there will be $ \frac{1}{64}$ shaded.

So $ \frac{7}{16}\plus{}\frac{9}{64}\equal{}\frac{37}{64}$, which incidentially is $ \frac{6^2\minus{}5^2}{6^2}$. See if you can prove this always works!


modular arithmetic
number theory
relatively prime
Diophantine equation
Zsigmondy

$(x,y,z,w) = (1,1,1,0)$, $(2,2,1,1,)$, $(1,0,0,0)$ or $(3,0,0,1)$

If $x = 0$, then we get odd-odd=odd, contradiction

If $x\geq 3$, $y \geq 1$, then mod 3, we get $z$ is odd, mod 8, we get $z$ is even, contradiction.

if $y \geq 1$, $2\geq x \geq 1$ then [

if $x=1$, then if $z\geq 1$, $w \geq 1$[
mod 7, we find that $y=4$ (mod 6), which means that $y$ is even. mod 5, we find that $y=1$ mod 4, which means that $y$ is odd, contradiction.

]else if $z = 0$, $w \geq 0$[ If $y=0$, then we get the solution $(1,0,0,0)$. we check that $y=1$ dosen't work, so mod 9, we get a contradiction.

]else if $z > 0$, $w = 0$[ If $y \leq 1$, we check to find the solution $(1,1,1,0)$. Otherwise, mod 9, and we find that $z$ is divisible by 3. However since $5^{3k} +1$ is divisible by 7, we get a contradiction.
]

if $x=2$, then [ we find the solution $(2,2,1,1)$. Mod 7 or mod 5, we find that y is even. Let  $y= 2g$. If y is larger than 4 then $5^z7^w = 4*3^y -1 =  (2*3^g-1)(2*3^g+1)$. We find that $(2*3^g-1)$ must equal to $5^z$ or $7^w$, where $g>1$. We have already proven that this is impossible. (this is equivalent to the previous two cases.)
]
]

Now, consider the case when $y=0$. Mod 7, we find that $x$ is divisible by 3. If $z \geq 1$, then mod 5, we find that $x$ is divisible by 4. However, since $2^{2k}-1$ is divisible by 3, thus, contradiction. Finally, we are left with the case that $y=0$, $z=0$. If $x\leq3$, then we check and find the solution $(3,0,0,1)$, and if $x>3$, mod 16 would yield a contradiction.

Q.E.D.
P.S. There got to be a better solution!

i have the other solution
first we easy to prove that x  <= 2 then we have two case:
if x=1 then easu to prove that the equation hasn't root (use modular)
if x=2 then we mast use Pell's equation and we have the result (for the case two it's two long and i am very lazy to write it) :D  :D  :D

maybe I'm blind, but how do you particularly reach pell's equation in here. And even if u do I don't think it gets easier than the ordinary solution.

[quote="orl"]Find all nonnegative integer solutions $ (x,y,z,w)$ of the equation
\[ 2^x\cdot3^y \minus{} 5^z\cdot7^w \equal{} 1.
\]
[/quote]
It is clear that $ x\geq 1$,because otherwise left side of equation is even.
Suppose $ y \equal{} 0$:
Then if $ z\geq 1$,we conclude that $ 2^{x}\equiv 1$ mod $ 5$,but it means that $ 4|x$,hence $ 3|2^{x} \minus{} 1$,so $ 3|5^{z}\cdot 7^{w}$,contradiction.
Therefore,$ z \equal{} 0$ and $ 2^{x} \equal{} 1 \plus{} 7^{w}$.
Now if $ x\geq 4$,it follows that $ 7^{w}\equiv 1$ mod $ 16$,but as one can check,it is impossible.
Therefore,$ x\leq 3$.Checking cases $ x \equal{} 1,3$,it gives that $ \boxed{(1,0,0,0)}$ and $ \boxed{(3,0,0,1)}$ are solutions.
Now we can assume $ y > 0$.Consider two cases:

[b]1[/b]-If $ x \equal{} 1$:
Then considering our equation modulo $ 3$,we conclude that $ z$ is odd,thus $ z\geq 1$,and after that modulo $ 5$,which gives $ y\equiv 1$ mod $ 4$.
Suppose $ w > 0$,it follows that $ 2\cdot 3^{y}\equiv 1$ mod $ 7$,but last congruence is impossible,as $ y\equiv 1$ mod $ 4$.
Therefore,$ w \equal{} 0$,and our equation can be rewritten,as follows:
$ 2\cdot 3^{y} \minus{} 5^{z} \equal{} 1$.
As we've observed $ y\equiv 1$ mod $ 4$,hence if $ y \equal{} 1$:
Then $ z \equal{} 1$,a new solution is $ \boxed{(1,1,1,0)}$.
If $ y\geq 5$,then
$ 5^{z}\equiv \minus{} 1$ mod $ 9$,hence $ z \equal{} 6k \plus{} 3$,or
$ 5^3 \plus{} 1|5^{z} \plus{} 1|2\cdot 3^{y}$,contradiction.

[b]2[/b]-If $ x\geq 2$:
Consider our equation modulo $ 4$ and modulo $ 3$:
$ 7^{w}\equiv \minus{} 1$ mod $ 4$,hence $ w$ is odd.
$ 5^{z}\equiv \minus{} 1$ mod $ 3$,hence $ z$ is odd.
So $ w$ and $ z$ are odd,hence
$ 2^{x}\cdot 3^{y}\equiv 5^{z}\cdot 7^{w} \plus{} 1\equiv 5\cdot 7 \plus{} 1\equiv 4$ mod $ 8$.
Therefore $ x \equal{} 2$ and $ y$ is even.
Consider our equation modulo $ 5$ and $ 7$:
$ 4\cdot 3^{y}\equiv 1$ mod $ 5$ and $ 7$,hence $ \boxed{y \equal{} 12k \plus{} 2}$.
We can rewrite our equation as:
$ (2\cdot 3^{6k \plus{} 1} \minus{} 1)(2\cdot 3^{6k \plus{} 1} \plus{} 1) \equal{} 5^{z}\cdot 7^{w}$.
As $ \gcd(2\cdot 3^{6k \plus{} 1} \minus{} 1,2\cdot 3^{6k \plus{} 1} \plus{} 1) \equal{} 2$ and $ 7\not{|}2\cdot 3^{6k \plus{} 1} \minus{} 1$,it follows that
$ 2\cdot 3^{6k \plus{} 1} \plus{} 1 \equal{} 7^{w}$ and $ 2\cdot 3^{6k \plus{} 1} \minus{} 1 \equal{} 5^{z}$
But the second equation was considered in first case,hence $ k \equal{} 0$ and $ y \equal{} 2$.
So the last solution is $ \boxed{(2,2,1,1)}$.(all solutions were written in form $ (x,y,z,w)$.

The answer is $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$ and $(2,2,1,1)$.
-------------------
[color=blue][b]Zsigmondy approach[/b][/color] First, repeated applications of Zsigmondy's theorem solve the problem immediately in the situation where $abcd = 0$, so from now on we assume $a,b,c,d > 0$. (This is not too difficult even without Zsigmondy, but involves several cases.)
[list] [*]Taking modulo $3$ gives $1 + 2^c \equiv 0 \pmod 3$ so it follows $c$ is odd. [*]Taking modulo $8$ gives $2^a \cdot 3^b \equiv 1 + 5 \cdot 7^d \pmod 8$ from which it follows that $a < 3$. So either $a = 1$ or $a = 2$. [*]Taking modulo $5$ gives $1 \equiv 2^a 3^b \equiv 2^{a-b} \pmod 5$, so it follows that $a \equiv b \pmod 4$. [*]Taking modulo $7$ gives $2^a 3^b \equiv 1 \pmod 7$. Since $a$ and $b$ have the same parity, this proves $a \neq 1$ (since otherwise $6M^2 \equiv -1 \pmod 7$). [/list]
So we are reduced to the case where $a = 2$ and $b \equiv 2 \pmod 4$. Let $b = 2e$. Then a difference of squares gives \[ \left( 2 \cdot 3^e - 1 \right)\left( 2 \cdot 3^e + 1 \right) = 5^c \cdot 7^d. \] Since the two factors are relatively prime and the first is $2 \pmod 3$, it follows $2 \cdot 3^e = 5^c + 1$ and $2 \cdot 3^e = 7^d - 1$. Another application of Zsigmondy gives $c=d=e=1$, so $(a,b,c,d) = (2,2,1,1)$.
-------------------
[color=blue][b]Longer approach without Zsigmondy[/b][/color] Here is a solution without Zsigmondy.
It's clear that $a \ge 1$, otherwise the left-hand side is even. The remainder of the solution involves several cases. [list] [*]First, suppose $b=0$. [list] [*]If $c \ge 1$, then modulo $5$ we discover $2^a \equiv 1 \pmod 5$ and hence $4 \mid a$. But then modulo $3$ this gives $-5^c 7^d \equiv 0$, which is a contradiction. [*]Hence assume $c=0$. Then this becomes $2^a-7^d=1$. This implies $1+7^d \equiv 2^a \pmod{16}$, and hence $a \le 3$. Exhausting the possible values of $a=0,1,2,3$ we discover that $(3,0,0,1)$ and $(1,0,0,0)$ are solutions. [/list] [*]Henceforth suppose $b > 0$. Taking modulo $3$, we discover that $5^c \equiv -1 \pmod 3$, so $c$ must be odd and in particular not equal to zero. Then, taking modulo $5$ we find that \[ 1 \equiv 2^a3^b \equiv 2^{a-b} \pmod 5. \] Thus, $a \equiv b \pmod 4$. Now we again have several cases. [list] [*]First, suppose $d=0$. Then $2^a3^b = 5^c+1$. Taking modulo $4$, we see that $a=1$ is necessary, so $b \equiv 1 \pmod 4$. Clearly we have a solution $(1,1,1,0)$ here. If $b \ge 2$, however, then taking modulo $9$ gives $5^c \equiv -1 \pmod 9$, which occurs only if $c \equiv 0 \pmod 3$. But then $5^3+1 = 126$ divides $5^c+1 = 2^a3^b$, which is impossible.
[*]Now suppose $d \neq 0$ and $a$, $b$ are odd. Then $6M^2 \equiv 1 \pmod 7$, where $M = 2^{\frac{a-1}{2}}3^{\frac{b-1}{2}}$ is an integer. Hence $M^2 \equiv -1 \pmod 7$, but this is not true for any integer $M$.
[*]Finally, suppose $b, c, d \neq 0$, and $a=2x$, $b=2y$ are even integers with $x \equiv y \pmod 2$, and that $c$ is odd. Let $M = 2^x 3^y$. We obtain $(M-1)(M+1) = 5^c 7^d$. As $\gcd(M-1, M+1) \le 2$, this can only occur in two situations. [list] [*]In one case, $M-1=5^c$ and $M+1=7^d$. Then $5^c+1 \equiv 2^x 3^y$. We have already discussed this equation; it is valid only when $x=y=1$ and $c=1$, so $(a,b,c,d) = (2,2,1,1)$. [*]In the other case, $M+1=5^c$ and $M-1=7^d$. Taking the first relation modulo $3$, we obtain that $M \equiv 2^c-1 \equiv 1 \pmod 3$. Hence $y=0$, and $x$ is even. Now $2^x+1 = 5^c$ and $2^x-1=7^d$. But if $x$ is even then $3 = 2^2-1 \mid 2^x-1 \mid 7^d$, which is impossible. Hence there are no solutions here. [/list] [/list] [/list] In summary, the only solutions are $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$ and $(2,2,1,1)$.

can anyone explain to me how does Zsigmondy work here? doesn't it state,that $ a^n - b^n $ has at least one prime factor that does not divide $ a^k - b^k $ (yes it has some exceptions)? or does this theorem has some corollaries?

[quote="v_Enhance"]First let's make sure we agree on the statement of Zsigmondy, since there are a few variants.
[b]Zsig[/b]: Let $a < b$ be relatively prime positive integers. Then
[list][*] $a^n-b^n$ has a primitive prime divisor for all $n$, except when $n=2$ and $a+b=2^k$, or $(a,b,n)=(1,2,6)$.
[*] $a^n+b^n$ has a primitive prime divisor for all $n$, except when $(a,b,n)=(1,2,3)$.[/list]

Basically what this means is that once we pin down a Diophantine equation to something like $c^n \pm 1 = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$ for some fixed primes, we are immediately done modulo small cases. For a concrete example, take the Diophantine equation \[ 19^n - 1 = 2^a3^b5^c7^d11^e13^f17^g. \] Let $n$ run from $1$ to $\infty$. You can check by hand that all the primes appear by the time you reach $n=16$. So for $n \ge 17$ there are no solutions by Zsig, since there must be a prime $p$ dividing $19^n-1$ that we haven't seen before.[/quote]
thank you so much,that's clear after all (:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=587398&p=3481126#p3481126

My solution is very straightforward and it doesn't take many lines to write it.

However, I want to point out that there is also a solution using Stormer's Theorem. It will take me a great amount of time to write it, but it's completely mechanic.

[hide=Solution][quote=China NMO 2005 P6]Find all nonnegative integer solutions $(x,y,z,w)$ of the equation\[2^x\cdot3^y-5^z\cdot7^w=1.\][/quote]
[b][color=#000]Solution:[/color][/b] Clearly, $x \geq 1$. When $w=0$
$$2^x3^y=1+5^z$$
If $z=0$ $\implies$ $x=1,y=0$ & If $z=1$ $\implies$ $x=1, y=1$. If $x \geq 2$, taking $\pmod{4}$ gives us contradiction! If $x=1$, $y \geq 2$
$$2 \times 3^y=1+5^z$$
Taking, $\pmod{9}$ followed by $\pmod{7}$ $\implies$ No Solutions! Suppose $w=1$
$$2^x3^y=1+7 \times 5^z$$
If $z=0$ $\implies$ $x=3,y=0$. If $z=1$ $\implies$ $x=2,y=2$. Let $z \geq 2$. If $y=0$,
$$2^x=1+7 \times 5^z$$
Taking, $\pmod{7}$ & $\pmod{5}$ $\implies$ $12 \mid x$. Hence, $7 \times 5^z \equiv 0 \pmod{9}$, a contradiction! So, $ y \geq 1$. Clearly, $z$ should be odd.
$$2^x3^y=1+7 \times 5^z$$
If $y \geq 2$ $\implies$ $1+7 \times 5^z \equiv 0 \pmod{9} \implies z \equiv 1 \pmod{6}$. Hence, if $x \geq 3$ $\implies$ $1+7 \times 5^z$ $\equiv$ $0 \pmod{8}$ which is not possible. If $y=1$ $\implies$
$$3 \times 2^x = 1+ 7 \times 5^z$$
Taking, $\pmod{5}$ $\implies$ $x \equiv 1 \pmod{4}$. Hence, $ 2^x \equiv 5 \pmod{7}$, Contradiction! Hence now we may assume $ w \geq 2$
$$2^x \times 3^y= 1+ 5^z \times 7^w$$
For $x \geq 3$, $w$ is odd from $\pmod{4}$ and $5^z \equiv 1 \pmod{8}$. For $y \geq 1$ $\implies$ $z$ is odd, hence, $8$ does not divide $5^z-1$. If $y=0$, we similarly, get contradiction from $12 \mid x$ and considering $\pmod{9}$. For $x=2$,
$$4 \times 3^y=1+5^z \times 7^w$$
Taking, $\pmod{49}$ $\implies$ $32 \mid y$. Hence, $\pmod{5}$ $\implies$ $\text{ LHS }$ is $4 \pmod{5}$ a contradiction! If $x=1$,
$$2 \times 3^y = 1+5^z \times 7^w$$
Hence, $w$ is even & $z$ is odd. Hence, for $y \geq 2$, taking $\pmod{9}$ gives a contradiction! Clearly, for $w \geq 2$, $y \le 1$ is not possible. Hence we have run out of all possible cases. $\boxed{(x,y,z,w)=(1,1,1,0),(2,2,1,1),(1,0,0,0),(3,0,0,1)}$ $\qquad \blacksquare$
[/hide]

The answers are $(1,0,0,0)$, $(1,1,1,0)$, $(2,2,1,1)$, $(3,0,0,1)$.

[b]Claim.[/b]     If $x,y,z,w>0$, then $(x,y,z,w)=(2,2,1,1)$.

[i]Proof.[/i]     First assume $x\ge3$. Taking modulo $8$, we have $5^z7^w\equiv-1\pmod8$, however $5^z\in\{1,5\}$ and $7^w\in\{1,7\}$, so we must have $(5^z,7^w)=(1,7)$. It follows that $z$ is even and $w$ is odd. However taking modulo $6$, we have $-(-1)^z\equiv1\pmod6$, so $z$ is odd, contradiction.

    Hence we may take the following two cases:
    [list]
        [*]Assume $x=1$. Then $5^z7^w=2\cdot3^b-1$, so $2\cdot3^y\equiv1\pmod{35}$. This modulo $5$ gives $y\equiv1\pmod4$, and this modulo $7$ gives $y\equiv4\pmod6$, absurd.
        [*]Assume $x=2$. Then $5^z7^w=4\cdot3^y-1$, so $4\cdot3^y\equiv1\pmod{35}$. Modulo $5$ gives $y$ is even, so \[5^z7^w=(2\cdot3^{y/2}-1)(2\cdot3^{y+2}+1).\]
            The two terms are relatively prime, so they must be $5^z$ and $7^w$ in some order.

            That is, either $2\cdot3^{y/2}=5^z+1$ or $5^z-1$. By Zsigmondy theorem, $z\le2$, and it is easy to check that only $(2,2,1,1)$ works.
    [/list]
    This exhausts all cases, thus proving the claim. $\blacksquare$

To finish, note the following:
[list]
    [*]If $x=0$, then the left-hand expression is even, so no solutions.
    [*]If $y=0$, then $5^z7^w=2^x-1$, so by Zsigmondy theorem we need $x\le4$ or $x=6$, for which only $(1,0,0,0)$ and $(3,0,0,1)$ work.
    [*]If $z=0$, then $2^x3^y=7^w+1$, but the right-hand expression is always $2\pmod3$, so $y=0$, and by Mih\u ailescu theorem only $(1,0,0,0)$ and $(3,0,0,1)$ work.
    [*]If $w=0$, then $2^x3^y=5^z+1$, so by Zsigmondy theorem, $z\le2$, thus only $(1,0,0,0)$ and $(1,1,1,0)$ work.
[/list]
This completes the proof.

Eyed has a solution; I think it's really well-written:
[quote=Eyed]The only solution is \[(3,0,0,1), (2,2,1,1), (1,1,1,0), (1,0,0,0)\]

Case 1: $a\geq 3, b\geq 1$

If $a\geq3$ and $b\geq 1$, then taking $\mod 24$, we get $5^{c}7^{d}\equiv -1\mod 24$. Taking $\mod 8$ we get $5^{c}7^{d}\equiv -1\mod 8$, which implies $c$ is even and $d$ is odd. Then, if $c$ is even, $5^{c}\equiv 1\mod 24$, and if $d$ is odd, $7^{d}\equiv 7\mod 24$, so $5^{c}7^{d}\equiv 7\mod 24\not\equiv-1\mod 24$, a contradiction. There are no solutions for this case.

Case 2: $a\geq 3$, $b = 0$

We get $2^{a} - 5^{c}7^{d} =1$. If $c\geq 1$, then taking $\mod 5$ results in $a$ even, taking $\mod 8$ results in $c$ even, and taking $\mod 3$ gives us $1-1\equiv 1\mod3$, a contradiction. Otherwise, assume $c = 1$. Then, we get $2^{a} - 7^{d} = 1$. If $d\geq 2$, then by Catalan's conjecture there are no solutions. Thus, we only need to check $d = 0,1$, in which $(a,d) = (3,1)$ works, which is the only solution for this case.

Case 3: $b \geq1$ and $a = 2$.

We have
\[4\cdot 3^{b} -1 = 5^{c}7^{d}\]
Taking $\mod 3$ results in $c$ is odd, so $c\geq 1$. Taking $\mod 5$ results in $b\equiv2\mod4$. Taking $\mod 4$ also reveals that $d$ is odd, so $d\geq 1$. Taking $\mod7$ results in $b\equiv2\mod 6$, so $b\equiv2\mod12$. Then, we let $b = 2b_{1}$. Factoring, we get
\[(2\cdot3^{b_{1}} - 1)(2\cdot3^{b_{1}} + 1) = 5^{c}7^{d}\]
Either $b_{1}\equiv 1\mod 12$ or $b_{1}\equiv 7\mod 12$. In both cases, we can't have one factor be divisible by both $7$ and $5$, so one factor is equal to $5^{c}$ and the other $7^{d}$. If $2\cdot 3^{b_{1}} + 1 = 5^{c}$, taking $\mod 4$ results in no solutions. If $2\cdot 3^{b_{1}} + 1= 7^{d}$, then
\[b_{1} = v_{3}(2\cdot 3^{b_{1}}) = v_{3}(7^{d} - 1) = v_{3}(6) + v_{3}(d) = 1 + v_{3}(d)\]
so $v_{3}(d) = b_{1} - 1$, and $d\geq 3^{b_{1} - 1}$. It is easy to see that, for any $b_{1}>1$, the RHS is much larger than the LHS. The only solution is therefore $b_{1} = 1, d = 1$. For this case, the only solution is $b = 2, c = 1, d = 1$.

Case 4: $b \geq 1, a= 1$

Our equation becomes \[2\cdot 3^{b}- 5^{c}7^{d} = 1\]
Taking $\mod 3$, we get that $c$ is odd, so $c\geq 1$. Then, taking $\mod 5$, we get that $b$ is odd. If $d\geq 1$, then taking $\mod 7$ we have $c\equiv 4\mod 6$, a contradiction, so $d = 0$. Our equation now becomes $2\cdot3^{b} = 5^{c} + 1$. Then, we have $b = v_{3}(5^{c} + 1) = v_{3}(6) + v_{3}(c)$, so $v_{3}(c) = b-1 \Rightarrow c\geq 3^{b-1}$. Then, for all $b\geq 1$, the RHS would be much greater than the LHS. Thus, the only solution for this case is $b = 1, c = 1, d = 0$.

Case 5: $b\geq 1, a = 0$

Our equation becomes \[3^{b} - 5^{c}7^{d} = 1\]

Taking $\mod 3$ we get $c$ is odd, so $c\geq 1$. Then, taking $\mod 5$, we have $b\equiv 0\mod 4$. Let $b = 2b_{1}$. Then we get
\[(3^{b_{1}} - 1)(3^{b_{1}} + 1) = 5^{c}7^{d}\]
If one of the factors is divisible by both $5$ and $7$, then the other factor is divisible by neither, so the other factor must be $1$, which is impossible. Thus, assume each factor is a power of $5$ or $7$. If $3^{b_{1}} - 1 = 5^{c}$, for any $c\geq2$ this is impossible by Catalan's conjecture, and for $c = 0,1$ we can check it also doesn't work. If $3^{b_{1}}-1 = 7^{d}$, again if $d\geq 2$ this is impossible by Catalan's, and if $d = 0,1$ we can again check it does not work. There are no solutions for this case.

Case 6: $b<1, a<3$
We can hand check all the values of $a$ and $b$ then, which are $(a,b) = (2,0), (1,0), (0,0)$. The only values that work are $(a,b,c,d) = (1,0,0,0)$.

Now, since we've covered all the cases, we can assure that the only solutions are $(a,b,c,d) = $
\[(3,0,0,1), (2,2,1,1), (1,1,1,0), (1,0,0,0)\][/quote]

First we deal with the edge cases where one of the variables is 0. Suppose $a=0$, then we have $3^b-1=5^c7^d$. We can check $b\in [0, 6]$ don't yield any solutions, and for $b\ge 7$, we can apply Zsigmondy to get a contradiction. Next, suppose $b=0$ and $a>0$, so $2^a-1=5^c7^d$. We see $a=1, c=d=0$ and $a=3, c=0, d=1$ work, and $a=2, 4, 6$ don't work, and for all other $a$ we can apply Zsigmondy to get a contradiction. Next, suppose $c=0$ and $a, b>0$, so $2^a3^b=7^d+1$. Taking mod 3 immediately yields a contradiction. Finally, suppose $d=0$ and $a, b, c>0$, so $2^a3^b=5^c+1$. We see $c=1, a=b=1$ works, and for $c\ge 2$ we can apply Zsigmondy to get a contradiction. Putting this all together, we find that the solutions we get when one of the variables is 0 are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0)$. \\

Now, suppose $a, b, c, d\ge 1$. Taking mod 3 we get $c$ is odd. We proceed by splitting into cases based on the value of $a$. If $a\ge 3$, then taking mod 4 we get $d$ is odd, but then taking mod 8 we get $5^c\equiv 1\pmod{8}$, which is a contradiction since $c$ is odd. If $a=1$, then taking mod 5 we get $b\equiv 1\pmod{4}$, but then taking mod 7 we get $3^b\equiv 4\pmod{7}$, which is a contradiction since $b$ is odd. Finally, we deal with $a=2$; first note $(2, 2, 1, 1)$ is a solution. Now, taking mod 5 we get $b$ even, so set $b=2b_1$, and we have the equation $(2\cdot 3^{b_1}+1)(2\cdot 3^{b_1}-1)=5^c7^d$.  The two factors on the LHS are relatively prime, so we either have $7^d+1=2\cdot 3^{b_1}$, or $7^d-1=2\cdot 3^{b_1}$. The former case dies mod 3. For the latter, note $d=1$ gives us the solution we already have, and $d\ge 2$ gets killed by Zsigmondy.\\

Hence, putting everything together, our solutions for $(a, b, c, d)$ are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$.

Mod bash because I don't like Zsigmondy lol (jk I'm just dumb and forgot): We just consider many cases which I won't go into too much detail.

If $b = d = 0$, notice that $a = 1$ a solution when $c = 0$ so $(1, 0, 0, 0)$ is a solution, but when $a \ge 2$ mod $4$ gives no solutions.

If $b = c = 0$, notice by mod $64$ that there does not exist an integer $d$ where $7^d \equiv -1 \pmod {64}$, and also notice by mod $7$ that $a \equiv 0 \pmod 3$. When $a = 0$ we get the solution we already had. When $a = 3$ we get $d = 1$ so $(3, 0, 0, 1)$ is a solution, but when $a \ge 6$ we get mod $64$ contradiction.

If $a = 0$ then mod $2$ gives no solutions.

If $b = 0$ but $c, d$ are nonzero, then mod $5$ gives $a \equiv 0 \pmod 3$ and mod $7$ gives $a \equiv 0 \pmod 4$ so $a \equiv 0 \pmod {12}$, but mod $13$ means $5^c7^d \equiv 0 \pmod {13}$, which is false, so no solutions.

If $d = 0$ and $a, b, c$ are nonzero, if $a = 1$ then we get $(1, 1, 1, 0)$ as a solution. By mod $4$ we see $a \le 1$ so $a = 1$ so consider when $c \ge 2$. By mod $9$ we have $c \equiv 3 \pmod 6$ and by mod $7$ this means $5^c = -1 \pmod 7$ so we need $2\cdot 3^b \equiv 0 \pmod 7$ which is false, so no solutions.

If $c = 0$ and $b$ is nonzero, mod $3$ gives no solutions.

Finally we consider when $a, b, c, d$ are all nonzero. Mod $5$ gives $a \equiv b \pmod 4$, so in addition with mod $7$ gives $a \equiv b \equiv 2, 4, 0 \pmod 6$. This means $a, b$ are even so we let $2x = a, 2y = b$ for positive integers $x, y$. Using difference of squares, we get $5^c7^d = (2^x3^y-1)(2^x3^y+1)$ so either $2^x3^y-1 = 5^c$ and $2^x3^y+1 = 7^d$, or the other way around so we have two cases.

If $5^c-2 = 7^d$, then mod $7$ gives $c \equiv 4 \pmod 6$ so mod $9$ gives $7^d \equiv 5^c-2 \equiv 2 \pmod 9$. However, there does not exist positive integer $d$ where $7^d \equiv 2 \pmod 9$, so there are no solutions for this case.

If $7^d-2 = 5^c$, if $c = 1$ then $d = 1$ so we get a solution $(2, 2, 1, 1)$. Otherwise $c \ge 2$, so by mod $25$ we see that there $7^d \equiv 7, 49, 43, 1 \ne 2 \pmod {25}$ which gives no solutions.

Therefore, the solutions are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$. $\blacksquare$

The cases where $abcd=0$ can be dealt with by simple mods and Zsigmondy. Henceforth assume $abcd \neq 0$.

[i]First Step.[/i] If $a \geq 3$, then notice the following:
[list]
[*]By taking mod $3$, $c$ must be odd;
[*]By taking mod $8$, $c$ must be even and $d$ odd.
[/list]
This is an obvious contradiction.

[i]Second Step.[/i] Now assume $a=2$. This is perhaps the hardest case of all. Similar analysis of mod $3$ and mod $8$ yields $c$ and $d$ are both odd. Mod $35$ yields $b$ is even. Thus we have $$5^c 7^d = (2 \cdot 3^{b/2} + 1)(2 \cdot 3^{b/2} - 1).$$ This means that $\{5^c, 7^d\} = \{2 \cdot 3^{b/2} + 1, 2 \cdot 3^{b/2}- 1\}$. Now Zsigmondy works again (!).

[i]Third Step.[/i] Finally, assume $a=1$. The congruence collapses to mod $8$ and mod $5$ now.

Collecting all the edge cases, the final solutions are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$.

would doing [hide=this]
If $y=0$, then $2^x-1=5^z\cdot 7^w$, and because $2^3-1=7$ and $2^4-1=3\cdot 5$, so there are no solutions for $x>4$. This gives $(3,0,0,1)$ and $(1,0,0,0)$.

If $z=0$, then $7^w+1=2^x\cdot 3^y$, but because $7^w+1\equiv 2\pmod{3}$, we must have $y=0$. But because $7^1+1=2^3$, there will be no more solutions for $w>1$. If $w=1$, then $x=3$, so we have $(3,0,0,1)$. If $w=0$, then $x=1$, so we have $(1,0,0,0)$.

If $w=0$, then $5^z+1=2^x\cdot 3^y$. $5^1+1=2\cdot 3$, so there are no solutions for $z>1$. When $z=0$, we obtain $(1,0,0,0)$, and when $z=1$, we obtain $(1,1,1,0)$.
[/hide] for the case when $wxyz=0$ work?

If $a \ge 3$, then by $\pmod 8$ and $\pmod 3$ we receive a contradiction. Therefore $a < 3$. Notice that $a=0$ obviously doesn't work. If $a=1$ and $c, d > 0$, then we have $3^b = 18 \pmod{35}$ which has no solutions. Therefore one of $c,d$ has to be $0$ and these cases are easy to check. Now if $a = 2$, notice that by $\pmod{35}$, $b = 2 \pmod{12}$, so $b$ is even. Let $b = 2k$. $4 \cdot 3^b - 1 = (2 \cdot 3^k + 1)(2 \cdot 3^k - 1) = 5^c7^d$. Now since they are relatively prime to each other, each one is equal to either $5^c$ or $7^d$. Now, Zsigmondy's finishes.

Use Zsigmondy if any of a,b,c,d are 0 (for example $b=0$ means one of the primes in the set $\{5,7\}\mid5^c7^d=2^a-1$ is s.t. $5\nmid2^k-1\forall k<a\implies a=4$ and then solve, or $7\nmid2^k-1\forall k<a\implies a=3$ and solve). Henceforth assume all of them are positive, and call the assertion of taking mod x f(x), and taking mod 2 of x g(x).

    We have $\{f(8),f(3)\}\rightarrow\{(g(d)=1,g(c)=0),(g(c)=1)\}$, contradiction, so $a < 3$. $a=1\implies3^b = 18 \pmod{35}$ which has no solutions. $$a = 2\stackrel{f(35)}{\implies}b = 2 \pmod{12}\implies g(b) = 0,3^b4 - 1 = (3^{b'}2 + 1)(3^{b'}2 - 1) = 5^c7^d;$$ since they are relatively prime to each other, each one is equal to either $5^c$ or $7^d$; Zsigmondy also finishes here.

    We conclude the answers are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$.

We can handle all of the cases where $abcd = 0$.
$\newline$

If $d = 0$, then we have $2^{a}3^{b} = 5^c + 1$. By Zsigmondy, the only solution here is $(a, b, c, d) = (1, 1, 1, 0)$ and $(1, 0, 0, 0)$.
$\newline$

If $c = 0$ then we have $2^{a}3^{b} = 7^d + 1$. Using Zsigmondy again, we have $(a, b, c, d) = (3, 0, 0, 1)$ and $(1, 0, 0, 0)$.
$\newline$

If $b = 0$ then $2^a = 1 + 5^{c}7^{d}$. Then using Zsigmondy, we have $(3, 0, 0, 1)$.

$\newline$
If $a = 0$ then $3^b = 1 + 5^{c}7^{d}$. There are no solutions here.

$\newline$
Now, taking$\pmod{8}$ for $a \geq 3$ we get $5^{c}7^d \equiv -1\pmod{8}$. Then $c$ is even and $d$ is odd.
Taking$\pmod{3}$, we get $5^{c}7^d = -1\pmod{3}$, if $c$ even and $d$ odd, then it is not true, so we have a contradiction.

Therefore, $a < 3$.

$\newline$
If $a = 1$, then by$\pmod{5}$ $23^b \equiv 1\pmod{5}$
Then $b \equiv 1\pmod{4}$. From our previous results with$\pmod{8}$, this is impossible, so $a$ must equal $2$.

$\newline$.
For $a = 2$, $(b, c, d) = (2, 1, 1)$ is clearly a solution.
There aren't any other solutions for $(c, d) \leq 1$.

Taking$\pmod{35}$, we have $4(3^b) \equiv 1\pmod{35}$, so $b = 0\pmod{2}$
Then by difference of squares we have
$5^{c}7^d = (2 \cdot 3^{\frac{b}{2}} - 1)(2 \cdot 3^{\frac{b}{2}} + 1)$.
$\newline$
Since the two terms on the RHS are relatively prime, one of them must equal to $5^c$ and the other $7^d$.
For $5^c =  2 \cdot 3^{\frac{b}{2}} - 1$, $(a, b, c, d) = (2, 2, 1, 1)$ is clearly possible, and by Zsig there are no other solutions.

For $5^c = 2 \cdot 3^{\frac{b}{2}} + 1$, we have $5^c - 1 = 2 \cdot 3^{\frac{b}{2}}$. Since $4$ is a factor of $5^c - 1$, this is impossible.
$\newline$
So all of our solutions are $(2, 2, 1, 1)$, $(3, 0, 0, 1)$, $(1, 0, 0, 0)$, $(1, 1, 1, 0)$.

The answer is $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$, and $(2,2,1,1)$.\\

If any of $a,b,c,d$ are zero, we can use Zsigmondy to show that the first three are the only solutions. For example, when $d=0$, we need $5^c+1$ to only have 2 and 3 as prime factors, and $c=1$ already has both 2 and 3, so by Zsigmondy, everything above $c=1$ has at least one other prime factor, so only $c=0$ and $c=1$ can possibly work. We can use a similar approach for $a=0$, $b=0$, and $c=0$.\\

From now, assume $a,b,c,d\geq 1$. The core idea of this problem is repeatedly taking suitable moduli.\\

Claim: $a\leq 2$. Suppose otherwise. Taking mod 8 reveals that $$5^c7^d\equiv 7\pmod{8}.$$ This means that $c$ is even and $d$ is odd. However, this means that $5^c\equiv 1\pmod{3}$ and of course $7^d\equiv1\pmod 3$. However, since $2^a3^b$ is a multiple of 3, taking the original equation mod 3 gives $$0-1\equiv 1\pmod{3},$$ contradiction.\\

Case 1: $a=1$. Then, we have $$2\cdot 3^b-5^c\cdot 7^d=1.$$ Taking mod 5, we have $3^b\equiv 3\pmod{5}$, which means that $b\equiv1\pmod{4}$. However, taking mod 7, we have that $3^b\equiv 4\pmod{7}$, so $b\equiv 4\pmod{6}$, but this contradicts with out previous conclusion. Therefore, no more solutions in this case.\\

Case 2: $a=2$. Taking mod 5 gives $3^b\equiv 4\pmod{5}$. $b\equiv 2\pmod{4}$. In particular, $b$ is even. Thus, factor as a difference of squares $$(2\cdot 3^{b/2}-1)(2\cdot 3^{b/2}+1)=5^c\cdot 7^d.$$ Note that each term on the left side can only have 5 and 7 as prime factors. However, they only differ by 2, so it is not possible for both of them to be divisible by 5 or both to be divisible by 7, the only possibility is that one of them is a power of 5 and one of them is a power of 7.\\

Claim 2: The only way a power of 5 and a power of 7 differ by 2 is 5 and 7. Consider $5^n\pm 2=7^m$. If $n\geq 2$, then taking mod 25 yields that $$7^m\equiv \pm2\pmod{25},$$ neither of which is possible, as the cycle for 7 is $1,7,-1,-7,1,\dots$

Thus, we must have $2\cdot 3^{b/2}-1=5,$ so $b=2$. This gives the solution $(2,2,1,1)$.

nice   :)
[color=blue][b]Case 1.$ abcd=0$[/color] By Zsigmondy, we want only two primes in the $u^n\pm 1$ factorization, so we can do only few cases giving us the answer here is $$\boxed {(a,b,c,d)=(1,0,0,0), (3,0,0,1), (1,1,1,0)}$$
[color=blue][b]Case 2. $a,b,c,d>0$ [/color]
[list]
[*]$\pmod 3$ gives $c$ odd
[*]$\pmod 8$ gives $a<3$
[*]$\pmod 5$ gives $2^a\cdot 3^b\equiv 1$ so $4\mid a-b$
[*]$\pmod 7$ gives $2^a3^b\equiv 1 \pmod 7$ and if $a=1$ then $b$ odd and $3^b\equiv 4\pmod 7$ impossible, so $a\neq 1$
 [/list]
This gives us $a=2$ and $b$ even, so
$$\left(2\cdot 3^{\frac{b}{2}}-1\right)\left(2\cdot 3^{\frac{b}{2}}+1\right)=5^c\cdot 7^d$$
Both factor are relatively primes and the first is $-1\pmod 3$ so
$$2\cdot 3^{\frac{b}{2}}=5^c+1 \hspace{0.5cm}\text{and}\hspace{0.5cm} 2\cdot 3^{\frac{b}{2}}=7^d-1$$
So we can do Zsigmondy again and $d=1$ so $b=a=2$ and $c=1$, so the answer is $$\boxed {(a,b,c,d)=(1,0,0,0), (3,0,0,1), (1,1,1,0),(2,2,1,1)}$$

The solution set is $\boxed{(a,b,c,d) = (3,0,0,1),(1,1,1,0),(2,2,1,1), (1,0,0,0)}$.

If $a=0$, modulo $2$ yields an obvious contradiction.

If $b=0$, we have

\[2^a-1 = 5^c7^d.\]

Realize that $2^{12}-1$ contains a factor of $5$ and $7$ in it, so $a \le 12$ by Zsigmondy. Manually testing each case yields the first solution.

If $c=0$, we have

\[7^d+1 = 2^a3^b.\]

The LHS cannot be a multiple of $3$, so we just have

\[7^d+1=2^a.\]

Now, Zsigmondy shows that the only possible solutions satisfy $d<2$, which yields the first and last solutions.

If $d=0$, we have

\[5^c+1 = 2^a3^b.\]

Zsigmondy shows that only $c=0$ or $c=1$ works, yielding the second and last solutions.

From this point on, assume that $abcd \neq 0$ as we have exhausted all of the cases where $abcd=0$.

For the sake of contradiction, suppose that $a \ge 3$:

[list]
    [*] Modulo $3$ gives $2^c \equiv -1 \pmod{3}$, so $c$ is odd.

    [*] Modulo $8$ gives

    \[5 \cdot (-1)^d = -1 \pmod{8}, \]

    which is impossible.
[/list]

Thus, $a <3$. Now, we will consider this equation through more mods:

[list]
    [*] Modulo $5$ yields

    \[2^a(3)^b \equiv 2^a \left(\frac{6}{2} \right)^b \equiv 2^a \cdot 2^{-b} \equiv 2^{a-b} \equiv 1 \pmod{5}.\]

    It follows that $4 \mid a-b$.

    [*] Modulo $7$ gives

    \[2^a3^b \equiv 1 \pmod{7}.\]
[/list]

If $a=1$, we have

\[6(2^{a_0}3^{b_0})^2 \equiv 1 \pmod{7},\]

for nonnegative integers $a_0$ and $b_0$ (note that $a$ and $b$ are the same parity). However, $6$ is not a quadratic residue modulo $7$, so there are no solutions in this case.

This means that $a=2$. It follows that $b$ is even as well, so let $b=2e$. We have

\[4 \cdot 3^{2e}-1 = 5^c7^d\]
\[\implies (2 \cdot 3^e-1)(2 \cdot 3^e+1) = 5^c7^d.\]

The two terms are relatively prime up to a power of $2$, so a modulo $3$ argument yields

\begin{align*}
2 \cdot 3^e-1 = 5^c \\
2 \cdot 3^e+1 = 7^d.
\end{align*}

Zsigmondy applies one last time and gives $c=d=e=1$, the final solution in our solution set.

We first handle the cases when at least one of $a$, $b$, $c$, $d$ are 0:
[list]
[*] Four 0's: No solutions.
[*] Three 0's: $(1,0,0,0)$.
[*] Two or one 0: Isolate (either) lone term and apply Zsigmondy to get the solutions $(3,0,0,1)$, $(1,1,1,0)$.
[/list]

Assume $\min(a,b,c,d) \ge 1$. Mod 35 gives $a$, $b$ even while mod 3, 4 gives us $c$, $d$ odd. Thus our equation can be rewritten as
\[x^2-35y^2 = 1, \operatorname{rad}(x) = 6, \operatorname{rad}(y) = 35 \implies x+y \sqrt{35} = (6+\sqrt{35})^k\]

using Pell. Expanding using Binomial Theorem and comparing integer and irrational parts, we find the only solution $(2,2,1,1)$, giving our four solutions
\[\boxed{(1,0,0,0), (3,0,0,1), (1,1,1,0), (2,2,1,1).}\]

I claim that the only solutions are the only solutions are $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$, and $(2,2,1,1)$. \\\\
First, note that if we rearrange the equation, we get that
\[2^a3^b=5^c7^d+1,\]
the RHS of which must be odd. Therefore, $a\geq1$. Now, we will take cases on $b$.

[color=#9a00ff][b]1. $b=0$.[/b][/color]
Writing this in, we have that
\[2^a-1=5^c7^d.\]
First, I claim that $c=0$. This is because if $c>0$, then we have that $5\mid 2^a-1$, which means that we must have $4\mid a$, since the order of $2$ mod $5$ is $4$. This implies that
\[2^4-1\mid 2^a-1 \iff 15\mid 2^a-1 \iff 3\mid 2^a-1=5^c7^d,\]
an obvious contradiction. Therefore, we must have that $c=0$. This gives us that
\[2^a-1=7^d \iff 2^a=7^d+1.\]
Clearly, $(a,d)=(1,0)$ and $(a,d)=(3,1)$ are solutions. However, if $d>1$, by Zsigmondy's, we have that there must exist a prime $p$ that divides $2^a=7^d+1$ but not $7^1+1=8$. This is a contradiction, since no prime other than $2$ can divide $2^a$. Therefore, in the entire case where $b=0$, our only solutions are $(a,b,c,d)=(1,0,0,0)$ and $(3,0,0,1)$.

[color=#9a00ff][b]2. $b>0$.[/b][/color]
First, I claim that $c$ must be odd. This is because we have that
\[2^a3^b=5^c7^d+1 \iff 5^c7^d\equiv 2\mod 3 \iff 5^c\equiv 2\mod 3,\]
which only occurs when $c$ is odd. Therefore, since $c$ is odd, we must have that $c\geq 1$. Additionally, this also means that $5^c \equiv 5 \mod 8$. Since $7^d$ is always either $1$ or $-1$ mod $8$, this implies that
\[2^a3^b=5^c7^d+1\equiv \pm 5^c+1\equiv \pm 5+1 \mod 8,\]
meaning that $2^a3^b$ is either $4$ or $6$ mod $8$. This means that $a=1$ or $a=2$. We now use this to take cases.

[color=#9a00ff][b]2a. $a=1$.[/b][/color]
Writing this in, we have that
\[2(3^b)-1=5^c7^d \iff 2(3^b)=5^c7^d+1.\]
First, I claim that $d$ must be $0$. Note that since $c$ is odd and therefore $\geq 1$, we have that
\[5\mid 2(3^b)-1 \iff 3^b\equiv 3 \mod 5,\]
which only occurs when $b$ is $1$ mod $4$. However, if $d>0$, we have that
\[7\mid 2(3^b)-1 \iff 3^b\equiv 4 \mod 7,\]
which only occurs when $b$ is $4$ mod $6$, which implies that $b$ must be even. However, from before, we had that $c$ must be $1$ mod $4$, implying that $c$ must be odd, a contradiction. Therefore $d$ must be $0$. \\\\
Now, writing this in, we get that
\[2(3^b)-1=5^c \iff 2(3^b)=5^c+1.\]
Clearly, $(b,c)=(1,1)$ is a solution. However, if $c>1$, note that by Zsigmondy's, we have that there must exist a prime $p$ that divides $2(3^b)=5^c+1$ but not $5^1+1=6$. This is a contradiction, since no primes other than $2$ or $3$ can divide $2(3^b)$. Therefore, since we must also have that $b>0$, there exist no other solutions in this case other than $(1,1,1,0)$.

[color=#9a00ff][b]2b. $a=2$.[/b][/color]
Again writing this in, we have that
\[4(3^b)-1=5^c7^d.\]
Now, note that since $c$ is odd and therefore $\geq 1$, we have that
\[5\mid 4(3^b)-1 \iff 4(3^b)\equiv 1 \mod 5 \iff 3^b \equiv 4 \mod 5,\]
which only occurs when $b$ is $2$ mod $4$. Therefore $b$ must be even. Let $b=2k$. Now by difference of squares, we have that
\[4(3^b)-1=5^c7^d\iff 2(3^k)-1 \mid 5^c7^d \iff 2(3^k)-1=5^m7^n,\]
for some non-negative integers $m$ and $n$. However, this is the exact same equation as [color=#9a00ff][b](2a)[/b][/color], which gave us that $k$ must be $1$. Using this, we get that
\[5^c7^d=4(3^{2k})-1=4(3^2)-1,\]
which gives us our only solution in this case of $(a,b,c,d)=(2,2,1,1)$. \\\\
Finally, we have exhausted all cases, giving us that the only solutions are $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$, and $(2,2,1,1)$, finishing the problem.