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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
GCD of terms in a sequence
BBNoDollar   0
an hour ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
an hour ago
0 replies
Number Theory
fasttrust_12-mn   13
N an hour ago by KTYC
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
13 replies
fasttrust_12-mn
Aug 15, 2024
KTYC
an hour ago
GCD of terms in a sequence
BBNoDollar   0
an hour ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
an hour ago
0 replies
Aime type Geo
ehuseyinyigit   3
N an hour ago by sami1618
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
3 replies
ehuseyinyigit
Yesterday at 9:04 PM
sami1618
an hour ago
minimizing sum
gggzul   1
N 2 hours ago by RedFireTruck
Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$. Find
$$min\{12x-4y-3z\}.$$
1 reply
gggzul
3 hours ago
RedFireTruck
2 hours ago
Equilateral Triangle inside Equilateral Triangles.
abhisruta03   2
N 2 hours ago by Reacheddreams
Source: ISI 2021 P6
If a given equilateral triangle $\Delta$ of side length $a$ lies in the union of five equilateral triangles of side length $b$, show that there exist four equilateral triangles of side length $b$ whose union contains $\Delta$.
2 replies
abhisruta03
Jul 18, 2021
Reacheddreams
2 hours ago
USAMO 1984 Problem 5 - Polynomial of degree 3n
Binomial-theorem   8
N 2 hours ago by Assassino9931
Source: USAMO 1984 Problem 5
$P(x)$ is a polynomial of degree $3n$ such that

\begin{eqnarray*}
P(0) = P(3) = \cdots &=& P(3n) = 2, \\
P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\
P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\
&& P(3n+1) = 730.\end{eqnarray*}

Determine $n$.
8 replies
Binomial-theorem
Aug 16, 2011
Assassino9931
2 hours ago
Finding positive integers with good divisors
nAalniaOMliO   2
N 2 hours ago by KTYC
Source: Belarusian National Olympiad 2025
For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$.
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$.
2 replies
1 viewing
nAalniaOMliO
Mar 28, 2025
KTYC
2 hours ago
Balkan MO 2025 p1
Mamadi   1
N 3 hours ago by KevinYang2.71
Source: Balkan MO 2025
An integer \( n > 1 \) is called good if there exists a permutation \( a_1, a_2, \dots, a_n \) of the numbers \( 1, 2, 3, \dots, n \), such that:

\( a_i \) and \( a_{i+1} \) have different parities for every \( 1 \le i \le n - 1 \)

the sum \( a_1 + a_2 + \dots + a_k \) is a quadratic residue modulo \( n \) for every \( 1 \le k \le n \)

Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.

Remark: Here an integer \( z \) is considered a quadratic residue modulo \( n \) if there exists an integer \( y \) such that \( y^2 \equiv z \pmod{n} \).
1 reply
Mamadi
4 hours ago
KevinYang2.71
3 hours ago
Random Points = Problem
kingu   4
N 3 hours ago by zuat.e
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
4 replies
kingu
Apr 27, 2024
zuat.e
3 hours ago
CooL geo
Pomegranat   2
N 3 hours ago by Curious_Droid
Source: Idk

In triangle \( ABC \), \( D \) is the midpoint of \( BC \). \( E \) is an arbitrary point on \( AC \). Let \( S \) be the intersection of \( AD \) and \( BE \). The line \( CS \) intersects with the circumcircle of \( ACD \), for the second time at \( K \). \( P \) is the circumcenter of triangle \( ABE \). Prove that \( PK \perp CK \).
2 replies
Pomegranat
Yesterday at 5:57 AM
Curious_Droid
3 hours ago
Unique Rational Number Representation
abhisruta03   18
N 3 hours ago by Reacheddreams
Source: ISI 2021 P3
Prove that every positive rational number can be expressed uniquely as a finite sum of the form $$a_1+\frac{a_2}{2!}+\frac{a_3}{3!}+\dots+\frac{a_n}{n!},$$where $a_n$ are integers such that $0 \leq a_n \leq n-1$ for all $n > 1$.
18 replies
abhisruta03
Jul 18, 2021
Reacheddreams
3 hours ago
Math solution
Techno0-8   1
N 3 hours ago by jasperE3
Solution
1 reply
Techno0-8
6 hours ago
jasperE3
3 hours ago
D1027 : Super Schoof
Dattier   1
N 3 hours ago by Dattier
Source: les dattes à Dattier
Let $p>11$ a prime number with $a=\text{card}\{(x,y) \in \mathbb Z/ p \mathbb Z: y^2=x^3+1\}$ and $b=\dfrac 1 {((p-1)/2)! \times ((p-1)/3)! \times ((p-1)/6)!} \mod p$ when $p \mod 3=1$.



Is it true that if $p \mod 3=1$ then $a \in \{b,p-b, \min\{b,p-b\}+p\}$ else $A=p$.
1 reply
Dattier
Today at 5:15 PM
Dattier
3 hours ago
IMO ShortList 2008, Number Theory problem 3
April   24
N Apr 29, 2025 by sansgankrsngupta
Source: IMO ShortList 2008, Number Theory problem 3
Let $ a_0$, $ a_1$, $ a_2$, $ \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $ \gcd (a_i, a_{i + 1}) > a_{i - 1}$. Prove that $ a_n\ge 2^n$ for all $ n\ge 0$.

Proposed by Morteza Saghafian, Iran
24 replies
April
Jul 9, 2009
sansgankrsngupta
Apr 29, 2025
IMO ShortList 2008, Number Theory problem 3
G H J
Source: IMO ShortList 2008, Number Theory problem 3
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April
1270 posts
#1 • 5 Y
Y by Davi-8191, anantmudgal09, Adventure10, Mango247, kiyoras_2001
Let $ a_0$, $ a_1$, $ a_2$, $ \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $ \gcd (a_i, a_{i + 1}) > a_{i - 1}$. Prove that $ a_n\ge 2^n$ for all $ n\ge 0$.

Proposed by Morteza Saghafian, Iran
Z K Y
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Allnames
925 posts
#2 • 2 Y
Y by Adventure10, Mango247
April wrote:
Let $ a_0$, $ a_1$, $ a_2$, $ \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $ \gcd (a_i, a_{i + 1}) > a_{i - 1}$. Prove that $ a_n\ge 2^n$ for all $ n\ge 0$.
I use $ (a;b)$ as $ gcd(a;b)$.
I prove this nice problem by induction.
It is obvious to see that $ a_n > a_{n - 1}\forall n\in N$
Firstly, $ a_0\ge 1$. Let $ (a_1;a_2) = d > a_0\ge 2$. Thus $ a_1\ge 2$, $ a_2 = kd \ge 4$ because $ k\ge 2$.
Suppose that this problem has been right until $ n - 1$. It means $ a_i\ge 2^i \forall i\le n - 1$.
Assume that $ 2^{n - 1}\le a_{n - 1} < a_n < 2^n$. Let $ (a_n;a_{n - 1}) = d > a_{n - 2}\ge 2^{n - 2}$ and $ a_n = kd < 2^n;k\ge 2$.
Because $ d\ge 2^{n - 2}$, if $ k\ge 4$ then $ a_n = kd\ge 2^n$ contr.It implies $ k = 2$ or $ k = 3$.
Case I: $ k = 2$ then $ a_{n - 1} = d\ge 2^{n - 1}$(induction hypothesis) and $ a_n = 2d\ge 2^n$ which is contradiction !.
Case II: $ k = 3$ then if $ a_{n - 1} = d$ by the same argument above, we have the contradiction .
It means $ a_n = 3d;a_{n - 1} = 2d$.Let $ (a_{n - 2};2d) = T > a_n - 3\ge 2^{n - 3}$. Hence $ 2d = aT,a\ge 2$
$ a\ge 2$ implies $ T < 2^{n - 1}$(since $ 2^{n - 1}\le 2d = aT < 2^n$). If $ a\ge 4$ then $ aT\ge 2^{n - 1}$.It is impossible.
Then $ a = 2$ or $ a = 3$.
Case1: $ a = 2$ then $ d = T$ then $ a_{n - 2} = T\ge 2^{n - 2}$ then $ d\ge 2^{n - 1}$. Impossible!.
Case 2: $ k = 3$. If $ a_{n - 2} = T$ then $ a_{n - 1} = 3T\ge 3.2^{n - 2}$ then $ a_n = 1,5 a_{n - 1}\ge 4,5.2^{n - 2} > 2^n$ contradiction.
It means $ 2d = 3T$ and $ a_{n - 2} = 2T$ . Thus $ d = 1,5T < a_{n - 2}$. But $ (a_n;a_{n - 1}) = d > a_{n - 2}$ since initial hypothesis.
So that assumption is wrong then $ a_n\ge 2^n$.
I hope I don't make any stupid mistakes.
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mr bui
13 posts
#3 • 1 Y
Y by Adventure10
Sorry sir!!!!
But we have a sequence isn't satisfy above condition:
$ a_0=1,a_1=2,a_2=6,a_3=6$, but $ a_3<2^3=8$.....
I thinks we have other condition in this problem.
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bboypa
469 posts
#5 • 4 Y
Y by liimr, rtsiamis, Adventure10, Mango247
April wrote:
Let $ a_0$, $ a_1$, $ a_2$, $ \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $ \gcd (a_i, a_{i + 1}) > a_{i - 1}$. Prove that $ a_n\ge 2^n$ for all $ n\ge 0$.

Our aim $ a_n \ge 2^n$(*) is true for $ n \in \{0,1\}$, in fact $ a_1=1$ would imply $ 1=(a_2,a_1)>a_0 \in \mathbb{N}_0$, absurd. It is also true that $ \{a_n\}_{n \in \mathbb{N}}$ is strictly incrasing since $ \min\{a_{n+1},a_{n+2}\}$ $ \ge (a_{n+1},a_{n+2})$ $ >a_n$. Now if (*) is true for $ n \in \{0,1,\ldots,n-1\}$ then it is true also for $ n$: in fact we have $ a_n-a_{n-1}$ $ \ge (a_n-a_{n-1},a_{n-1})$ $ =(a_n,a_{n-1}) >a_{n-2} \implies$ $ a_n > a_{n-1}+a_{n-2}$. Now if $ \frac{a_{n-1}}{3} \ge (a_{n-1},a_n)> a_{n-2}$ then we are done since $ a_n>a_{n-1}+a_{n-2} \ge 4a_{n-2} \ge 2^n$. Otherwise $ \frac{a_{n-1}}{2}=(a_n,a_{n-1})$. Now if $ a_n \ge 2a_{n-1}$ we are done, otherwise it means that $ 2 \mid a_{n-1}$, $ 3 \nmid a_{n-1}$ and $ a_n=\frac{3}{2}a_{n-1}$. Now if $ \frac{a_{n-1}}{4} \ge a_{n-2} \implies a_n>a_{n-1} \ge 2^n$, in the last last case (since $ 3 \nmid a_{n-1}$) we must have $ (a_{n-2},a_{n-1})=\frac{a_{n-1}}{2}$, but $ \frac{a_{n-1}}{2}=(a_n,a_{n-1})>a_{n-2}=\frac{a_{n-1}}{2}$, contradiction.
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liimr
34 posts
#6 • 1 Y
Y by Adventure10
bboypa wrote:
in the last last case (since $ 3 \nmid a_{n-1}$) we must have $ (a_{n-2},a_{n-1})=\frac{a_{n-1}}{2}$, but $ \frac{a_{n-1}}{2}=(a_n,a_{n-1})>a_{n-2}=\frac{a_{n-1}}{2}$, contradiction.
this part i think has problem
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JuanOrtiz
366 posts
#7 • 4 Y
Y by mathocean97, HolyMath, Adventure10, Mango247
It is a pity that there are no correct solutions here.

Here is my solution. Basically I used $3^2 \textgreater 2^3$, and other similar inequalities.

We use induction. Trivially $a$ is increasing, and we can check by hand that $a_i \ge 2^i$ for $i \le 4$.

Now assume $a_m \ge 2^m$ for $i \le n$, and assume $a_{n+1} \textless 2^{n+1}$. If $a_n | a_{n+1}$ we achieve a contradiction since $a_{n+1} \ge 2a_n$. Therefore $a_n$ doesn't divide $a_{n+1}$. Let $d=(a_n, a_{n+1}) \textgreater a_{n-1}$. If $a_{n+1} \ge 4d$ we're done, and if $a_{n+1} \le 2d$ then $a_n | a_{n+1}$. Therefore $a_{n+1}=3d$ and so $a_n=2d$.

Let $e=(2d,a_{n-1}) \textgreater a_{n-2} \ge 2^{n-2}$. Recall $d \textgreater a_{n-1}$. If $a_{n-1} | 2d$ then $a_{n-1} \le \frac{2d}{3}$ we see $3d \ge 9(2^{n-2})$ so we're done. So we get $a_{n-1} \neq e$, so $a_{n-1} \ge 2e$. But if $a_{n-1} \ge 3e$ then $3d \textgreater 3a_{n-1} \ge 9e \textgreater 9a_{n-2} \ge 9(2^{n-2})$ so we're done. From this we get $a_{n-1}=2e$ and so $e \ge 2^{n-2}$. If $2d \ge 6e$ then $3d \ge 9e \ge 9(2^{n-2})$ so we're done. Therefore $3e \le 2d \le 5e$. Since $d=(a_{n+1},a_n) \textgreater a_{n-1}=2e$ we get $2d \ge 4e$. Therefore $2d=5e$. So we can write $a_{n+1}=15f$, $a_n=10f$, $a_{n-1}=4f$.

Let $X = a_{n-2} \textless (a_n, a_{n-1})=2f$. Redefine $d$ so that $d=(a_{n-1},a_{n-2}) \textgreater a_{n-3} \ge 2^{n-3}$. Let $T=4f/d$. If $T \ge 5$ then $4f \ge 5d \ge 5(2^{n-3})$ and so $2^{n+1} \textgreater 15f \ge 5d(15/4) \ge 2^{n-3}(75/4) \textgreater 2^{n+1}$, contradiction. Since $X \textless 2f$ then $d \textless 2f$ and so $T \ge 3$. So $d=f$ or $4f/3$ and $d | X\le 2f$ and so we see $X=d$. So $d \ge 2^{n-2}$ and from this, $4f = Td \ge 3d \ge 3(2^{n-2})$ and so $15f \ge 3(2^{n-2})(\frac{15}{4}) \textgreater 2^{n+1}$ so we're done.

Such an ugly problem :(
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xdiegolazarox
42 posts
#9 • 2 Y
Y by Adventure10, Mango247
My solution :) :
It is obvious to see that $ a_n > a_{n - 1}\forall n\in N$.It is easy to see that true for $n = 0$,$1$,$2$,$3$,$4$. Now suppose that is true for $n = 0$,$1$,$2$ ... $k$ , $ k\ge 4$ .We assume that $ a_{k+1}<2^{k + 1}$ ,then it is easy to see that $ a_{k+1}< {2a_{k},4a_{k-1}}$, also $a_{k-1} < (a_{k}; a_{k+1}) \mid  a_{k+1} - a_{k}$, also $a_{k} = p(a_{k}; a_{k+1})$ , then $4a_{k-1} >a_{k+1} \ge (p+1) (a_{k}; a_{k+1})> (p+1) (a_{k-1})$, then $ p \le 2$, but $p=1$ implies that $a_{k+1} \ge 2 a_{k}$ (contradiction) , then $p=2$ and $2a_{k+1}=3a_{k}$, in particular $a_{k+1}=3R$ and $a_{k}=2R$ , $R= (a_{k}; a_{k+1})$, but $ a_{k+1}<4a_{k-1} \implies a_{k-1} >\frac{3}{4}R$ , also $a_{k-1}<R$ ,let $Q= (a_{k}; a_{k-1}) \implies 2^{k-2}<Q=\frac{a_{k}-a_{k-1}}{X} < \frac{5R}{4X} \implies 2^{k+1}>3R>\frac{2^{k}3X}{5} \implies X\le3$ ,but $ R > a_{k-1}\ge Q=\frac{a_{k}-a_{k-1}}{X} \implies R>\frac{a_{k}}{X+1} = \frac{2R}{X+1} \implies X\ge2$ , but $X=2 \implies Q=\frac{a_{k}-a_{k-1}}{2} < \frac{a_{k}}{2} $ and $ a_{k-1}<R=\frac{a_{k}}{2} \implies \frac{a_{k}-a_{k-1}}{2} >  \frac{a_{k}}{4} \implies  \frac{a_{k}}{3}=\frac{a_{k}-a_{k-1}}{2} \implies \frac{3}{4}R<a_{k-1}=\frac{a_{k}}{3}=\frac{2R}{3}$ (contradiction), then $X=3$ ,also $Q=\frac{2R-a_{k-1}}{3}  > \frac{2R}{6}$ and $a_{k-1} >\frac{3}{4}R>\frac{R}{2} \implies Q=\frac{2R-a_{k-1}}{3}< \frac{2R}{4} \implies Q=\frac{2R}{5}\implies a_{k-1}=\frac{4R}{5}$ ,let $T=(a_{k-1}; a_{k-2})=\frac{a_{k-2}}{Y} \implies\frac{2(2^{k+1})}{15}>\frac{2R}{5}>a_{k-2}>Y(a_{k-3})>Y(2^{k-3}) \implies 32>15Y \implies Y\le2$ ,but $Y=1 \implies T=a_{k-2} \implies a_{k-2}\mid \frac{4R}{5}$ and $a_{k-2}<\frac{2R}{5}  \implies 2^{k-2}< a_{k-2} \le  \frac{4R}{15}< \frac{4(2^{k+1})}{45} $ (contradiction),then $Y=2 \implies \frac{a_{k-2}}{2} \mid \frac{4R}{5} \implies  a_{k-2} \mid \frac{8R}{5}$,but $  a_{k-2}\textless \frac{2R}{5}=\frac{8R}{(4)5} \implies  2^{k-2}\textless a_{k-2}\le \frac{8R}{25}\textless  \frac{8(2^{k+1})}{75}$ (contradiction) ,then $a_{k+1}\ge 2^{k + 1}$ ,with what would be complete induction . :D
This post has been edited 1 time. Last edited by xdiegolazarox, Feb 12, 2016, 4:14 AM
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Stranger8
238 posts
#10 • 1 Y
Y by Adventure10
mr bui wrote:
Sorry sir!!!!
But we have a sequence isn't satisfy above condition:
$ a_0=1,a_1=2,a_2=6,a_3=6$, but $ a_3<2^3=8$.....
I thinks we have other condition in this problem.


All terms in this sequence satisfy this condition not part of them ,directly you won't find an integer $a_4$ that satisfies this condition following your sequence
This post has been edited 1 time. Last edited by Stranger8, Sep 26, 2016, 1:21 AM
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bobthesmartypants
4337 posts
#11 • 2 Y
Y by Anar24, Adventure10
solution
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cip999
3645 posts
#12 • 32 Y
Y by Anar24, nmd27082001, rmtf1111, anantmudgal09, Naysh, Arc_archer, pavel kozlov, toto1234567890, Phie11, MathbugAOPS, JasperL, MNJ2357, HolyMath, Supercali, MarkBcc168, rashah76, XbenX, parola, Elyson, Mathematicsislovely, guptaamitu1, Wizard0001, tapir1729, VicKmath7, Infinityfun, Adventure10, Mango247, math_comb01, SerdarBozdag, IMUKAT, puffypundo, megarnie
I think this problem deserves more than it seems, so here's a proof not going through some dreadful induction and massive casework.

First, let $\frac{a_{n + 1}}{a_n} = \frac{x_n}{y_n}$, where $x_n$, $y_n$ are positive integers and $(x_n, \: y_n) = 1$. Then we get $$(a_{n + 1}, \: a_{n + 2}) = \frac{a_{n + 1}}{y_{n + 1}} > a_n \implies y_{n + 1} < \frac{a_{n + 1}}{a_n} = \frac{x_n}{y_n} \implies \frac{x_n}{y_n} \ge y_{n + 1} + \frac{1}{y_n}$$Now observe that $$a_n = a_0 \cdot \frac{a_1}{a_0} \cdots \frac{a_n}{a_{n - 1}} = a_0 \cdot \frac{x_0}{y_0} \cdots \frac{x_{n - 1}}{y_{n - 1}} \ge a_0\left(y_1 + \frac{1}{y_0}\right)\cdots\left(y_n + \frac{1}{y_{n - 1}}\right)$$By AM-GM $y_{i + 1} + \frac{1}{y_i} \ge 2\sqrt{\frac{y_{i + 1}}{y_i}}$; furthermore, since $a_1 = \frac{x_0}{y_0}a_0$, $y_0 \mid a_0 \implies a_0 \ge y_0$. Thus we have $$a_n \ge a_0\left(y_1 + \frac{1}{y_0}\right)\cdots\left(y_n + \frac{1}{y_{n - 1}}\right) \ge y_0\left(2\sqrt{\frac{y_1}{y_0}}\right)\cdots\left(2\sqrt{\frac{y_n}{y_{n - 1}}}\right) = 2^n\sqrt{y_0y_n} \ge 2^n$$which ends the proof.
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anantmudgal09
1980 posts
#13 • 3 Y
Y by Saikat002, Adventure10, Mango247
I'm writing this late at night so please let me know if you spot any errors :)
April wrote:
Let $ a_0$, $ a_1$, $ a_2$, $ \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $ \gcd (a_i, a_{i + 1}) > a_{i - 1}$. Prove that $ a_n\ge 2^n$ for all $ n\ge 0$.

Proposed by Morteza Saghafian, Iran

Note that $a_k \ge (a_k,a_{k+1})>a_{k-1}$ for all $k>0$. Note $a_0 \ge 1$ and $a_1>a_0\ge 1$ hence $a_1 \ge 2$. Now if $a_2=3$ then $(a_2,a_1)=1$; false! So $a_2 \ge 4$. Now pick the minimal $k>1$ with $a_{k+1}<2^{k+1}$. Then $2^k \le a_k<a_{k+1}<2^{k+1}$ and $k>1$ and $(a_{k+1}, a_k)>a_{k-1} \ge 2^{k-1}$.

Notice that $a_k \nmid a_{k+1}$ hence $(a_{k+1}, a_k) \le \frac{a_k}{2}$ so $a_k> 2a_{k-1}$. Also $a_{k+1}-a_k \ge (a_{k+1}, a_k)>2^{k-1}$ hence $a_k<3 \cdot 2^{k-1}$. Now if $(a_{k-1}, a_k) \ne a_k-2a_{k-1}$ then $(a_{k-1}, a_k)\le \frac{a_k-2a_{k-1}}{2} \le \frac{2^{k-1}}{2}<a_{k-2}$ a contradiction! Hence $a_k=n_1(a_k-2a_{k-1})$ for some integer $n_1>1$. Then $a_k=\frac{2n_1}{n_1-1}a_{k-1}$ and $a_k<3a_{k-1} \implies n_1>3$.

Now $(a_{k-1}, a_k)=\frac{a_{k-1}}{n_1-1}$ if $n_1-1$ is odd else $(a_{k-1}, a_k)=\frac{2a_{k-1}}{n_1-1}$ if $n_1-1$ is even. Either way, $a_{k-1}>2a_{k-2}$. Now $a_k<3 \cdot 2^{k-1} \implies a_{k-1}<3 \cdot 2^{k-2}$ and $2a_{k-2}<a_{k-1}<3a_{k-2}$. Thus, applying the previous argument again, we can find $n_2$ with $a_{k-1}=\frac{2n_2}{n_2-1}a_{k-2}$. Note that this procedure can be continued to obtain integers $n_3, \dots, n_{k-1}>3$ with similar relations; so $$3 \cdot 2^{j-1}>a_j=2^{j-1}\cdot \left(\frac{n_1 \cdot \dots \cdot n_{j-1}}{(n_1-1)\cdot \dots \cdot (n_{j-1}-1)}\right) \cdot a_1$$and $a_j$ is an integer for all $j \le k$. Thus, $a_1=2$ and $a_2=\frac{2n_{k-1}}{n_{k-1}-1}a_1$ is an integer so $a_2=5$ and so $5 \mid a_3$ but $a_3=\frac{2n_{k-2}}{n_{k-2}-1}a_2$ hence $a_3=11$ or $a_3=12$; contradiction!
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Rik786
36 posts
#15 • 3 Y
Y by cip999, Adventure10, Mango247
cip999 wrote:
I think this problem deserves more than it seems, so here's a proof not going through some dreadful induction and massive casework.

First, let $\frac{a_{n + 1}}{a_n} = \frac{x_n}{y_n}$, where $x_n$, $y_n$ are positive integers and $(x_n, \: y_n) = 1$. Then we get $$(a_{n + 1}, \: a_{n + 2}) = \frac{a_{n + 1}}{y_{n + 1}} > a_n \implies y_{n + 1} < \frac{a_{n + 1}}{a_n} = \frac{x_n}{y_n} \implies \frac{x_n}{y_n} \ge y_{n + 1} + \frac{1}{y_n}$$Now observe that $$a_n = a_0 \cdot \frac{a_1}{a_0} \cdots \frac{a_n}{a_{n - 1}} = a_0 \cdot \frac{x_0}{y_0} \cdots \frac{x_{n - 1}}{y_{n - 1}} \ge a_0\left(y_1 + \frac{1}{y_0}\right)\cdots\left(y_n + \frac{1}{y_{n - 1}}\right)$$By AM-GM $y_{i + 1} + \frac{1}{y_i} \ge 2\sqrt{\frac{y_{i + 1}}{y_i}}$; furthermore, since $a_1 = \frac{x_0}{y_0}a_0$, $y_0 \mid a_0 \implies a_0 \ge y_0$. Thus we have $$a_n \ge a_0\left(y_1 + \frac{1}{y_0}\right)\cdots\left(y_n + \frac{1}{y_{n - 1}}\right) \ge y_0\left(2\sqrt{\frac{y_1}{y_0}}\right)\cdots\left(2\sqrt{\frac{y_n}{y_{n - 1}}}\right) = 2^n\sqrt{y_0y_n} \ge 2^n$$which ends the proof.

Done a very good job and innovative
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mastermind.hk16
143 posts
#17 • 2 Y
Y by Adventure10, Mango247
First, note that $ a_i \geq (a_i, a_{i+1}) > a_{i-1}$.

Let us prove the result by strong induction on $n$.
Base case: $n=0,1,2,3$
Clearly, $a_0 \geq 1$ and $a_1 \geq 2$ because $a_1>a_0$.
$(a_2,a_1) \geq 2 \longrightarrow a_2 \geq a_1 +2 \geq 4$.
$(a_3,a_2) \geq 3 \longrightarrow a_3 \geq a_2 +3 \geq 7$. But if $a_3 =7$, then $(a_3,a_2)=1$. Contradiction. So $a_3 \geq 8$.

Assume for all $i \leq n, \ a_i \geq 2_i$.
Consider $a_{n+1}$. We have $a_{n+1} \geq (a_{n+2},a_{n+1})> a_n$.
Also, $d=(a_{n+1},a_{n}) > 2^{n-1}$ by the induction hypothesis.

Let $d = \frac{a_n}{m}$ and assume $m \geq 3$. Since $d \mid a_{n+1},$
Then $a_{n+1} \geq d(m+1) \geq \frac{a_n}{m}  (m+1) > (m+1) \cdot 2^{n-1} \geq 4 \cdot 2^{n-1}=2^{n+1}$

So now $m=1$. Then $a_{n+1} > a_n =d \Longrightarrow a_{n+1} \geq 2a_n \geq 2^{n+1}$.

Lastly, $m=2$. Then $a_{n+1} \geq \frac{3}{2} a_n$. Write $a_i = 2^i + b_i \ \ \forall i \leq n,  \ (b_i \geq 0)$
Since $d = \frac{a_n}{2} > a_{n-1} \longrightarrow b_n > 2b_{n-1}$.
If $b_n \geq 2^{n-1}$, then $a_{n+1} \geq 3(2^{n-1} +2^{n-2}) > 2^{n+1}$. The last inequality simplifies to $9>8$. So now assume $b_n < 2^{n-1} \dots (*)$

Consider $e=(a_n, a_{n-1})$ and $f=(a_{n-1}, a_{n-2})$.
If $e=a_{n-1}$, then $a_{n-1} \mid a_n \Longleftrightarrow a_{n-1} \mid b_n -2b_{n-1}$. But $0 < b_n -2b_{n-1} < 2^{n-1} \leq a_{n-1} \leq b_n -2b_{n-1}$. Contradiction.
So $ a_{n-2}<e \leq \frac{a_{n-1}}{2} \longrightarrow b_{n-1}-2b_{n-2}>0$.
Also $$2^{n-2}<e = (b_n -2b_{n-1}, a_{n-1}) \leq b_n -2b_{n-1}$$Hence $a_n =2^n +b_n > 2^n +2^{n-2} + 2b_{n-1}$.
If $b_{n-1} \geq 2^{n-3}$, we are done from $(*)$. So assume $b_{n-1} < 2^{n-3}$. Repeating the same argument as before we get that $f \leq \frac{a_{n-2}}{2}$. But $$2^{n-3}<f = (b_{n-1}-2b_{n-2}, a_{n-2}) \leq b_{n-1} -2b_{n-2} \leq b_{n-1}$$This is a contradiction. And our induction is complete.
This post has been edited 3 times. Last edited by mastermind.hk16, Aug 4, 2019, 4:29 AM
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ubermensch
820 posts
#18 • 1 Y
Y by Adventure10
Yay, we're back yet again with another keep-bashing-mindlessly-until-you-get-something problem!

Looking at the problem, there's nothing which immediately strikes as obviously helping to solve the problem, and the condition really calls for an inductive argument, so here we have it-

First let's observe, quite obviously, that the sequence is a strictly increasing one, as $a_i \geq gcd(a_i,a_{i+1})>a_{i-1}$.

Now, assume for all $i \in [1,2,3,...,i-1]$, $a_i \geq 2^i$ (strong inductive argument).
For the sake of contradiction, also assume that $a_i<2^i =>$ write $a_i=2^{i-1}+k, k<2^{i-1}$.

As $gcd(a_i,a_{i-1})>a_2$, and by assumption, write $a_{i-1}=2^{i-1}+l, l<2^{i-1} =>gcd(2^{i-1}+k,2^{i-1}+l)>2^{i-2}$
$=>gcd(k,2^{i-1}+l-k)>2^{i-2}=>k \mid 2^{i-1}+l-k$ (using $k<2^{i-1}$) and $k>2^{i-2}$

Hence $k \mid 2^{i-1}+l$. Writing $k=2^{i-2}+m$, we get $2^{i-2}+m \mid 2^{i-1}+l => 2^{i-2}+m \mid l-2m$.
As $l<2^{i-1},m<2^{i-2}=>|l-2m| <2^{i-1}=> 2m-l =2^{i-2}+m (\neq$ possible as $=>m>2^{i-2}=>k>2^{i-1}$) or $l=3m+2^{i-2}$.

When $l=3m+2^{i-2}$, we get $a_i=2^{i-1}+2^{i-2}+m$ and $a_{i-1}=2^{i-1}+2^{i-2}+3m$ - I was actually feeling a little despondent after getting this, thinking that I had done all of the above for nothing, until I realised it's a direct violation of monotonicity! $=>$ our assumption $a_i<2^i$ was indeed false.

And thus, we've finally wrapped up all the possible cases, and all that's left to establish is the base case(s). Notice that $a_0$ is automatically $ \geq 1$, and if $a_1<2=>a_1=1$, then $gcd(a_1,a_2)=1>a_3$ couldn't possibly hold true. Finally, as $gcd(a_2,a_3)>2$, if $a_2<4=>a_2=3$, and this is only possible if $a_1=2$. But as this would imply $gcd(a_1,a_2)=1$, $gcd(a_1,a_2)>a_0$ couldn't possibly be true. As we only used $i,i-1,i-2$ in our argument, we're done!
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v_Enhance
6877 posts
#19 • 5 Y
Y by WindTheorist, v4913, Adventure10, Mango247, MS_asdfgzxcvb
Note $a_i \ge \gcd(a_i, a_{i+1}) > a_{i-1}$ so that the sequence is strictly increasing. Thus let $r_i = \frac{a_{i+1}}{a_i} > 1$ for each $i$.

Let's rephrase the condition in terms of the $r_i$. Let $d_i$ denote the denominator of $r_i$ in lowest terms.

Claim: We always have $r_i > d_{i+1}$.

Proof. Extend $\gcd \colon {\mathbb Q}_{>0} \times {\mathbb Q}_{>0} \to {\mathbb Q}_{>0}$ by $\gcd(a,b) = \prod_p p^{\min(\nu_p(a), \nu_p(b))}$. Then the condition is $ \gcd(r_i a_i, r_i r_{i+1} a_i) > a_i 		\iff \frac{1}{d_{i+1}} = 		\gcd(1, r_{i+1}) > \frac{1}{r_i}$ as needed. $\blacksquare$

Let's say an index is critical if $r_i < 2$ and hence $d_i \ge 2$.

Claim: If $i \ge 3$ is a critical index, there exists $e \le 3$ such that $r_{i-1}$, \dots, $r_{i-e}$ are not critical and $r_{i-e} \dots r_i > 2^{e+1}$.

Proof. Choose a critical index $i$. We consider several cases.
  • If $d_i \ge 3$ then $r_{i-1} > d_i$ and we have $r_{i-1} r_i 		> d_i \cdot \frac{d_i+1}{d_i} \ge 4 = 2^2. $
  • Suppose $d_i = 2$ and $d_{i-1} = 1$. Then $r_{i-1} r_i 		> 3 \cdot \frac 32 = \frac 92 > 2^2$.
  • Suppose $d_i = 2$ and $d_{i-1} = 2$. There are two sub-cases. If $d_{i-2} = 1$ then $ r_{i-2} r_{i-1} r_i 		> 3 \cdot \frac 52 \cdot \frac 32 > 2^3.  $
    On the other hand if $d_{i-2} \ge 2$ then $ r_{i-3} r_{i-2} r_{i-1} r_i 		> d_{i-2} \cdot \frac{2d_{i-2}+1}{d_{i-2}} 		\cdot \frac 52 \cdot \frac 32 		= \frac{15}{4}(2d_{i-2}+1) 		\ge \frac{75}{4} > 2^4. $
  • Suppose $d_i = 2$ but $d_{i-1} \ge 3$. Then $r_{i-2} > d_{i-1}$ and we have $ r_{i-2} r_{i-1} r_i 		> d_{i-1} \cdot \frac{2d_{i-1}+1}{d_{i-1}} 			\cdot \frac 32 		= 7 \cdot \frac 32 > 2^3.$
This completes the proof. $\blacksquare$

Thus by induction it suffices to show $a_i \ge 2^i$ only for $i=0,1,2,3$, which is routine.
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yayups
1614 posts
#20 • 1 Y
Y by Adventure10
Note that \[a_i\ge \gcd(a_i,a_{i+1})>a_{i-1},\]so the sequence is strictly increasing. We'll be implicitly using this fact over and over in the proof.

We prove the result by induction on $i$. We begin with the base case.

Claim: [Induction Base Case] We have $a_i\ge 2^i$ for $i\in\{0,1,2\}$.

Proof: We have $a_0\ge 1$, and $a_1>a_0\ge 1$, so $a_1\ge 2$. It suffices to show that $a_2\ge 4$. Note that $a_2\ge a_1+1$ and $a_1\ge 2$, so the only case in which $a_2\le 3$ is if $a_2=2$ and $a_3=3$. This doesn't work as $\gcd(a_2,a_3)=1\not>a_1$. $\blacksquare$

Before proving the inductive step, we have the following useful lemma.

Lemma: If $a_i\le 2a_{i-1}$, then $a_i\mid a_{i+1}$.

Proof: Note that $\gcd(a_i,a_{i+1})=a_i$ or $\gcd(a_i,a_{i+1})\le a_i/2\le a_{i-1}$. The latter case can't happen by the problem condition, so we have $\gcd(a_i,a_{i+1})=a_i$, or $a_i\mid a_{i+1}$. $\blacksquare$

Claim: [Inductive Step] Suppose $i\ge 2$. Then, if $a_j\ge 2^j$ for all $0\le j\le i$, then $a_{i+1}\ge 2^i$.

Proof: Let $a_{i-2}=b$, $a_{i-1}=d\alpha$, $a_i=d\beta$, and $a_{i+1}=a$, where $\gcd(\alpha,\beta)=1$. The problem condition gives us $d>b$ and $\gcd(a,d\beta)>d\alpha$. Furthermore, we know that $b\ge 2^{i-2}$, $d\alpha\ge 2^{i-1}$, and $d\beta\ge 2^i$. Our goal is to show that $a\ge 2^{i+1}$. We have the following cases.

Case 1: Suppose $\beta\ge 8$. Then, \[a>d\beta\ge 8d>8b\ge 8\cdot 2^{i-2}= 2^{i+1},\]as desired.

Case 2: Suppose $\beta\le 7$ and $\beta\le 2\alpha$. Then, we have $d\beta\le 2(d\alpha)$, so by the lemma, we have $d\beta\mid a$. We also know that $a>d\beta$, so we have \[a\ge 2d\beta\ge 2\cdot 2^i=2^{i+1},\]as desired.

Case 3: Suppose $\beta\le 7$ and $\beta\ge 4\alpha$. Then, we have \[a>d\beta\ge 4d\alpha\ge 4\cdot 2^{i-1}=2^{i+1},\]as desired.

Case 4: Suppose $\beta\le 7$ and $2\alpha<\beta<4\alpha$. Since $\gcd(\alpha,\beta)=1$, we have the following subcases.
  • Case 4.1: Suppose $\alpha=1$ and $\beta=3$. Then, $\gcd(a,3d)>d$, so we have $\gcd(a,3d)\in\{3d,3d/2\}$. Thus, $\tfrac{3d}{2}\mid a$ and $a>3d$, so \[a\ge\frac{9d}{2}>4d=4d\alpha\ge 4\cdot 2^{i-1}=2^{i+1},\]as desired.
  • Case 4.2: Suppose $\alpha=2$ and $\beta=5$. Then, $\gcd(a,5d)>2d$, so $\gcd(a,5d)\in\{5d,5d/2\}$. Thus, $\tfrac{5d}{2}\mid a$ and $a>5d$ so either \[a\ge 10d=2\cdot(d\beta)\ge 2\cdot 2^i=2^{i+1},\]or $a=\tfrac{15d}{2}$. So it suffices to look at the case where $a_{i-2}=b$, $a_{i-1}=2d$, $a_i=5d$, $a_{i+1}=\tfrac{15d}{2}$. In this case, we have $d$ even, so $a_{i+1}\ge 15$. So we're already done if $i=2$, so we may assume $i\ge 3$. Because of this, we may define $c=a_{i-3}$, noting that $c\ge 2^{i-3}$. Recall that in this case, our sequence contains \[c,b,2d,5d,\frac{15d}{2}\]as a subsequence. We have the following two subcases.
    • Case 4.2.1: Suppose $b>2c$. Then, $\gcd(2d,b)>c$ so $\gcd(b,2(d-b))>c$. Since $d>b$, this implies that $2(d-b)>c$, so $2d>2b+c>5c$. Thus, \[a=\frac{15d}{2}>\frac{15}{4}\cdot 5c\ge \frac{75}{4}\cdot 2^{i-3}>2^{i+1},\]as desired.
    • Case 4.2.3: Suppose $b\le 2c$. By the lemma, we have that $b\mid 2d$ and $d>b$, so $2d\ge 3b$. Thus, \[a=\frac{15d}{2}\ge\frac{15}{4}\cdot 3b\ge\frac{45}{4}\cdot 2^{i-2}\ge 2^{i+1},\]as desired.
  • Case 4.4: Suppose $\alpha=2$ and $\beta=7$. Then, $\gcd(a,7d)\ge 2d$, so $\gcd(a,7d)\in\{7d,7d/2,7d/3\}$. Thus $a$ is a multiple of $7d/2$ or $7d/3$, and since $a>7d$, we have that \[a\ge \frac{4}{3}\cdot 7d\ge \frac{28}{3}\cdot 2^{i-2}\ge 2^{i+1},\]as desired.
  • Case 4.4: Suppose $\alpha=3$ and $\beta=7$. Then, $\gcd(a,7d)\ge 2d$, so $\gcd(a,7d)\in\{7d,7d/2\}$. The proof then proceeds exactly as in the previous case.
Thus, we've shown that $a\ge 2^{i+1}$ in all cases, proving the claim. $\blacksquare$

Combining the base case and the inductive step completes the proof.
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aops29
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#21 • 2 Y
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Solution to 2008 N3
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awesomeming327.
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#22
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We proceed with induction. It is easy to verify that $a_0\ge 1$ and $a_1\ge 2$. Suppose $a_n\ge 2^n$ and $a_{n+1}\ge 2^{n+1}$, then
\begin{align*}
\text{gcd}(a_{n+2}, a_{n+1}) &> a_n \\
\text{gcd}(a_{n+3}, a_{n+2}) &> a_{n+1}
\end{align*}Let $a_{n+2}=k\text{gcd}(a_{n+2}, a_{n+1}) = l\text{gcd}(a_{n+3}, a_{n+2})$. If $k=1$ then $a_{n+2}\mid a_{n+1}$, impossible by the second one. If $k=2$ then $a_{n+2}=2a_{n+1}\ge 2^{n+2}$ as desired. If $k=3$ then either $a_{n+2}=3a_{n+1}$ which implies the result or
\[a_{n+2}=\frac32 a_{n+1}\]but that'll imply $a_{n+3}=a_{n+2}$ which similarly to $a_{n+2}\mid a_{n+1}$ will not work. If $k=4$ then $a_{n+2}>4a_n\ge 2^{n+2}$ as desired.
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megarnie
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#23
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Suppose this was false.

Clearly this is true for $0$ (since $a_0 \ge 1 = 2^0$). Now let $n$ be the smallest nonnegative integer for which $a_{n+1} < 2^{n+1}$. This implies $a_i \ge 2^i$ for all $0\le i\le n$.

First notice that the sequence must be strictly increasing.

Let $d = \gcd(a_n, a_{n+1})$. We have $d > a_{n-1} \ge 2^{n-1}$, therefore $\frac{a_{n+1}}{d} < 4$.

Now, this implies that both $\frac{a_n}{d}$ and $\frac{a_{n+1}}{d}$ are positive integers under $4$. If $a_n = d$, then $a_n \mid a_{n+1}$, but since $a_{n+1} > a_n$, we have $a_{n+1} \ge 2a_n \ge 2^{n+1}$, contradiction. Thus, $a_n = 2d$ and $a_{n+1} = 3d$ must hold and $d > a_{n-1}$. If $n = 1$, then $a_2 = 3d < 4$, so $d = 1$, absurd since $d > a_0$. Thus, $n > 1$.

Therefore, we have $3d < 2^{n+1} \implies d < \frac{2^{n+1}}{3}$. Looking at $\gcd(a_{n-1}, a_n)$, we see $\gcd(a_{n-1}, 2d) > a_{n-2} \ge 2^{n-2}$.

We have \[ \frac{a_{n-1}}{\gcd(a_{n-1}, 2d)} < \frac{a_{n-1}}{2^{n-2}} \le \frac{d}{2^{n-2}} \le \frac 83, \]so the fraction must be either $1$ or $2$. If it is $1$, then $a_{n-1} \mid 2d$. Now note that $2d < \frac{2^{n+2}}{3} \le a_{n-1} \cdot \frac{8}{3}$, meaning that either $a_{n-1} = 2d$ or $2a_{n-1} = 2d$, both contradict $a_{n-1} < d$. Hence we must have $\frac{a_{n-1}}{\gcd(a_{n-1}, 2d) } = 2$, so $\gcd(a_{n-1}, 2d) = \frac{a_{n-1}}{2}$.

Now, let $2d = k \cdot \frac{a_{n-1}}{2}$. Since $a_{n-1} < d$, $k > 4$. We have $k \cdot 2^{n-2} \le k \cdot \frac{a_{n-1}}{2} < \frac{2^{n+2}}{3}$, so $k < \frac{16}{3}$. Thus, $k = 5$ must hold. Therefore $d$ is a multiple of $5$, so let $d = 5x$.

We see that $a_{n-1} = 4x, a_n = 10x, a_{n+1} = 15x$. If $n = 2$ held, then $a_3 = 15x < 8$, which is impossible. Now we look at $\gcd(a_{n-2}, 4x)$. Call it $m$. It must be greater than $a_{n-3} \ge 2^{n-3}$.

We have $\frac{2^{n-1}}{5} < x < \frac{2^{n+1}}{15}$. Notice that $\gcd(a_{n-1}, a_n) > a_{n -2}$, so $\gcd(4x, 10x) > a_{n-2}$ implying that $a_{n-2} < 2x$. We see that $ \frac{a_{n-2}}{m} < \frac{2x}{m} < \frac{2^{n+2}}{15 \cdot 2^{n-3}} < 3$, so $\frac{a_{n-2}}{m}$ must be either $1$ or $2$. Since $a_{n-2} > a_{n-3}$, we cannot have $a_{n-2} = m$, so we must have $a_{n-2} = 2m$. Thus, we must have $m < x$.

Now, we have $\frac{4x}{m} > 4$, but \[ \frac{4x}{m} < \frac{4x}{2^{n-3}} < \frac{64}{15} < 5 ,\]so $\frac{4x}{m}$ is not an integer.

Hence we have a contradiction, so $a_n \ge 2^n$ for all nonnegative integers $n$.
This post has been edited 1 time. Last edited by megarnie, Feb 3, 2024, 7:17 PM
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vsamc
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#24
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Solution
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i3435
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#25
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Note that $a_n> a_{n-1}$ since $\text{gcd} (a_n,a_{n+1})\le a_n$.

I show that for $n\ge 3$, if $a_n<2a_{n-1},4a_{n-2},8a_{n-3}$ then $15\mid a_n$ and if $n\ge 4$ then $16a_{n-4}<a_n$. This will solve the problem because
  • for $n=0$ $a_n\ge 1$,
  • for $n=1$ $a_1>a_0$ so $a_1\ge 2$,
  • for $n=2$ $a_2>a_1>a_0$ so either $a_2\ge 4$ or $a_2=3,a_2=2,a_2=1$ which is impossible
  • for $n=3$ $a_3\ge 8$ or $15\mid a_3$
  • for $n\ge 4$ strong induction works.

    Suppose for $n\ge 3$ that $a_n<2a_{n-1},4a_{n-2},8a_{n-3}$. In what follows, we use the property that $\text{gcd}(a_i,a_{i+1})$ is less than the absolute value of any nonzero linear combination of $a_i$ and $a_{i+1}$. $a_n-a_{n-1}>a_{n-2}$, so $a_{n-1}\le \frac{3}{4} a_n$ since $a_{n-2}>\frac{1}{4}a_n$. $2a_{n-1}-a_n>a_{n-2}>\frac{a_n}{4}$ so $a_{n-1}>\frac{5}{8}a_n$. $\left|3a_{n-1}-2{a_n}\right|$ is either $0$ or greater than $\frac{a_n}{4}$, which is impossible unless $a_{n-1}=\frac{2}{3}a_n$.

    $a_{n-2}<a_n-a_{n-1}$ so $a_{n-2}<\frac{a_n}{3}$. $\frac{a_n}{8}<a_{n-3}<a_{n-1}-2a_{n-2}$ so $a_{n-2}<\frac{13}{48}a_n$. $\left|2a_{n-1}-5a_{n-2}\right|$ is either $0$ or greater than $\frac{a_n}{8}$, so it equals $0$ and $a_{n-2}=\frac{4}{15}a_n$. Thus $15\mid a_n$.

    If $n\ge 4$, then $a_{n-3}<a_{n-1}-2a_{n-2}$ so $a_{n-3}<\frac{2}{15}a_n$. $a_{n-4}<a_{n-2}-2a_{n-3}<\frac{a_n}{60}$, as desired.
This post has been edited 1 time. Last edited by i3435, Jul 8, 2024, 1:16 PM
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lnzhonglp
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#26
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We use strong induction on $n$. The base cases up to $n=2$ are easy to verify. It is easy to check that the sequence must be strictly increasing. Now suppose $a_n \geq 2^n$ for up to $n$, and let $\gcd(a_n, a_{n+1}) = m$ and $a_{n+1} = km > ka_{n-1}$. If $k \geq 4$, then $a_{n+1} \geq 4a_{n-1} \geq 2^{n+1}$ and we are done.

If $k=3$, then we get $\gcd(a_n, 3m) = m > a_{n-1}$ and $a_n$ must be either $m$ or $2m$. Let $a_{n-1} = m-b$. If $a_n = m$, then $a_{n+1} = 3m \geq 3 \cdot 2^{n} > 2^{n+1}$. If $a_n = 2m$, then $$\frac 12 \gcd(2m, 2m-2b) =  \gcd(m, m-b) = \gcd(m, b) \leq \gcd(2m, m-b) = \gcd(a_n, a_{n-1}) \leq a_{n-2} = 2^{n-2},$$so $b \geq 2^{n-2}$ and $m \geq 2^{n-2} + 2^{n-1}$, so $a_{n+1} = 3m \geq 2^{n+1} + 2^{n-2} > 2^{n+1}$, as desired.

If $k = 2$, then $a_{n+1} \geq 2a_n \geq 2^{n+1}$ and we are done. We have checked all cases so the induction is complete.
This post has been edited 2 times. Last edited by lnzhonglp, Jul 24, 2024, 9:48 PM
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Assassino9931
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#27
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awesomeming327. wrote:
which implies the result or
\[a_{n+2}=\frac32 a_{n+1}\]but that'll imply $a_{n+3}=a_{n+2}$ which similarly to $a_{n+2}\mid a_{n+1}$ will not work. If $k=4$ then $a_{n+2}>4a_n\ge 2^{n+2}$ as desired.
How does $a_{n+3} = a_{n+2}$ follow? I think this very direct induction approach (similarly to some other posts) does not work.
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L13832
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#29
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got deleted by mistake ig

solution
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sansgankrsngupta
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#30
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OG! $a_i \geq gcd(a_{i+1}, a_i) > a_{i-1} \implies$ the sequence is strictly increasing, denote $g_i = gcd(a_{i+1}, a_i)$. Now we proceed by strong induction on $n$
Base Cases: $n=0,1,2,3$
Will write ;ater
Inductive step: Assume the problem statement is true for all $n \in \{0,1,2 \cdots k\}$
Now for $n =k+1$, assume FTSOC that $a_{k+1}< 2^{k+1}$;
$2^{k-2}  <  g_{k}|a_{k+1}$ also, $a_{k+1}>a_k \geq g_{k}$. Hence $a_{k+1}= 2g_k$ or $3g_k$.

$\blacksquare$ If $a_{k+1} = 2g_k$, then since $g_k| a_k  $ and $a_k <a_{k+1}= 2g_k$, we have that $a_k=g_k$. But then, $$a_{k+1}=2g_k= 2a_{k} \geq 2(2^{k}) =2^{k+1}$$produces a contradiction!
$\blacksquare$ If, $a_{k+1}= 3g_{k}$, then since $g_k| a_k  $ and $a_k <a_{k+1}= 3g_k$, , we have that $a_k=g_k$ or $2g_k$.
$\hspace{1cm}$ $\bullet$ If $a_k=g_k$, then $$a_{k+1}=3g_k= 3a_{k} \geq 3(2^{k}) >2^{k+1}$$
$\hspace{1cm}$ $\bullet$ If $a_k=2g_k$, then $a_k= 3g_k< 2^{k+1} \implies $$$ 2^{k-1}\leq a_{k-1}< g_k< \frac23 2^k < \frac43 2^{k-1} \leq \frac43 a_{k-1} \implies a_{k-1}> \frac34 g_k $$.
$\hspace{1cm}$ We have that $g_{k-1}  \mid  a_{k-1}$. $\hspace{1.5cm}$ If $g_{k-1} \leq \frac{a_{k-1}}{3}  \implies  a_{k-2}<g_{k-1} \leq \frac{a_{k-1}}{3} <  \frac29 2^k \leq 2^{k-2}$ which is a contradiction.

$\hspace{1cm}$ Hence, $\hspace{1cm} $ $g_{k-1}= a_{k-1}$ or $\frac{a_{k-1}}2$. If $a_{k-1}= g_{k-1}$, then $a_{k-1}| a_k = 2g_k$ since, $$\frac{2g_k}{3}< \frac{3}{4} g_k <a_{k-1} < \frac{2g_k}{2}$$$\hspace{1cm}$,this is a contradiction! Hence, $g_{k-1}= \frac{a_{k-1}}{2} \mid a_k = 2g_k \implies a_{k-1}|4g_k$. Since $ \frac46 g_k< \frac34g_k< a_{k-1}<\frac44 g_k$. Hence, $a_{k-1} = \frac45 g_k$. Will write the remaining later
This post has been edited 1 time. Last edited by sansgankrsngupta, Apr 29, 2025, 8:15 AM
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