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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
limit of function at 5pi
RenheMiResembleRice   0
3 minutes ago
Find f(x)
0 replies
RenheMiResembleRice
3 minutes ago
0 replies
Slovenia 2019 TST2 P2
pj294   5
N 10 minutes ago by jasperE3
Source: 2019 Slovenia 2nd TST Problem 2
Determine all non-negative real numbers $a$, for which $f(a)=0$ for all functions $f: \mathbb{R}_{\ge 0}\to \mathbb{R}_{\ge 0} $, who satisfy the equation $f(f(x) + f(y)) = yf(1 + yf(x))$ for all non-negative real numbers $x$ and $y$.
5 replies
pj294
Feb 14, 2019
jasperE3
10 minutes ago
"Mistakes were made" -Luke Rbotaille
a1267ab   9
N 31 minutes ago by asbodke
Source: USA TST 2025
Let $a_1, a_2, \dots$ and $b_1, b_2, \dots$ be sequences of real numbers for which $a_1 > b_1$ and
\begin{align*}
    a_{n+1} &= a_n^2 - 2b_n\\
    b_{n+1} &= b_n^2 - 2a_n
\end{align*}for all positive integers $n$. Prove that $a_1, a_2, \dots$ is eventually increasing (that is, there exists a positive integer $N$ for which $a_k < a_{k+1}$ for all $k > N$).

Holden Mui
9 replies
a1267ab
Dec 14, 2024
asbodke
31 minutes ago
Yet another circle!
Rushil   7
N an hour ago by FireMaths
Source: INMO 1999 Problem 4
Let $\Gamma$ and $\Gamma'$ be two concentric circles. Let $ABC$ and $A'B'C'$ be any two equilateral triangles inscribed in $\Gamma$ and $\Gamma'$ respectively. If $P$ and $P'$ are any two points on $\Gamma$ and $\Gamma'$ respectively, show that \[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]
7 replies
Rushil
Oct 7, 2005
FireMaths
an hour ago
No more topics!
Two Recurrences
rkm0959   15
N Mar 1, 2025 by jaescl
Source: 2015 Final Korean Mathematical Olympiad Day 2 Problem 5
For a fixed positive integer $k$, there are two sequences $A_n$ and $B_n$.
They are defined inductively, by the following recurrences.
$A_1 = k$, $A_2 = k$, $A_{n+2} = A_{n}A_{n+1}$
$B_1 = 1$, $B_2 = k$, $B_{n+2} = \frac{B^3_{n+1}+1}{B_{n}}$
Prove that for all positive integers $n$, $A_{2n}B_{n+3}$ is an integer.
15 replies
rkm0959
Mar 22, 2015
jaescl
Mar 1, 2025
Two Recurrences
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G H BBookmark kLocked kLocked NReply
Source: 2015 Final Korean Mathematical Olympiad Day 2 Problem 5
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rkm0959
1721 posts
#1 • 4 Y
Y by doxuanlong15052000, Davi-8191, Adventure10, Mango247
For a fixed positive integer $k$, there are two sequences $A_n$ and $B_n$.
They are defined inductively, by the following recurrences.
$A_1 = k$, $A_2 = k$, $A_{n+2} = A_{n}A_{n+1}$
$B_1 = 1$, $B_2 = k$, $B_{n+2} = \frac{B^3_{n+1}+1}{B_{n}}$
Prove that for all positive integers $n$, $A_{2n}B_{n+3}$ is an integer.
This post has been edited 4 times. Last edited by rkm0959, Jun 6, 2015, 2:13 PM
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rkm0959
1721 posts
#2 • 2 Y
Y by Adventure10, Mango247
Hints
This post has been edited 2 times. Last edited by rkm0959, Jun 6, 2015, 5:36 PM
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rkm0959
1721 posts
#3 • 4 Y
Y by shinichiman, Exista, Adventure10, Mango247
Wrong Solution
This was my solution at the contest, can you guys verify this?
(My solution was more rigorous than this in the actual contest though)
This post has been edited 4 times. Last edited by rkm0959, Mar 13, 2016, 10:24 AM
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rkm0959
1721 posts
#4 • 4 Y
Y by VyasR, neverlose, Adventure10, Mango247
This took me an hour to write... Need to practice with LATEX more....
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VyasR
21 posts
#5 • 3 Y
Y by neverlose, Adventure10, Mango247
WOW! NICE LATEX!!!!!
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polya78
105 posts
#6 • 3 Y
Y by toto1234567890, Adventure10, Mango247
Let $B_n(x)=P_n(x)/Q_n(x)$. Then,
$B_n(x)=\frac{Q_n(x)}{Q_{n+1}^3(x)}\frac{P_{n+1}^3(x)+Q_{n+1}^3(x)}{P_n(x)}$

If $n\ge 3$ then if $B_n(x)=0, B_{n-1}(x)=v$, where $v^3+1=0$, and $B_{n+1}(x)=-v^2$, also a third root of $-1$. So if $z$ is a kth root of $P_n(x)=0$, $z$ is a kth root of $P_{n-1}(x)-vQ_{n-1}(x)=0$, for one of the three possible values of $v$, and also a kth root of $P_{n+1}(x)+v^2Q_{n+1}(x)=0$. Hence $P_n(x)$ divides $P_{n+1}^3(x)+Q_{n+1}^3(x)$.

I won't bother to repeat the induction argument that shows that both the Q's and A's are powers of $x$ with Fibonacci exponents.
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shinichiman
3212 posts
#7 • 2 Y
Y by Adventure10, Mango247
rkm0959 wrote:
Claim 5. If $D_{i}$ are all integers for all $i\le n$, the gcd of $D_{i}$ and $D_{i-1}$ is $1$ for all $i\le n$.
Proof of Claim 5. From Claim 4, it suffices to show that gcd of $D_{i}$ and $k$ is 1.
This is straightforward by induction and Claim 4.
Check that $D_3 = k^3+1$ and the recurrence for $D_{n}$ will finish the proof. $\blacksquare$
I don't think it's true. Take $\gcd \left(  D_i,D_{i-1} \right)=2$ with $2 \mid k$ and it still satisfies the condition. We will have $2 \nmid D_3$ but $2 \mid D_i$ with $i \ge 4$.
We can prove that $\gcd \left( D_i,D_{i-1} \right) \in \{ 1,2 \}$.
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rkm0959
1721 posts
#8 • 1 Y
Y by Adventure10
Yep.. I realized that my proof was flawed very recently :(
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shinichiman
3212 posts
#9 • 4 Y
Y by rkm0959, Exista, Adventure10, Mango247
rkm0959 wrote:
Yep.. I realized that my proof was flawed very recently :(

Need time to check this
This post has been edited 5 times. Last edited by shinichiman, Feb 5, 2017, 12:28 PM
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freefrom_i
41 posts
#10 • 2 Y
Y by Adventure10, Mango247
Instead of $k$ we can put $x $ and assume that $D_i$ is a polynomial with integer coefficient for each $i$ then we have $\gcd (D_i, D_{i-1})=1$
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ScienceHouse
6 posts
#11 • 3 Y
Y by shinichiman, Adventure10, Mango247
shinichiman wrote:
Claim 7. If $\gcd (D_i,D_{i-1})=2^w$ for $i \ge 5$ then $v_2(D_{i+1})=3v_2(D_i)-v_2(D_{i-1})$ and $v_2(D_{i+1})< C_{i+1}$.
Proof.

I think $D_5$ should be $(k^3+1)^8+3(k^3+1)^5+3(k^3+1)^2+1$, so $v_2(D_5)$ is at least $3$.
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for63434
105 posts
#12 • 2 Y
Y by Adventure10, Mango247
My solution...
It is easy to see that $A_n=k^{F_n}$
step 1. Let $B_t=C_t/k^{F_{2t-3}}$,, then we only need to show that $C_n$ is an integer
step 2. using the recurrence formular of $B_n$, find the recurrence formular of $C_n$
step 3. using the formular, find $gcd(C_n,C_{n-1})=1$
step 4. Now, by induction we can see that $C_n$ is an integer using the property we proved in step 3..
(It is kind of a bash, but not much... I solved this problem in 30 minutes including bashing)
This post has been edited 1 time. Last edited by for63434, Mar 19, 2017, 3:42 PM
Reason: typo
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EagleEye
497 posts
#13
Y by
rkm0959 wrote:
Yep.. I realized that my proof was flawed very recently :(
shinichiman wrote:
Claim 7. If $\gcd (D_i,D_{i-1})=2^w$ for $i \ge 5$ then $v_2(D_{i+1})=3v_2(D_i)-v_2(D_{i-1})$ and $v_2(D_{i+1})< C_{i+1}$.

Some nasty problems with $2$ can be easily solved by dropping $2$ from $k$. More precisely,
if we let $k = 2^l m$, $l\geq 0$, $m$ is odd, and $D_n = m^{F_{2n}} B_{n+3}$. Then we get
$$D_{n+1} = \frac{D_n^3 + m^{3F_{2n}}}{D_{n-1}}$$Since $m$ is odd, we can prove that $\gcd(D_n,D_{n-1}) = \gcd(D_n,m)=1$ for every non-negative integer $n$ by "rkm0959"'s original arguments.

freefrom_i wrote:
Instead of $k$ we can put $x $ and assume that $D_i$ is a polynomial with integer coefficient for each $i$ then we have $\gcd (D_i, D_{i-1})=1$
This idea does not work unless additional arguments are provided. To apply this idea on "rkm0959"'s proof directly, we need $P, Q\in\mathbb{Z}[x]$ s.t. $PD_i + QD_{i-1}=1$. But we cannot guarantee that, even though $D_i$ and $D_{i-1}$ don't have any common factor.
This post has been edited 4 times. Last edited by EagleEye, Jul 7, 2020, 8:58 PM
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yayups
1614 posts
#14
Y by
Solved with pinetree1.

Let $f_n(X)$ be the sequence of rational functions satisfying $f_1=1$, $f_2=X$, and \[f_{n+2}=\frac{f_{n+1}^3+1}{f_n}.\]We have the following key claim.

Claim: For each $n$, we have that $f_n$ is actually a Laurent polynomial (i.e. a finite integer linear combination of $X^m$ where $m$ ranges over all integers). Furthermore, $X^Nf_n$ and $X^Nf_{n-1}$ share no non-monomial polynomial factor where $N$ is large enough such that $X^Nf_n$ and $X^N f_{n-1}$ are integer polynomials.

Proof: The proof is by induction, with the base cases of $n\le 4$ verified by hand.

Now suppose it is true for $1,\ldots,n$, we'll show it for $n+1$. Indeed, we have \[f_{n+1} = \frac{f_n^3+1}{f_{n-1}} = \frac{(f_{n-1}^3+1)^3+f_{n-2}^3}{f_{n-1}f_{n-2}^3}=:\frac{P}{Q}.\]To show that this is a Laurent polynomial, it suffices to show that any polynomial factor of the numerator of $Q$ that is an irreducible polynomial to some power divides the numerator of $P$. Indeed, since the numerators of $f_{n-1}$ and $f_{n-2}$ don't share any factors (by the inductive hypothesis), it suffices to show that $P/f_{n-1}$ and $P/f_{n-2}^3$ are Laurent polynomials.

Indeed, we have \[P/f_{n-1} = f_{n-1}^2(3+3f_{n-1}^3+f_{n-1}^6) + \frac{1+f_{n-2}^3}{f_{n-1}} = f_{n-1}^2(3+3f_{n-1}^3+f_{n-1}^6)+f_{n-3},\]which is clearly a Laurent polynomial. Furthermore, \[P/f_{n-2}^3 = 1+f_n^3,\]which is also clearly a Laurent polynomial. Thus, $f_{n+1}$ is a Laurent polynomial.

Now we show that $f_{n+1}$ and $f_n$ share no non-monomial factors in their numerator. Indeed, suppose they shared a factor $g$, which is an integer polynomial. We see that \[1=f_{n+1}f_{n-1}-f_n^3,\]so $g$ must divide $1$, which is not possible. This proves the claim entirely. $\blacksquare$

Given the claim, we have the following bound.

Claim: Let $a_n$ be the least non-negative integer such that $x^{a_n}f_n$ is an integer polynomial. Then, \[a_{n+1}\le 3a_n-a_{n-1}.\]
Proof: We have \[f_{n+1}f_{n-1}=f_n^3+1,\]so \[(f_{n+1}\cdot x^{3a_n-a_{n-1}})\cdot(f_{n-1}\cdot x^{a_{n-1}})=(f_n\cdot x^{a_n})^3 + 1.\]Note that $f_{n-1}\cdot x^{a_{n-1}}$ has nonzero constant term, so if $f_{n+1}\cdot x^{3a_n-a_{n-1}}$ is not a polynomial, then neither is $(f_n\cdot x^{a_n})^3 + 1$, which is clearly false, so $f_{n+1}\cdot x^{3a_n-a_{n-1}}$ is a polynomial. Thus, \[a_{n+1}\le 3a_n-a_{n-1},\]as desired. $\blacksquare$

Now by plugging in $k$ into $f_n$, we see that we'd be done as long as $a_n\le F_{2n}$. We show this now. The main claim is that we can iterate the inequality from the claim to get that \[a_{n+1}\le F_{2k} a_{n-k+2} - F_{2k-2}a_{n-k+1}\]for all $2\le k\le n-2$. Indeed, taking $k=n-2$ gives the desired result since $a_4=1$ and $a_3=0$, so in fact, we have $a_{n+1}\le F_{2n-4}$, so $a_n\le F_{2n-6}$, so we're done.
This post has been edited 1 time. Last edited by yayups, Jul 17, 2020, 1:54 AM
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mathaddiction
308 posts
#15 • 2 Y
Y by Mango247, Mango247
Let $C_n=A_{2n}B_n$. Now
$$\frac{C_{n+2}}{A_{2(n+2)}}=\frac{\left(\frac{C_{n+1}}{A_{2(n+1)}}\right)^3+1}{\frac{C_{n}}{A_{2n}}}$$Notice that $A_{2n+4}=\frac{A_{2n+2}^3}{A_{2n}}$. Hence the terms with $A$ cancelled. and we have
$$C_{n+2}=\frac{C_{n+1}^3+A_{2n+2}^3}{C_n}\qquad (1)$$Now we have
$$C_{n+1}=\frac{C_{n}^3+A_{2n}^3}{C_{n-1}}\qquad (2)$$and
$$C_{n}=\frac{C_{n-1}^3+A_{2n-2}^3}{C_{n-2}}\qquad(3)$$Now assume $C_1,C_2,...,C_{n+1}$ are all integers. Then from $(2)$
$$C_{n+1}^3\equiv A_{2n}^9C_{n-1}^{-3}\pmod {C_n}$$From $(3)$, $C_{n-1}^3\equiv -A_{2n-2}^3\pmod{C_n}$
Hence $C_{n+1}^3+A_{2n+2}^3\equiv -A_{2n+2}^3+A_{2n+2}^3\equiv 0\pmod{C_n}$
This compeltes the proof.
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jaescl
11 posts
#16
Y by
Note that the sequence denoted by the letter $B$ shares the same recurrence relation as the sequence defined in the notes of Darij Grinberg (Page 126 and $r=3$) but with a different value for $B_2$. Hence some results are also valid for the sequence of this thread, as for example:
$$B_{n}=B_{n-4}B_{n-1}^{3}-\left(\frac{(B_{n-2}^{3}+1)^3-1}{B_{n-2}}\right)$$Which can be used to prove the claim of the first post of this thread or other statements such as the following:

The expressions $k^{F_{2n}}B_{n-2}B_{n+1}^{3}$ and $k^{F_{2n}}\left(\frac{(B_{n}^{3}+1)^3-1}{B_{n}}\right)$ leave the same remainder when divided by $k^{F_{2n}-F_{2(n-1)}}$ for every integer $n\geq 3$ (Where $F_{n}$ is the nth Fibonacci number)

See also: https://artofproblemsolving.com/community/c6h428645
This post has been edited 1 time. Last edited by jaescl, Mar 2, 2025, 5:48 PM
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