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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.

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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
Inspired by RMO 2006
sqing   1
N a minute ago by sqing
Source: Own
Let $ a,b>0 $ and $ \frac { a}{b} +\frac {4b}{a}=5. $ Prove that
$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+k}{a}  \geq \frac { \sqrt{5(21k+28)}}{6}  $$Where $ k\geq 0.138. $
$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+1}{a}  \geq \frac {7\sqrt 5}{6}  $$$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+8}{a}  \geq \frac {7\sqrt 5}{3}  $$$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+\frac{1}{4}}{a}  \geq \frac {\sqrt {665}}{12}  $$
1 reply
+1 w
sqing
an hour ago
sqing
a minute ago
Square NT FE
Why_rF   13
N 7 minutes ago by MathLuis
Source: own
Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that
\[f(m)+n \mid m^2f(m) - f(n) \]for all positive integers $m$ and $n$.
13 replies
+1 w
Why_rF
Apr 14, 2021
MathLuis
7 minutes ago
Russian Diophantine Equation
LeYohan   0
7 minutes ago
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
0 replies
LeYohan
7 minutes ago
0 replies
Nice orthocenter config
Rijul saini   12
N 10 minutes ago by Commander_Anta78
Source: India IMOTC 2024 Day 4 Problem 3
Let $ABC$ be an acute-angled triangle with $AB<AC$, and let $O,H$ be its circumcentre and orthocentre respectively. Points $Z,Y$ lie on segments $AB,AC$ respectively, such that \[\angle ZOB=\angle YOC = 90^{\circ}.\]The perpendicular line from $H$ to line $YZ$ meets lines $BO$ and $CO$ at $Q,R$ respectively. Let the tangents to the circumcircle of $\triangle AYZ$ at points $Y$ and $Z$ meet at point $T$. Prove that $Q, R, O, T$ are concyclic.

Proposed by Kazi Aryan Amin and K.V. Sudharshan
12 replies
Rijul saini
May 31, 2024
Commander_Anta78
10 minutes ago
Arrange marbles
FunGuy1   3
N 12 minutes ago by FunGuy1
Source: Own?
Anna has $200$ marbles in $25$ colors such that there are exactly $8$ marbles of each color. She wants to arrange them on $50$ shelves, $4$ marbles on each shelf such that for any $2$ colors there is a shelf that has marbles of those colors.
Can Anna achieve her goal?
3 replies
FunGuy1
3 hours ago
FunGuy1
12 minutes ago
Projective geometry
definite_denny   1
N 14 minutes ago by Funcshun840
Source: IDK
Let ABC be a triangle and let DEF be the tangency point of incircirle with sides BC,CA,AB. Points P,Q are chosen on sides AB,AC such that PQ is parallel to BC and PQ is tangent to the incircle. Let M denote the midpoint of PQ. Let EF intersect BC at T. Prove that TM is tangent to the incircle
1 reply
1 viewing
definite_denny
4 hours ago
Funcshun840
14 minutes ago
Nice problem of concurrency
deraxenrovalo   1
N 33 minutes ago by Funcshun840
Let $(I)$ be an inscribed circle of $\triangle$$ABC$ and touching $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Let $EE'$ and $FF'$ be diameters of $(I)$. Let $X$ and $Y$ be the pole of $DE'$ and $DF'$ with respect to $(I)$, respectively. $BE$ cuts $(I)$ again at $K$. $CF$ cuts $(I)$ again at $L$. The tangent at $K$ of $(I)$ cuts $AX$ at $M$. The tangent at $L$ of $(I)$ cuts $AY$ at $N$. Let $U$ and $V$ be midpoint of $IM$ and $IN$, respectively.

Show that : $UV$, $E'F'$ and perpendicular bisector of $ID$ are concurrent.
1 reply
deraxenrovalo
Today at 4:39 AM
Funcshun840
33 minutes ago
Inspired by old results
sqing   1
N an hour ago by ytChen
Source: Own
Let $  a, b> 0,a + 2b= 1. $ Prove that
$$ \sqrt{a + b^2} +2 \sqrt{b+ a^2} +  |a - b| \geq 2$$Let $  a, b> 0,a + 2b= \frac{3}{4}. $ Prove that
$$ \sqrt{a + (b - \frac{1}{4})^2} +2 \sqrt{b + (a-  \frac{1}{4})^2} + \sqrt{ (a - b)^2+ \frac{1}{4}}  \geq 2$$
1 reply
sqing
May 20, 2025
ytChen
an hour ago
D1036 : Composition of polynomials
Dattier   0
an hour ago
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
0 replies
Dattier
an hour ago
0 replies
inequality
NTssu   4
N an hour ago by Oksutok
Source: Peking University Mathematics Autumn Camp
For given real number $\theta_1, \theta_2, ......, \theta_l$, prove there exists positive integer $k$ and positive real number $a_1, a_2, ......, a_k$, such that $a_1+a_2+ ......+ a_k=1$, for any $n \leq k$, $m \in \{1,2,......,l\}$, $\left| \sum_{j=1}^n a_j sin(j \theta_m ) \right|< \frac{1}{2018n} $ holds.
4 replies
NTssu
Oct 11, 2019
Oksutok
an hour ago
Nice geometry
gggzul   0
an hour ago
Let $ABC$ be a acute triangle with $\angle BAC=60^{\circ}$. $H, O$ are the orthocenter and excenter. Let $D$ be a point on the same side of $OH$ as $A$, such that $HDO$ is equilateral. Let $P$ be a point on the same side of $BD$ as $A$, such that $BDP$ is equilateral. Let $Q$ be a point on the same side of $CD$ as $A$, such that $CDP$ is equilateral. Let $M$ be the midpoint of $AD$. Prove that $P, M, Q$ are collinear.
0 replies
gggzul
an hour ago
0 replies
Inspired by 2025 KMO
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=0 $ and $ a^2+b^2+c^2+d^2= 6 .$ Prove that $$ -\frac{3}{4} \leq abcd\leq\frac{9}{4}$$Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=6 $ and $ a^2+b^2+c^2+d^2= 18 .$ Prove that $$ -\frac{9(2\sqrt{3}+3)}{4} \leq abcd\leq\frac{9(2\sqrt{3}-3)}{4}$$
3 replies
sqing
Yesterday at 2:39 PM
sqing
2 hours ago
Reflections and midpoints in triangle
TUAN2k8   0
2 hours ago
Source: Own
Given an triangle $ABC$ and a line $\ell$ in the plane.Let $A_1,B_1,C_1$ be reflections of $A,B,C$ across the line $\ell$, respectively.Let $D,E,F$ be the midpoints of $B_1C_1,C_1A_1,A_1B_1$, respectively.Let $A_2,B_2,C_2$ be the reflections of $A,B,C$ across $D,E,F$, respectively.Prove that the points $A_2,B_2,C_2$ lie on a line parallel to $\ell$.
0 replies
TUAN2k8
2 hours ago
0 replies
a exhaustive question
shrayagarwal   19
N 2 hours ago by SomeonecoolLovesMaths
Source: number theory
If $ a$ and $ b$ are natural numbers such that $ a+13b$ is divisible by $ 11$ and $ a+11b$ is divisible by $ 13$, then find the least possible value of $ a+b$.
19 replies
shrayagarwal
Dec 4, 2006
SomeonecoolLovesMaths
2 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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