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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
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
interesting set problem
Dr.Poe98   1
N 4 minutes ago by americancheeseburger4281
Source: Brazil Cono Sur TST 2024 - T3/P3
For a pair of integers $a$ and $b$, with $0<a<b<1000$, a set $S\subset \begin{Bmatrix}1,2,3,...,2024\end{Bmatrix}$ $escapes$ the pair $(a,b)$ if for any elements $s_1,s_2\in S$ we have $\left|s_1-s_2\right| \notin \begin{Bmatrix}a,b\end{Bmatrix}$. Let $f(a,b)$ be the greatest possible number of elements of a set that escapes the pair $(a,b)$. Find the maximum and minimum values of $f$.
1 reply
Dr.Poe98
Oct 21, 2024
americancheeseburger4281
4 minutes ago
Reflection lies on incircle
MP8148   5
N 8 minutes ago by deraxenrovalo
Source: GOWACA Mock Geoly P3
In triangle $ABC$ with incircle $\omega$, let $I$ be the incenter and $D$ be the point where $\omega$ touches $\overline{BC}$. Let $S$ be the point on $(ABC)$ with $\angle ASI = 90^\circ$ and $H$ be the orthocenter of $\triangle BIC$, so that $Q \ne S$ on $\overline{HS}$ also satisfies $\angle AQI = 90^\circ$. Prove that $X$, the reflection of $I$ over the midpoint of $\overline{DQ}$, lies on $\omega$.
5 replies
MP8148
Aug 6, 2021
deraxenrovalo
8 minutes ago
Symmetric inequality FTW
Kimchiks926   20
N 42 minutes ago by Marcus_Zhang
Source: Latvian TST for Baltic Way 2020 P1
Prove that for positive reals $a,b,c$ satisfying $a+b+c=3$ the following inequality holds:
$$ \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} \ge 1 $$
20 replies
Kimchiks926
Oct 17, 2020
Marcus_Zhang
42 minutes ago
Interesting problem
V-217   0
an hour ago
On the side $(BC)$ of the triangle $ABC$ consider a mobile point $M$. Let $B'$ the orthogonal projection of $B$ on $AM$. If the mobile points $N\in (BB'$ and $P\in (AM$ are such that $ANPC$ is a paralellogram, find the locus of point $P$ when $M$ goes through $BC$.
0 replies
V-217
an hour ago
0 replies
Equilateral triangle fun
navi_09220114   6
N an hour ago by wassupevery1
Source: Own. Malaysian IMO TST 2025 P8
Let $ABC$ be an equilateral triangle, and $P$ is a point on its incircle. Let $\omega_a$ be the circle tangent to $AB$ passing through $P$ and $A$. Similarly, let $\omega_b$ be the circle tangent to $BC$ passing through $P$ and $B$, and $\omega_c$ be the circle tangent to $CA$ passing through $P$ and $C$.

Prove that the circles $\omega_a$, $\omega_b$, $\omega_c$ has a common tangent line.

Proposed by Ivan Chan Kai Chin
6 replies
navi_09220114
Today at 1:05 PM
wassupevery1
an hour ago
circle geometry solvable by many ways
Dr.Poe98   4
N an hour ago by americancheeseburger4281
Source: Brazil Cono Sur TST 2024 - T3/P4
Let $ABC$ be a triangle, $O$ its circumcenter and $\Gamma$ its circumcircle. Let $E$ and $F$ be points on $AB$ and $AC$, respectively, such that $O$ is the midpoint of $EF$. Let $A'=AO\cap \Gamma$, with $A'\ne A$. Finally, let $P$ be the point on line $EF$ such that $A'P\perp EF$. Prove that the lines $EF,BC$ and the tangent to $\Gamma$ at $A'$ are concurrent and that $\angle BPA' = \angle CPA'$.
4 replies
Dr.Poe98
Oct 21, 2024
americancheeseburger4281
an hour ago
Dealing with Multiple Circles
Wildabandon   4
N an hour ago by Double07
Source: PEMNAS Brawijaya University Senior High School Semifinal 2023 P4
A non-isosceles triangle $ABC$ and $\ell$ is tangent to the circumcircle of triangle $ABC$ through point $C$. Points $D$ and $E$ are the midpoints of segments $BC$ and $CA$ respectively, then line $AD$ and line $BE$ intersect $\ell$ at points $A_1$ and $B_1$ respectively. Line $AB_1$ and line $BA_1$ intersect the circumcircle of triangle $ABC$ at points $X$ and $Y$ respectively. Prove that $X$, $Y$, $D$ and $E$ concyclic.
4 replies
Wildabandon
Dec 1, 2024
Double07
an hour ago
Thanks u!
Ruji2018252   1
N an hour ago by pco
Jqkrjfđrfffffff
1 reply
Ruji2018252
2 hours ago
pco
an hour ago
funny title
nguyenvana   1
N an hour ago by pco
Source: no from book
Find all the functions f: R+ to R+ which satisfy the functional equation:
f(2f(x)+f(y)+xy)=xy+2x+y (x,y R+)
1 reply
nguyenvana
3 hours ago
pco
an hour ago
subsets of subset has same sum
61plus   3
N an hour ago by sttsmet
Source: 2015 China TST 2 Day 2 Q2
Set $S$ to be a subset of size $68$ of $\{1,2,...,2015\}$. Prove that there exist $3$ pairwise disjoint, non-empty subsets $A,B,C$ such that $|A|=|B|=|C|$ and $\sum_{a\in A}a=\sum_{b\in B}b=\sum_{c\in C}c$
3 replies
61plus
Mar 19, 2015
sttsmet
an hour ago
Dear Sqing: So Many Inequalities...
hashtagmath   23
N an hour ago by MTA_2024
I have noticed thousands upon thousands of inequalities that you have posted to HSO and was wondering where you get the inspiration, imagination, and even the validation that such inequalities are true? Also, what do you find particularly appealing and important about specifically inequalities rather than other branches of mathematics? Thank you :)
23 replies
hashtagmath
Oct 30, 2024
MTA_2024
an hour ago
(ab)^2 + (bc)^2 + (ca)^2
GorgonMathDota   13
N an hour ago by ektorasmiliotis
Source: Shortlist BMO 2019, A5
Let $a,b,c$ be positive real numbers, such that $(ab)^2 + (bc)^2 + (ca)^2 = 3$. Prove that
\[ (a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \ge 1. \]
Proposed by Florin Stanescu (wer), România
13 replies
GorgonMathDota
Nov 7, 2020
ektorasmiliotis
an hour ago
weird combinatorics/algebra
Dr.Poe98   1
N 2 hours ago by americancheeseburger4281
Source: Brazil Cono Sur TST 2024 - T3/P2
For each natural number $n\ge3$, let $m(n)$ be the maximum number of points inside or on the sides of a regular $n$-agon of side $1$ such that the distance between any two points is greater than $1$. Prove that $m(n)\ge n$ for $n>6$.
1 reply
Dr.Poe98
Oct 21, 2024
americancheeseburger4281
2 hours ago
A nice problem
hanzo.ei   0
2 hours ago

Given a nonzero real number \(a\) and a polynomial \(P(x)\) with real coefficients of degree \(n\) (\(n > 1\)) such that \(P(x)\) has no real roots. Prove that the polynomial
\[
Q(x) \;=\; P(x) \;+\; a\,P'(x) \;+\; a^2\,P''(x) \;+\; \dots \;+\; a^n\,P^{(n)}(x)
\]has no real roots.
0 replies
hanzo.ei
2 hours ago
0 replies
m=4k^2-5
Zorro   33
N Dec 14, 2023 by bhan2025
Source: Poland 2005, IMO Shortlist 2004, number theory problem 4
Let $k$ be a fixed integer greater than 1, and let ${m=4k^2-5}$. Show that there exist positive integers $a$ and $b$ such that the sequence $(x_n)$ defined by \[x_0=a,\quad x_1=b,\quad x_{n+2}=x_{n+1}+x_n\quad\text{for}\quad n=0,1,2,\dots,\] has all of its terms relatively prime to $m$.

Proposed by Jaroslaw Wroblewski, Poland
33 replies
Zorro
Apr 16, 2005
bhan2025
Dec 14, 2023
m=4k^2-5
G H J
Source: Poland 2005, IMO Shortlist 2004, number theory problem 4
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Zorro
92 posts
#1 • 4 Y
Y by Com10atorics, Adventure10, Mango247, and 1 other user
Let $k$ be a fixed integer greater than 1, and let ${m=4k^2-5}$. Show that there exist positive integers $a$ and $b$ such that the sequence $(x_n)$ defined by \[x_0=a,\quad x_1=b,\quad x_{n+2}=x_{n+1}+x_n\quad\text{for}\quad n=0,1,2,\dots,\] has all of its terms relatively prime to $m$.

Proposed by Jaroslaw Wroblewski, Poland
This post has been edited 1 time. Last edited by djmathman, Aug 1, 2015, 2:54 AM
Reason: formatting
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grobber
7849 posts
#2 • 5 Y
Y by phiReKaLk6781, numbertheorist17, Adventure10, Mango247, and 1 other user
Let $u,\bar u$ be the roots of the equation $x^2-x-1=0$ (the characteristic equation of the sequence). Since every prime divisor of $m$ has $5$ as a quadratic residue, if we fix $p$, we can regard $u,\bar u$ as residues $\pmod p$, and carry out the computations in this manner.

The general term of the sequence is $x_n=\alpha u^n+\beta \bar u^n$ (where I changed the notations a bit and took $x_0=a,x_1=b$), and we must show that we can pick $\alpha,\beta\pmod p$ s.t. $\alpha u^n+\beta \bar u^n\ne 0,\ \forall n$ in $\mathbb Z_p$. We can do this by taking $au\equiv b\pmod p$, which translates (after doing the computations) to $\beta\equiv 0\pmod p$.
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zibi
94 posts
#3 • 2 Y
Y by Adventure10, Mango247
Quote:
Since every prime divisor of m has 5 as a quadratic residue, if we fix p, we can regard $u,\bar u$ as residues $\pmod p$, and carry out the computations in this manner.
I' m beginner in "mathematic english" :blush: Can you explain me what does "quadratic residue" mean ?
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Zorro
92 posts
#4 • 2 Y
Y by Adventure10, Mango247
jak sama nazwa wskazuje, jest to reszta kwadratowa
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namanhams
46 posts
#5 • 2 Y
Y by Adventure10, Mango247
Can you give me a more clearly solution ,Grobber ?
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Pascual2005
1160 posts
#6 • 2 Y
Y by Adventure10, Mango247
Nice solution Grobber, it is also what I had!
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Leva1980
173 posts
#7 • 3 Y
Y by Adventure10, Mango247, and 1 other user
It is a very nice solution!
One can make it a little bit more elementary.
If we take $ x = 2k + 1 + m $ then $ x^{2} - 2x - 4 \equiv
(2k+1)^{2} - 2(2k+1) - 4 = 4k^{2} + 4k + 1 - 4k - 2 - 4 = 4k^{2}-5 = m \equiv 0 \ (mod \ m) $.
Therefore because of $ x $ being even we can take $ y = x/2 $ and then
$ x^{2} - 2x - 4 = 4(y^{2} - y - 1) \equiv 0 \ (mod \ m) $, and because $ m $
is odd, we get that $ y^{2} - y - 1 \equiv 0 \ (mod \ m) $. Also we have
$ gcd(y,m)= 1 $, because $ 2y \equiv 2k+1 \ (mod \ m) $, so
$ 2(2k-1)y \equiv 4k^{2} - 1 \equiv 4 \ (mod \ m) $, and $ m $ is odd.
Therefore also $ y^{n} $ is co-prime to $ m $. Now if we take
$ x_{1} = y $, $ x_{2} = y^{2} $ and define $ x_{n} $ by recursion
$ x_{n+2} = x_{n+1} + x_{n} $, then of course $ x_{n+2} \equiv x_{n+1} + x_{n} \ (mod \ m) $,
but also because of $ y $ satisfying $ y^{2} - y - 1 \equiv 0 \ (mod \ m) $, we get that
$ y^{n+2} \equiv y^{n+1} + y^{n} \ (mod \ m) $, and by induction one can show that
$ x_{n} \equiv y^{n} \ (mod \ m) $ for every $ n \geqslant 0 $. Now,
we have seen that $ y^{n} $ is co-prime to $ m $, so $ x_{n} $ is also co-prime to $ n $.
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k2c901_1
146 posts
#8 • 2 Y
Y by Adventure10, Mango247
Incidentally, this was also problem 6 of the 2nd TST of Taiwan 2005.
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beta
3001 posts
#9 • 1 Y
Y by Adventure10
Suppose $k$ is not a multiple of 5, then let prime factorization $m=p_1^{a_1}p_2^{a_2}...p_r^{a_r}$.
Define the Fibonacci sequence to be $F_0=1, F_1=1, F_n=F_{n-1}+F_{n-2}$.
By Binet's formula,
$F_n = \frac{1}{\sqrt{5}} \left ( \left (\frac{1+\sqrt{5}}{2} \right )^{n+1} - \left ( \frac{1-\sqrt{5}}{2} \right )^{n+1} \right )$

Lemma 1: $x^2 = q^2(\mod p)$ has at two solutions $x$ modulo $p^a$ if $(p, q)=1$ and $p$ is odd prime.

Proof: It factors to $(x+q)(x-q)=0(\mod p)$. then $p$ divides either $x+q$ or $x-q$ the desired lemma follows.

Hence from Lemma 1 we know $\sqrt{5} = \pm 2k (mod p_i)$ for $i=1, 2, ..., r$ since we know $p_i$ is not 2 or 5.

Lemma 2: $F_n \neq \left ( \frac{1+\sqrt{5}}{2} \right)F_{n-1} (\mod p_i)$.

Proof: Assume it's true, then we know $\sqrt{5}=\pm 2k$ and it's relatively prime to $p_i$ so WLOG we let it be $\sqrt{5}=2k$ we substitute $F_n$ and $F_{n+1}$ and simplify we have

$(1-2k)^n (4k) = 0(\mod p_i)$. but if $p_i|1-2k$ then $p_i|m+(1+2k)(1-2k)$ => $p_i | 4$=> $p_i=2$ so contradiction. If $p_i | 4k$ then $p_i |5$ also contradiction.
The desired lemma follows(you could just make a simple argument with limits, but I'm not sure how rigorous that is)

Now let
$a=2k^2-k-3$
$a = -\frac{1+2k}{2} (\mod m)$ and so $a$ is relatively prime to $m$ since
$4k^2-5 - (2k+1)(2k-1) = -4$ and $(4, 4k^2-5)=1$

$b=1$ . Now $x_n=F_{n-1}a + F_{n})b=-F_{n-1} \phi + F_n(\mod p_i)$ and it's never 0 relatively prime to all $p_i$ so relatively prime to $m$ as desired.

If $k$ is a multiple of 5 it's not big deal, as 5 is the largest power of 5 that divides m so $m=5 p_1^{a_1} p_2^{a_2}...p_r^{a_r}$ so we can just say

$a=2k^2-k-3$
$a = 2(\mod 5)$ and $a = -\frac{1+2k}{2} (\mod \frac{m}{5})$

$b=1$, and we know $x_n$ is relatively prime to all $p_i$ and is relatively prime to 5 (modulo 5 $x_n$ sequence is 2, 1, 3, 4, 2, 1, 3, 4....). So $x_n$ and $m$ are relatively prime too.

QED
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abdurashidjon
119 posts
#10 • 1 Y
Y by Adventure10
Is Leva 1980's solution is true? I could not translate to any of $a$ or $b$.
Abdurashid
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shfdfzhjj
347 posts
#11 • 2 Y
Y by Adventure10, Mango247
grobber wrote:
The general term of the sequence is $ x_n = \alpha u^n + \beta \bar u^n$ (where I changed the notations a bit and took $ x_0 = a,x_1 = b$), and we must show that we can pick $ \alpha,\beta\pmod p$ s.t. $ \alpha u^n + \beta \bar u^n\ne 0,\ \forall n$ in $ \mathbb Z_p$. We can do this by taking $ au\equiv b\pmod p$, which translates (after doing the computations) to $ \beta\equiv 0\pmod p$.

Can you explain in detail? :maybe:
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Mewto55555
4210 posts
#12 • 2 Y
Y by Adventure10, Mango247
Here's another solution:

Solution
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yunxiu
571 posts
#13 • 2 Y
Y by Adventure10, Mango247
abdurashidjon wrote:
Is Leva 1980's solution is true? I could not translate to any of $a$ or $b$.
Abdurashid

$a=1$ and $b=y$.
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sayantanchakraborty
505 posts
#14 • 2 Y
Y by Adventure10, Mango247
Another possible approach




Bye...

Sayantan...
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rkm0959
1721 posts
#15 • 4 Y
Y by hakN, Resolut1on07, Adventure10, Mango247
Claim. There is a natural number $b$ such that $b^2=b+1 \pmod{m}$.
Proof of Claim. As $m$ is an odd number, it suffices to find a $b$ that $(2b-1)^2 \equiv 5 \pmod{m}$
Set $m=\prod_{i=1}^k p^{e_i}_i$. For each $p_i$, as $(2k)^2 \equiv 5 \pmod{p_i}$, we have $\left( \frac{5}{p} \right)=1$.
Now there exists an $x$ such that $x^2 \equiv 5 \pmod{p_i}$ for each $i$.
By Hensel's lemma on $f(x)=x^2-5$, we can see that there exists an $x$ that $x^2 \equiv 5 \pmod{p^{e_i}_i}$.
By CRT, we can ensure that there exists an $x$ that $x^2 \equiv 5 \pmod{m}$, and we conclude.

Now set $x_0=1$ and $x_1=b$. Then $x_n \equiv b^n \pmod{m}$, so $(x_n,m)=1$ as desired.
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pandadude
710 posts
#16 • 2 Y
Y by Adventure10, Mango247
umm... isn't this just $a=2,b=2k+1$?

Take any prime $p$ that divides $4k^2-5$
then $(2k)^2 \equiv 5 (mod p)$

using characteristic polynomials, we get $x_n \equiv 2*(\frac{1+2k}{2})^n (mod p)$

since $p$ is odd and obviously $2k$ is not $-1(modp)$ so $x_n$ coprime to $p$ so it is also coprime to $4k^2-5$ for all n.
This post has been edited 3 times. Last edited by pandadude, Aug 20, 2018, 12:01 AM
Reason: .
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Mathotsav
1508 posts
#18 • 4 Y
Y by teomihai, AlastorMoody, Adventure10, Mango247
Solution:
We can take $a=1, b=2k^2+k-2$.
Lemma: If we consider the sequence $x_1=1, x_2=s, x_n=x_{n-1}+x_{n-2}$ then all terms of $x_n$ are co-prime to $s^2-s-1$.
Proof: Any term of $x_n$ is in the form $sF_r+F_{r-1}$.
Now, note that $F_r^2(s^2-s-1)-(sF_r+F_{r-1})(sF_r-F_{r+1})=F_{r-1}F_{r+1}-F_r^2$ which is always of the form $\pm 1$ by Cassini's identity, thus any $x_n$ is coprime to $s^2-s-1$.
If we put $s=2k^2+k-2$ note that $(2k^2+k-2)(2k^2+k-3)-1=4k^4+4k^3-9k^2-5k^2+5=(4k^2-5)(k^2+k-1)$ thus for $a=1, b=2k^2+k-2$ the terms are always coprime to $4k^2-5$.
@3below(math_pi_rate): The lemma is not true for $p=3$
This post has been edited 1 time. Last edited by Mathotsav, Apr 13, 2020, 10:02 AM
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math_pi_rate
1218 posts
#19 • 1 Y
Y by amar_04
Nice problem! Here's a different solution (Hopefully correct!): Let $F_n$ denote the Fibonacci sequence with $F_{-1}=1$ and $F_0=0$. It's easy to see (using induction) that $$x_n=aF_{n-1}+bF_n \quad \forall n \in \mathbb{N} \cup \{0\}$$The crucial lemma is as follows:-

LEMMA For all primes $p$ such that $5$ is a quadratic residue modulo $p$, there exists some $C \in \mathbb{Z}/ p\mathbb{Z}$ such that $$F_n \not \equiv CF_{n-1} \pmod{p}$$for any $n \in \mathbb{N} \cup \{0\}$.

Proof of Claim If $p=5$, then one can easily check that $C=3$ works (since in this case, one finds that $F_n$ is periodic modulo $5$ with period $4$, which is not true). So assume $p>5$, and suppose the Lemma is not true. Then, for all $i \in \mathbb{Z}/ p\mathbb{Z}$, there is some index $n_i \geq 0$ such that $$F_{n_i} \equiv iF_{n_i-1} \pmod{p}$$For all such $i$, let $n_i$ be the smallest such index. Since $\left(\frac{5}{p} \right)=1$, so there exists some $y$ such that $y^2 \equiv 5 \pmod{p}$. Then, by Binet's Formula, \begin{align*} F_n &=\frac{1}{\sqrt{5}} \left(\left(\frac{1+\sqrt{5}}{2} \right)^n-\left(\frac{1-\sqrt{5}}{2} \right)^n \right) \\ &\equiv \frac{1}{y} \left(\left(\frac{1+y}{2} \right)^n-\left(\frac{1-y}{2} \right)^n \right) \pmod{p} \\ &\equiv \frac{1}{y} \left(\left(\frac{1+y}{2} \right)^{n+p-1}-\left(\frac{1-y}{2} \right)^{n+p-1} \right) \pmod{p} \\ &\equiv F_{n+p-1} \pmod{p} \\ \end{align*}which means that $F_n$ is periodic modulo $p$, with its period $T$ dividing $p-1$. In particular, this gives that $F_{n+T}=F_n$ for some $1 \leq T \leq p-1$. Since $n_0,n_1, \dots ,n_{p-1}$ are the smallest indices by definition, so due to periodicity after $p-1$ indices, we get $0 \leq n_i \leq p-2$. But then, by Pigeonhole Principle, two of these indices must coincide. Suppose $n_i=n_j$ with $i<j$. Then we get $$jF_{n_i-1}=jF_{n_j-1} \equiv F_{n_j} \equiv F_{n_i} \equiv iF_{n_i-1} \pmod{p} \Rightarrow (j-i)F_{n_i-1} \equiv 0 \pmod{p}$$As $0<j-i<p$, so we have $$p \mid F_{n_i-1} \Rightarrow F_{n_i} \equiv iF_{n_i-1} \equiv 0 \pmod{p} \Rightarrow p \mid \gcd(F_{n_i},F_{n_i-1})$$However, this contradicts the well known fact that $\gcd(F_r,F_{r-1})=1$ for all $r \in \mathbb{N} \cup \{0\}$. Thus, we have a contradiction, proving the Lemma. $\Box$

Return to the problem at hand. Let $p_1,p_2, \dots, p_s$ be all distinct prime divisors of $m$. Then $5 \equiv (2k)^2 \pmod{p_i}$ gives that $5$ is a quadratic residue modulo $p_i$ for all $i \in [s]$. By our Lemma, there exist constants $C_1,C_2, \dots ,C_s$ such that $$F_n \not \equiv C_iF_{n-1} \pmod{p_i} \quad \forall n \in \mathbb{N} \cup \{0\} \text{ and } i \in [s]$$Using CRT, there exists an integer $Z$ satisfying $$Z \equiv C_i \pmod{p_i} \quad \forall i \in [s]$$Choose $b=1$ and $a>0$ with $a \equiv -Z \pmod{m}$. Then, for all $i \in [s]$, we get $$x_n=aF_{n-1}+bF_n \equiv F_n-ZF_{n-1} \not \equiv 0 \pmod{p_i} \Rightarrow \gcd(x_n,m)=1 \quad \forall n \geq 0 \text{ } \blacksquare$$
This post has been edited 4 times. Last edited by math_pi_rate, Apr 12, 2020, 6:18 AM
Reason: $F_{r-1}$ instead of $F_r-1$
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Al3jandro0000
804 posts
#20
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@above your lemma is valid for any $p $, notice you're not using the fact $\left (\frac {5}{p}\right)=1$.
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math_pi_rate
1218 posts
#21 • 1 Y
Y by amar_04
Al3jandro0000 wrote:
@above your lemma is valid for any $p $, notice you're not using the fact $\left (\frac {5}{p}\right)=1$.

I am using it in the proof of the Lemma while writing $\sqrt{5}$ as $y$. Note that if $5$ was not a quadratic residue, then I would have to work in $\mathbb{Z}[\sqrt{5}]$, and so Fermat's Little Theorem won't be applicable (Lagrange's Theorem gives a higher bound on the order, which isn't compatible with my PHP usage).

On the orther hand, there might be a possibility of the Lemma being true for all primes. I wonder if someone can prove/disprove that?
This post has been edited 2 times. Last edited by math_pi_rate, Apr 11, 2020, 9:16 AM
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TheThor
222 posts
#22 • 1 Y
Y by Mango247
..................
This post has been edited 2 times. Last edited by TheThor, Jul 22, 2020, 7:16 PM
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math_pi_rate
1218 posts
#23 • 1 Y
Y by amar_04
Tbh I don't think the Lemma is true always (In fact, one can easily check that it fails for $p=3$). However, it can be interesting to determine for what values of $p$ it works. Here are some comments PMed to me by Superguy (I haven't read it thoroughly, but it seems ok at first glance).
Superguy in First PM wrote:
Okay so far I have approached problem a bit.
First of all I have neglected the $F_{-1}$ part which won't change much of the proof
Assume ftsoc that there doesn't exist such a $C$
By considering the group $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ we can see that if $\left (\frac {5}{p}\right)=-1$ then $p$ is a prime in $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$
Now using some tricks u can get that $a^{(p+1)}\equiv -1 \pmod{p}$ where $a$ is golden ratio.
Using this u can easily get
$F_{n+p+1}\equiv -F_{n}\pmod{p}$
We get that period is atmost $2p+2$
So considering the sets $(F_{1},F_{2},.....F_{p+1})$ and $(F_{p+2},F_{p+3}......F_{2p+2})$ and using the minimal index notation one can see that $2\leq n_{i}\leq p+1$
This means that there should be $p$ i's given by $p$ $n_{i}$'s which implies each member of first set should give a separate $i$

For instance in case of $p=3$ clearly no such constant exists while in the case $p=17$ we clearly have a constant as $F_{9}\equiv F_{18}\equiv 0\pmod{17}$
Edit:See below for a bit more detailed explanation.
Superguy in Second PM wrote:
Hmm a bit more
So clearly there should not be any index $0<i<p+1$ such that $p|F_{i}$ otherwise we can get our constant from above argument.
Now assuming for no such index $F_{i}$ is multiple of $p$.
Hence all $(\frac{F_{i}}{F_{i-1}})$ are uniquely defined modulo $p$
Now assuming that two of them are congruent then
$\frac{F_{i}}{F_{i-1}}\equiv \frac{F_{m}}{F_{m-1}}\pmod{p}$ for some $m>i$
Now manipulating terms we get
$(\frac{1+\sqrt{5}}{1-\sqrt{5}})^{m-j}\equiv 1\pmod{p}$.
Now clearly we can show that order divides $2(p+1)$ and hence if order is less than $(p+1)$ then we can clearly get $(m,j)$ and thus our desired constant however in case order is equal to $2(p+1)$ or $(p+1)$ we can't get our desired constant.

Others can comment on this if they find something interesting :)
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Superguy
354 posts
#24
Y by
Well last part can be reduced to the fact that such a constant exists only if $F_{p+1}$ doesn't has a primitive prime divisor as $p$ which I don't think is possible to prove.
For instance after $p=3$ next such prime for which no such constant exists is $p=7$
followed by $p=23$
This post has been edited 1 time. Last edited by Superguy, Apr 13, 2020, 12:53 PM
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mathaddiction
308 posts
#25
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Lemma. If $p$ is a prime such that $p|4k^2-5$ then we can pick $a,b$ such that for all $N\in\mathbb N$ we have $p\nmid x_n$.
Proof.
CASE I: $p=5$. It suffices to take $a=1$ and $b=3$
CASE II: $p\neq 5$
Now let $F_0=0, F_1=1$ and $F_n$ be the $n^{th}$ Fibonacci number. It is easy to show that
$$x_{n+2}=F_{n+1}a+F_nb$$CLAIM 1. $F_p\equiv 1\pmod p$
Proof.
Using the general term formular for Fibonacci sequence,
\begin{align*}
F_p&=\frac{1}{\sqrt 5}\left(\left(\frac{1+\sqrt 5}{2}\right)^p-\left(\frac{1-\sqrt{5}}{2}\right)^p\right)\\
&=\frac{2}{2^p\sqrt 5}(\sum_{i=0}^{\frac{p-1}{2}}\sqrt 5\binom{p}{2i+1}5^i)\\
&\equiv\frac{1}{2^{p-1}}5^{\frac{p-1}{2}}\\
&\equiv5^{\frac{p-1}{2}} \pmod{p}
\end{align*}The first equivalence follows from $p|\binom{p}{2i+1}$ for all $1\leq i\leq \frac{p-3}{2}$The last equvialence follow from $p\neq 2$ and Fermat little theorem. Now notice that $p|4k^2-5$, hence by Euler's criterion we have
$$5^{\frac{p-1}{2}}\equiv \left(\frac{5}{p}\right)=1\pmod p$$This proves CLAIM 1. $\blacksquare$
CLAIM 2. $F_{p-1}\equiv 0\pmod p$
Proof. The proof is similar to CLAIM 1. Again, using the general term formula for Fibonacci sequence,
\begin{align*}
F_{p-1}&=\frac{1}{\sqrt 5}\left(\left(\frac{1+\sqrt 5}{2}\right)^{p-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{p-1}\right)\\
&=\frac{1}{2^{p-2}}\sum_{i=1}^{\frac{p-3}{2}}\binom{p-1}{2i+1}5^i
\end{align*}Notice that
$$\binom{p-1}{2i+1}=\frac{(p-1)!}{(2i+1)!(p-2i+1)!}\equiv \frac{(p-1)!}{(-1)^{2i+1}(p-1)!}\equiv -1\pmod p$$Hence
$$F_{p-1}\equiv -\frac{1}{2^{p-2}}\sum_{i=1}^{\frac{p-3}{2}}5^i=-\frac{5^{\frac{p-1}{2}}-1}{4\cdot 2^{p-2}}\equiv 0\pmod p$$$\blacksquare$
Now this implies that $x_{n+p-1}\equiv x_n \pmod p$ for all $n\in\mathbb N$. We now take $a=1$. It suffices to find $b$ which satisfies
$$F_{i+1}+F_ib\not\equiv 0\pmod p$$for all $0\leq i\leq p-2$. Notice that there exists at most $1$ $b$ which satisfies
$$F_{i+1}+F_ib\equiv 0\pmod p$$Since there are only $p-1$ incongruences, each of which eliminates at most $1$ chocie of $b$, together with the fact that there are $p$ residue classes modulo $p$, which gives $p$ choices for $b$, we conclude that we can always pick such $p$. This proves the lemma. $\blacksquare$
Now let $p_1,...,p_n$ be the prime factors of $m$. Then from the lemma for each $i$ we can pick $a_i$ and $b_i$ so that the sequence defined by $x_0=a_i$, $x_1=b_i$ and
$$x_{n+2}=x_{n+1}+x_n$$has all its terms relatively prime to $p_i$. Now by Chinese Remanider Theorem there exists integers $a$ and $b$ satisfying
$$a\equiv a_i\pmod{p_i}$$$$b\equiv b_i\pmod{p_i}$$Now by taking this $a$ and $b$ the proof is completed.
This post has been edited 1 time. Last edited by mathaddiction, Aug 10, 2020, 12:20 PM
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yayups
1614 posts
#26
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By CRT, it suffices to show the problem for $p\mid 4k^2-5$ where $p$ is a prime. Note that
\[x_n = F_{n-1}a + F_n b,\]and since $(\tfrac{5}{p})=1$, we have that $F_{n+p-1}\equiv F_n\pmod{p}$ by Binet's thereom and Fermat's little theorem.

Let $t$ be the smallest positive integer such that $p\mid F_t$, so $t$ is the order of $\tfrac{1+\sqrt{5}}{1-\sqrt{5}}$ mod $p$, so $t\mid p-1$. Now, if $t\nmid n$, then $\tfrac{F_{n-1}}{F_n}$ can take at most $t-1$ values (since it repeats mod $t$), so it takes at most $p-2$ values, so there is some nonzero value mod $p$ it never achieves. Let this value be $c$.

Take $a=1$, $b\equiv -c\pmod{p}$. Then,
\[x_n = F_{n-1}a + F_n b\equiv F_{n-1}-F_nc\pmod{p}.\]If $t\mid n$, then $F_n\equiv 0\pmod{p}$ and $F_{n-1}\not\equiv 0\pmod{p}$ (else $p\mid F_m$ for all $m$ which is not true), so $p\nmid x_n$. If $t\nmid n$, then $F_n\not\equiv 0\pmod{p}$, so this is never $0$ by the selection of $c$. Thus, these values of $a$ and $b$ yield all $x_n$ relatively prime to $p$, as desired.
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Sprites
478 posts
#27 • 2 Y
Y by centslordm, Mango247
Note that \[x_n = F_{n-1}a + F_n b,\]Claim: There is a natural number $b$ such that $b^2=b+1 \pmod{m}$
Proof: We need $(2b-1)^2 \equiv 5 \pmod m$ i.e $5$ is a quadratic residue modulo $m$
Suppose that $m=\prod_{i=1}^{x} {p_i}^{e_i}$
By Hensel's Lemma,we can assure that each congruence has a solution and we can conclude by CRT.(here $f'(x)=8b-4$,and clearly we can choose such an $b$ s.t $ p \nmid 8b-4$ to get a solution.)
Now choose $a=1$ and clearly $x^n \equiv b^n \neq 0 \pmod m$ and $\gcd(x_n,m)=1$(by induction.
This post has been edited 1 time. Last edited by Sprites, Aug 28, 2021, 8:58 AM
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hakN
429 posts
#28 • 1 Y
Y by centslordm
Solution
This post has been edited 3 times. Last edited by hakN, Oct 2, 2021, 3:58 PM
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CyclicISLscelesTrapezoid
372 posts
#29 • 4 Y
Y by centslordm, Mango247, Mango247, Mango247
Is this right? This seems like a troll solution.

It's sufficient to construct a sequence mod all primes $p$ dividing $4k^2-5$, then combine them using CRT. If $p \mid 4k^2-5$, then $p$ is odd and $5$ is a quadratic residue $\!\!\!\mod p$. Let $a=1$, and choose $b$ such that $(2b-1)^2 \equiv 5 \pmod{p}$. It's easy to check that this condition forces $\frac{x_4}{x_3}=\frac{3b+2}{2b+1}=b=\frac{x_1}{x_0} \pmod{p}$ if $x_3$ is nonzero $\!\!\!\mod{p}$. Now, we check that $x_1,x_2,x_3 \not\equiv 0 \pmod{p}$:

If $x_1=b \equiv 0 \pmod{p}$, then $(2b-1)^2 \equiv 1 \equiv 5 \pmod{p}$, so $p=2$, a contradiction.
If $x_2=b+1 \equiv 0 \pmod{p}$, then $(2b-1)^2 \equiv 9 \equiv 5 \pmod{p}$, so $p=2$, a contradiction.
If $x_3=2b+1 \equiv 0 \pmod{p}$, then $(2b-1)^2 \equiv 4 \equiv 5 \pmod{p}$, so there are no choices of $p$, a contradiction.

Since $\frac{x_3}{x_0} \equiv \frac{x_4}{x_1} \pmod{p}$, we know that $\frac{x_{n+3}}{x_n} \equiv \frac{x_3}{x_0} \pmod{p}$. Therefore, we can write $x_n=x_i \cdot \left(\frac{x_3}{x_0}\right)^{\left\lfloor \frac{n}{3} \right\rfloor}$, where $i \in \{0,1,2\}$. This is nonzero $\!\!\!\mod{p}$, so we are done.
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awesomehuman
496 posts
#30 • 1 Y
Y by centslordm
Let $p$ be a prime and $p|m$. Note $p$ is odd.

Let $a\equiv 1\pmod{p}$ and $b\equiv k+2^{-1}\pmod{p}$. Assume towards a contradiction $b\equiv 0\pmod{p}$. Then, $2k\equiv -1\pmod{p}$. Then, $5\equiv 4k^2\equiv 1\pmod{p}$, so $p|4$, so $p=2$, a contradiction.

We have $$b^2\equiv k^2+k+2^{-2}\equiv 2^{-2}(4k^2+4k+1)\equiv 2^{-2}(5+4k+1)\equiv 2^{-1}(2k+3)\equiv k+2^{-1}+1\equiv b+1\pmod{p}.$$
We claim $x_n\equiv b^n\pmod{p}$. We will prove it with induction with $n=0, 1$ as the base case. We have
$$x_n\equiv x_{n-1}+x_{n-2}\equiv b^{n-1}+b^{n-2}\equiv b^{n-2}(b+1)\equiv b^n\pmod{p}.$$So, since $b\not \equiv 0\pmod{p}$, none of the $x_n$ are equivalent to $0\pmod{p}$. Thus the whole sequence is relatively prime for $p$.

So, for each prime $p|m$, there exists a residue $b_p$ such that when $x_0\equiv 1\pmod{p}$ and $x_1\equiv b_p\pmod{p}$, $x_i$ is never a multiple of $p$. By the Chinese Remainder Theorem, there exists $B$ such that $B\equiv b_p\pmod{p}$ for all $p|m$. When $a=1$ and $b=B$, the sequence will always be relatively prime to $m$.
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IAmTheHazard
5000 posts
#31 • 1 Y
Y by centslordm
Let $a=1$ and $b=2k^2+k-2$.
$$(2k+k-2)^2-(2k+k-2)-1=(4k^2-5)(k^2-k-1) \implies (2k+k-2)^2 \equiv (2k+k-2)+1 \pmod{m} \implies 2k^2+k-2 \perp m \text{ and } x_n \equiv (2k^2+k-2)^n \pmod{m}.~\blacksquare$$
This post has been edited 1 time. Last edited by IAmTheHazard, Mar 15, 2023, 7:26 PM
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IAmTheHazard
5000 posts
#32 • 1 Y
Y by centslordm
Ok so the motivation of the above solution comes from the fact that for each prime $p$ dividing $m$ we can write $\sqrt{5} \equiv 2k$ and the recursion can be characterized as usual in the form
$$x_n \equiv x\left(\frac{1+2k}{2}\right)^n+y\left(\frac{1-2k}{2}\right)^n$$and it clearly necessary that all of these should be nonzero mod $p$ (in fact by CRT it is also sufficient, so we've reduced the problem to this). The best way to deal with this is to do the stupidest thing possible and set $y=0$ which yields (after adding $4k^2-5$) the above solution.
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awesomeming327.
1669 posts
#33 • 1 Y
Y by centslordm
Note that $5$ is always a quadratic residue of any $p\mid m$. Let $(2k-1)^2\equiv 5\pmod p$ then $p\mid k^2-k-1$. Then pick $a=1$, $b\equiv k\pmod p$, and we see that $x_n\equiv k^n\pmod p$ and by CRT we're done.
This post has been edited 1 time. Last edited by awesomeming327., May 22, 2023, 10:44 PM
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darshpatel
187 posts
#34
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We show that for each prime divisor $p$ of $m$, we can find some $a_p$ and $b_p$ so that $\left(x_n\right)$ has all its terms nonzero modulo $p$. Let $p$ be a prime divisor of $m$.

If $5$ is a prime divisor of $m$ and $p=5$, we take $a_5=1$ and $b_5=3$. Then $\left(x_n\right)$ repeats $1, 3, 4, 2, 1, 3,\dots$ modulo $5$, with all terms being nonzero.

Now we consider $p\neq 5$. Then $k\not\equiv 0\pmod{p}$, since that would otherwise imply $p\mid 5$. We also note that $p$ is odd, since $2$ does not divide $4k^2-5$. Now let $\alpha\equiv \frac{1+2k}{2}$ motivation We have
\[\alpha^2\equiv \frac{1+4k+4k^2}{4}\equiv \frac{6+4k}{4}\equiv \frac{3+2k}{2}\equiv \alpha + 1\]This also implies $\alpha\not\equiv 0$. Now, let $a_p\equiv 1$ and $b_p\equiv \alpha$. We claim by induction that $x_n\equiv \alpha^n$. We have the base cases $n=0$ and $n=1$ already. Then for the induction step, we have for $n\geq 2$ that
\[x_n\equiv x_{n-1}+x_{n-2}\equiv \alpha^{n-1}+\alpha^{n-2}\equiv \alpha^{n-2}(\alpha+1)\equiv \alpha^{n-2}\alpha^2\equiv \alpha^n\]which completes the induction. Then, since $\alpha\not\equiv 0$, all terms of $\left(x_n\right)$ are nonzero modulo $p$.

Using the Chinese Remainder Theorem, we can guarantee the existence of $a$ and $b$ such that $a\equiv a_p\pmod{p}$ and $b\equiv b_p\pmod{p}$ for all prime divisors $p$ of $m$. This guarantees that $\left(x_n\right)$ will have all its terms not divisible by any prime divisor of $m$, and thus relatively prime to $m$.
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bhan2025
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#35
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An alternative continuation after the
$x_n = x(\frac{1+k}{2})^n + y(\frac{1-k}{2})^n$ construction:

There are at most $p-1$ possible unique modulo $p$ pairs for $((\frac{1+k}{2})^n,(\frac{1-k}{2})^n)$
Each pair eliminates at most $p$ different potential pairs $(x,y)$
We are done as there must be at least one pair leftover
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