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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
My Unsolved FE in R+
ZeltaQN2008   2
N 6 minutes ago by megarnie
Source: Ho Chi Minh TST 2017 - 2018
Find all functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all any $x,y\in (0,\infty):$
$$f(1+xf(y))=yf(x+y)$$
2 replies
ZeltaQN2008
an hour ago
megarnie
6 minutes ago
Something nice
KhuongTrang   32
N 22 minutes ago by arqady
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
32 replies
KhuongTrang
Nov 1, 2023
arqady
22 minutes ago
Infimum of decreasing sequence b_n/n^2
a1267ab   35
N 36 minutes ago by shendrew7
Source: USA Winter TST for IMO 2020, Problem 1 and TST for EGMO 2020, Problem 3, by Carl Schildkraut and Milan Haiman
Choose positive integers $b_1, b_2, \dotsc$ satisfying
\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]and let $r$ denote the largest real number satisfying $\tfrac{b_n}{n^2} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?

Carl Schildkraut and Milan Haiman
35 replies
1 viewing
a1267ab
Dec 16, 2019
shendrew7
36 minutes ago
IMO Genre Predictions
ohiorizzler1434   52
N an hour ago by justaguy_69
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
52 replies
ohiorizzler1434
May 3, 2025
justaguy_69
an hour ago
\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2 }+\sqrt{c^2+a^2+2}\ge 6
parmenides51   19
N 2 hours ago by NicoN9
Source: JBMO Shortlist 2017 A1
Let $a, b, c$ be positive real numbers such that $a + b + c + ab + bc + ca + abc = 7$. Prove
that $\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \ge 6$ .
19 replies
parmenides51
Jul 25, 2018
NicoN9
2 hours ago
Inspired by Austria 2025
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N 4 hours ago by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
4 hours ago
Inequality with a,b,c
GeoMorocco   7
N 4 hours ago by lele0305
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
7 replies
GeoMorocco
Apr 11, 2025
lele0305
4 hours ago
Property of the divisors of k^3 - 2
Scilyse   2
N 5 hours ago by Assassino9931
Source: KoMaL A. 892
Given two integers, $k$ and $d$ such that $d$ divides $k^3 - 2$. Show that there exists integers $a$, $b$, $c$ satisfying $d = a^3 + 2b^3 + 4c^3 - 6abc$.

Proposed by Csongor Beke and László Bence Simon, Cambridge
2 replies
Scilyse
Jan 13, 2025
Assassino9931
5 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   1
N 5 hours ago by sami1618
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
1 reply
BR1F1SZ
Yesterday at 9:45 PM
sami1618
5 hours ago
Nordic 2025 P3
anirbanbz   8
N 6 hours ago by lksb
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
anirbanbz
Mar 25, 2025
lksb
6 hours ago
another functional inequality?
Scilyse   32
N 6 hours ago by ihategeo_1969
Source: 2023 ISL A4
Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]for every $x, y \in \mathbb R_{>0}$.
32 replies
Scilyse
Jul 17, 2024
ihategeo_1969
6 hours ago
Mount Inequality erupts in all directions!
BR1F1SZ   1
N 6 hours ago by sami1618
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
1 reply
BR1F1SZ
Yesterday at 9:41 PM
sami1618
6 hours ago
Division involving difference of squares
BR1F1SZ   1
N Yesterday at 10:02 PM by grupyorum
Source: Austria National MO Part 1 Problem 4
Determine all integers $n$ that can be written in the form
\[
n = \frac{a^2 - b^2}{b},
\]where $a$ and $b$ are positive integers.

(Walther Janous)
1 reply
BR1F1SZ
Yesterday at 9:50 PM
grupyorum
Yesterday at 10:02 PM
IMO ShortList 2002, number theory problem 6
orl   30
N Apr 25, 2025 by Ilikeminecraft
Source: IMO ShortList 2002, number theory problem 6
Find all pairs of positive integers $m,n\geq3$ for which there exist infinitely many positive integers $a$ such that \[ \frac{a^m+a-1}{a^n+a^2-1}  \] is itself an integer.

Laurentiu Panaitopol, Romania
30 replies
orl
Sep 28, 2004
Ilikeminecraft
Apr 25, 2025
IMO ShortList 2002, number theory problem 6
G H J
Source: IMO ShortList 2002, number theory problem 6
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orl
3647 posts
#1 • 10 Y
Y by Davi-8191, JasperL, Adventure10, ImSh95, Mango247, bjump, Tastymooncake2, cubres, and 2 other users
Find all pairs of positive integers $m,n\geq3$ for which there exist infinitely many positive integers $a$ such that \[ \frac{a^m+a-1}{a^n+a^2-1}  \] is itself an integer.

Laurentiu Panaitopol, Romania
Attachments:
This post has been edited 6 times. Last edited by orl, Sep 27, 2005, 4:59 PM
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orl
3647 posts
#2 • 4 Y
Y by Adventure10, ImSh95, Mango247, cubres
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions :)
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gollywog
64 posts
#3 • 4 Y
Y by ImSh95, Adventure10, Mango247, cubres
Ljunggren's theorem kills it.

//my bad, there is answer m=5, n=3.
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Hawk Tiger
667 posts
#4 • 4 Y
Y by Adventure10, ImSh95, Mango247, cubres
May I know that what is"Ljunggren's theorem"?
Thanks.
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gollywog
64 posts
#5 • 4 Y
Y by Adventure10, ImSh95, Mango247, cubres
http://www.math.sc.edu/~filaseta/seminars/MSRI2002/MSRILecture42002.pdf
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hmm
9 posts
#6 • 20 Y
Y by Amir Hossein, JasperL, Pure_IQ, ValidName, Kobayashi, jeteagle, megarnie, ImSh95, Adventure10, Mango247, Stuffybear, cubres, and 8 other users
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Zhero
2043 posts
#7 • 6 Y
Y by Pure_IQ, ImSh95, Adventure10, Mango247, Stuffybear, cubres
hmm wrote:
Click to reveal hidden text
Wow, that was quite ingenious. :D I have another finish to this problem which I feel is slightly cleaner:

Since $ a^n + a^2 - 1 | a^m + a - 1$, $ a^n + a^2 - 1 | (a+1)(a^m + a - 1) - (a^n + a^2 - 1) = (a^{m+1} + a^m) + (a^2 - 1) - (a^n + a^2 - 1) = a^{m+1} + a^m - a^n = a^n(a^{m-n+1} + a^{m-n} - 1)$. Therefore, $ a^{n} + a^2 - 1 | a^{m-n+1} + a^{m-n} - 1$. Hence, $ n \leq m - n + 1$, so $ 2n - 1 \leq m$. But it was have shown that $ m \leq 2n - 1$, so we have that $ m = 2n - 1$. Therefore, $ a^n + a^2 - 1 | a^n + a^{n-1} - 1$, so $ a^n + a^2 - 1 | a^{n-1} - a^2$. The only way this can be is if $ a^{n-1} = a^2$, yielding $ n = 3$ and $ m = 5$, as desired.
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mathocean97
606 posts
#8 • 7 Y
Y by JasperL, ValidName, ImSh95, Adventure10, Mango247, cubres, and 1 other user
We first show that $f(a) = a^n+a^2-1$ has no double root. But it is easy to see that $\gcd(f(a), f'(a)) = \gcd(a^n+a^2-1, na^{n-1}+2a) = 1$ (not hard, I can show this more explicitly). Now let $r$ be a root of $f(a)$. Then $r^m+r-1 = 0, r^n+r^2-1 = 0$.

Rearranging the above equations gives $r^m = 1-r, r^n = (1-r)(1+r)$. Dividing these equations and rearranging gives $r^{m-n+1}+r^{m-n}-1 = 0$. But this holds for all roots $r$, so we get that $f(a) \vert a^{k+1}+a^k-1$, where $k = m-n$. By a rather famous result, $X^n-X-1$ is irreducible for all $n$, so if we reverse the coefficients, $X^n+X^{n-1}-1$ is also irreducible. Therefore, $f(a) = a^{k+1} + a^k - 1 \implies n = 3, k = 2, \implies m = 5, n = 3$.

This article seems to contain the proof that $X^n-X-1$ is irreducible in the beginning and much more later.
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Nanas
44 posts
#9 • 3 Y
Y by ImSh95, Adventure10, cubres
My solution is written in the language of modules of polynomials, though the key ideas are nearly the same. It is less elegant than solutions above (As usual). It may appear lengthy but it is just direct computations ( I didn't omit any details, it is just the same as in my scratch paper)

Denote $f(a)=\frac{a^m+a-1}{a^n+a^2-1}$. . We let $P_m(x)=x^m+x-1$ and $Q_n(x)=x^n+x^2-1$, we have since both polynomials are monics, we can use Euclidean theorem for monic integer polynomials, to conclude that there is $S,R\in \mathbb{Z}[x]$ where $ \deg R < n$, such that $ P_m(x)= Q_m(x) S(x) + R(x)$, we have then \[ f(x) = \frac{x^m+x-1}{x^n+x^2-1} = \frac {P_m(x)}{Q_n(x)}= S(x) + \frac{R(x)}{Q_n(x)} \] If $R(x)$ is non-zero, We get the $\frac{R(x)}{Q_n(x)} \rightarrow 0$, which imply that for all but finitely many integers $s$, $ |\frac{R(x)}{Q_n(x)}| <1 $. And as that term never achieves zero for large numbers, we can't have $f(a)$ integer for infinitely many $a$'s. So we must have $Q_n | P_m$ in $\mathbb{Z}[x]$. Easily one can see that we must have $m > n$ .

Now we prove that $Q_n | P_m$ iff $(m,n)=(5,3)$. We do module arithmetic in the Euclidean domain $\mathbb{Q}[x]$ to prove that. Note that $(x,Q_n(x))=1$ and $(x-1,Q_n(x))=1$ in $\mathbb{Q}[x]$, and hence \[ \begin{array}{lcl}
x^m+x-1 & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff x^m+x-x^n-x^2 & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff x^{m-1}+1-x^{n-1}-x & \equiv  & 0\pmod{x^n+x^2-1}  \\
\iff x^{n-1}(x^{m-n} - 1) + (1-x) & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff x^{m-2} + \cdots + x^{n-1} -1 & \equiv &0 \pmod{x^n+x^2-1}
\end{array}\]
If we let $m=n+i$ we get
\begin{align*}
x^{m-2} + \cdots + x^{n-1} -1 &\equiv x^{n+i-2}+ \cdots + x^{n-1} + x^{i} + x^{i-1} + \cdots + x^2 -1   \\
&\equiv  (x^{n+i-2}+ x^{i})+ (x^{n+i-3}+x^{i-1}) + \cdots + ( x^{n} + x^{2}) + x^{n-1} - x^{i} - x^{i-1} - \cdots - x^{2} -1 \\
&\equiv x^{i-2}+ x^{i-3}+ \cdots +1 + x^{n-1} - x^{i} - x^{i-1} - \cdots - x^{2} -1 \\
&\equiv x^{n-1} + x - x^{i} - x^{i-1} \pmod{x^n+x^2-1} 
\end{align*}
Now if $i<n$ we get $x^{n-1} + x - x^{i} - x^{i-1} = 0 $ which cannot happen unless $n-1=i$ and $i-1=1$ and hence $(n,i)=(3,2)$ and so we have the solution $(m,n) = (5,3)$. Now if $i=n+j ; j\geq0$ , then \[ \begin{array}{lcl}
x^{n-1} + x - x^{i} - x^{i-1} & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff - x^{n-1} - x + x^{n+j} + x^{n+j-1} & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff  x^{n+j-1} + x^{n+j-2} - x^{n-2} - 1 & \equiv  & 0\pmod{x^n+x^2-1}  \\
\iff  x^{n-2}(x^{j+1}-1) + x^{n+j-2} - 1 & \equiv & 0 \pmod{x^n+x^2-1} \\
\iff     x^{n+j-2} + 2x^{n+j-3} + \cdots + 2x^{n-2}+ x^{n} + \cdots + 1 & \equiv &0 \pmod{x^n+x^2-1}
\end{array}\]

But $x^{n+j-2} + 2x^{n+j-3} + \cdots + 2x^{n-2}+ x^{n} + \cdots + 1 \geq 1$ for $ x\geq0$ , while $s^n+s^2-1=0$ for some $s\in(0,1)$ since divisibility applies also in $\mathbb{R}[x]$ , we get $s^{n+j-2} + 2s^{n+j-3} + \cdots + 2s^{n-2}+ s^{n} + \cdots + 1 = 0$, getting a contradiction.
This post has been edited 2 times. Last edited by Nanas, Aug 4, 2015, 6:17 PM
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ABCDE
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#11 • 5 Y
Y by randomusername, ImSh95, Adventure10, Mango247, cubres
I found a pretty cool and unique solution to this problem. Can someone check for its validity?

Solution
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tree3
641 posts
#12 • 4 Y
Y by ImSh95, Adventure10, Mango247, cubres
Notice that $m>n$ since $m,n>3$ implies that if $m \leq n$, then $0<\frac{a^m+a-1}{a^n+a^2-1}<1$ for $a$ relatively large. First we claim that $\frac{a^m+a-1}{a^n+a^2-1}$ must always be an integer. Indeed, since $\frac{a^m+a-1}{a^n+a^2-1}=R(x)+\frac{Q(x)}{a^n+a^2-1}$ for polynomials $R,Q$, with $deg(Q)<deg(a^n+a^2+1)$, implying that for large $a$, $Q(x)<a^n+a^2-1$ occurs, meaning that the quotient cannot yield an integer unless $Q(x)=0$ for infinitely many values, implying the above claim. Now we can bound. Since
$a^n + a^2 - 1 | a^m + a - 1$, we get that $ a^n + a^2 - 1 | (a+1)(a^m + a - 1) - (a^n + a^2 - 1)$.
But $(a+1)(a^m + a - 1) - (a^n + a^2 - 1)= a^{m+1} + a^m - a^n = a^n(a^{m-n+1} + a^{m-n} - 1)$.
Therefore, $ a^{n} + a^2 - 1 | a^{m-n+1} + a^{m-n} - 1$. Hence, $ n \leq m - n + 1$, so $ 2n - 1 \leq m$. Plugging in $a=2$, we immediately get $2^n+3|2^m+1$ with $m<2n$. Then $gcd(2^n+3,2^m+1)=gcd(3(2^{m-n})-1,2^n+3)$. Since $3(2^{m-n})-1<3(2^{m-n})<3(2^n)$ and $3(2^{m-n})-1$ is odd, $3(2^{m-n})-1=2^n+3$. This rearranges to $3(2^k)=2^n+4$ for $k=m-n$. Since $n>2$, $k<2$, implying the only solution to be $m=5,n=3$.
This post has been edited 6 times. Last edited by tree3, Jan 20, 2018, 1:37 AM
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v_Enhance
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#13 • 4 Y
Y by v4913, ImSh95, Adventure10, Mango247
The condition is equivalent to $a^n+a^2-1$ dividing $a^m+a-1$ as polynomials. The big step is the following analytic one.

Claim: We must have $m \le 2n$.

Proof. Assume on contrary $m > 2n$ and let $0 < r < 1$ be the unique real number with $r^n+r^2 = 1$, hence $r^m+r = 1$. But now \begin{align*} 		0 &= r^m + r - 1 < r(r^n)^2 + r - 1 = r\left( (1-r^2)^2+1 \right) - 1 \\ 		&= -(1-r)\left( r^4+r^3-r^2-r+1 \right). 	\end{align*}As $1-r > 0$ and $r^4+r^3-r^2-r+1 > 0$, this is a contradiction $\blacksquare$

Now for the algebraic part. Obviously $m > n$. \begin{align*} 	a^n+a^2-1 &\mid a^m+a-1 \\ 	\iff a^n+a^2-1 &\mid (a^m+a-1)(a+1) = a^m(a+1) + (a^2-1)  \\ 	\iff a^n+a^2-1 &\mid a^m(a+1) - a^n \\ 	\iff a^n+a^2-1 &\mid a^{m-n}(a+1) - 1. \end{align*}The right-hand side has degree $m-n+1 \le n+1$, and the leading coefficients are both $+1$. So the only possible situations are \begin{align*} 	a^{m-n}(a+1) - 1 &= (a+1)\left( a^n+a^2-1 \right) \\ 	a^{m-n}(a+1) + 1 &=  a^n+a^2-1. \end{align*}The former fails by just taking $a=-1$; the latter implies $(m,n) = (5,3)$. As our work was reversible, this also implies $(m,n) = (5,3)$ works, done.
This post has been edited 1 time. Last edited by v_Enhance, Apr 8, 2019, 5:10 PM
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anantmudgal09
1980 posts
#14 • 5 Y
Y by ELMOliveslong, SatisfiedMagma, ImSh95, Adventure10, Mango247
orl wrote:
Find all pairs of positive integers $m,n\geq3$ for which there exist infinitely many positive integers $a$ such that \[ \frac{a^m+a-1}{a^n+a^2-1}  \]is itself an integer.

Laurentiu Panaitopol, Romania

It is standard to prove that $x^n+x^2-1 \mid x^m+x-1$ as polynomials, by the given condition. Now the former has different signs at $x=0$ and $x=1$, so pick a root $0<\alpha<1$. Notice that $\alpha^n+\alpha^2=1$ and $\alpha^m+\alpha=1$. Suppose $m \ge 2n$. Then $$\alpha^m \le \alpha^{2n} \implies 1=\alpha^2+\alpha^n<\alpha+\alpha^{2n} \iff \alpha^{2n}-\alpha^2>\alpha^n-\alpha \iff \alpha^n+\alpha<1=\alpha^2+\alpha^n \iff \alpha<\alpha^2 \iff \alpha>1$$which is a contradiction!

So $m<2n$. Now $x^n+x^2-1 \mid x^{m-n+1}+x^{m-n}-1$ by taking residues. Finally, the latter polynomial has degree $m-n+1 \le (2n-1)-(n-1)$ so in fact, these two polynomials are equal, hence $m-n=2, m=2n-1$ so $m=5, n=3$. Clearly, this pair works, so we're done.
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IndoMathXdZ
691 posts
#15 • 1 Y
Y by ImSh95
Kind of nice.
The condition forces the polynomial $a^n + a^2 - 1$ to divide $a^m + a - 1$.
$\textbf{Claim 01.}$ $m < 2n$.
$\textit{Proof.}$ Suppose otherwise, we have $m \ge 2n$.
Notice that $(0)^n + (0)^2 - 1 = -1$ and $(1)^n + (1)^2 - 1 = 1$. By the Intermediate Value Theorem, we have that the polynomial $a^n + a^2 - 1$ has a root $\alpha$ which satisfy $0 < \alpha < 1$. Therefore,
\[ \alpha^m + \alpha = \alpha^n + \alpha^2 \Rightarrow \alpha (1 - \alpha) = \alpha^n (1 - \alpha^{m - n}) = (1 - \alpha^2)(1 - \alpha^{m - n}) \]This gives us
\[ \alpha = (1 + \alpha)(1 - \alpha^{m - n}) \]which can be rewritten as $\alpha^{m - n} + \alpha^{m - n + 1} = 1$. Since $m - n + 1 > n \ge 3$ and $0 < \alpha < 1$. We have
\[1 =  \alpha^{m - n} + \alpha^{m - n + 1} \le \alpha^n + \alpha^{n - 1} \le \alpha^n + \alpha^2= 1\]results to a clear contradiction.

Now, we have
\[ x^n + x^2 - 1 | x^{m - n + 2} - x^{m - n} - x + 1 \]Notice that we have the bound $n \le m - n + 2 \le n + 1$.
We'll prove that we must have $m - n + 2 = n + 1$.
Assume otherwise, then we must have $LHS = RHS$ since they are both the polynomial in the same degree and both having leading coefficient $1$, which is a clear contradiction as $RHS$ has constant $+1$ and $LHS$ has constant $-1$.
Then we have $m - n + 2 = n + 1$.
By considering the constant term ad the largest coefficient, we deduce
\[ (x^n + x^2 - 1)(x - 1) = x^{n + 1} - x^{n - 1} - x + 1 \]But taking $x = 2$. We have $LHS = 2^n + 2^2 - 1 \le 2^{n} + 2^{n - 1} - 1$ since $n \ge 3$.
For equality to hold, we must have $n = 3$. This forces $m = 5$, and we could check that
\[ (x^3 + x^2 - 1)(x^2 - x + 1) = x^5 + x - 1 \]Therefore, the pair $(3,5)$ clearly satisfies.
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A_Math_Lover
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#16 • 4 Y
Y by thczarif, AlastorMoody, yayitsme, ImSh95
Lemma: For $ a, m, n\in \mathbb{N} $, we have
\[a^n+a^2-1\ |\ a^{m-i} + (-1)^i\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^i\left(a-1\right)\]
Proof: We proceed by induction on $ i $. For $ i=0, 1 $, it is true. So assume for some $i>2$, it is true.

\begin{align*}
	a^n+a^2-1\ &|\ a^{m-i} + (-1)^i\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^i\left(a-1\right)\\[1em]
	\implies a^n+a^2-1\ &|\ a^{m-i} + (-1)^i\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^i\left(a-1\right) - (-1)^i(a^n+a^2-1)\\[.5em]
	&=a^{m-i} +  (-1)^i\left(-a^n+ a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^{i+1}(a^2-a)\\[1em]
	\therefore a^n+a^2-1\ &|\ a^{m-i-1} + (-1)^{i+1}\left(a^{n-1} - a^{n-2} +\dots + (-1)^{i+1}a^{n-i}\right) + (-1)^{i+1}\left(a-1\right)\\
\end{align*}Which proves our lemma.

Now, let $ i=n $, we get,
\[a^n+a^2-1\ |\ a^{m-n}  + (-1)^n \left(\frac{a^n+1}{a+1}\right) + (-1)^n(a-1)\]Clearly, $ m\ge n $. Let $ m=nk+q $. If $ n $ is even,
\begin{align*}
a^n+a^2-1\ &|\ a^{m-n}+a-1 + \left(\frac{a^n+1}{a+1}\right)\\[.5em]
\implies a^n+a^2-1\ &|\ a^q+a-1 + k\left(\frac{a^n+1}{a+1}\right)
\end{align*}
Which is not possible since the polynomial on the right side has degree at most $ n-1 $, and can't be $ 0 $ for all $ a\in \mathbb{Z} $.

So, $ n $ is odd. Then we have,

\begin{align*}
a^n+a^2-1\ &|\ a^{m-n} + a-1 - \left(\frac{a^n+1}{a+1}\right) - 2(a-1)\\[.5em]
\implies a^n+a^2-1\ &|\ a^q + a -1 - k\left(\frac{a^n+1}{a+1}\right) -2k(a-1) = P(a)
\end{align*}
Since $ P(a) $ has degree at most $ n-1 $, $ P(a) =0 $ for all $ a\in \mathbb{Z} $.


\begin{align*}
	P(a) &= a^q + a -1 - k\left(\frac{a^n+1}{a+1}\right) -2k(a-1)\\
	\therefore a^q + a - 1 &= k\left(\frac{a^n+1}{a+1}\right)+2k(a-1)
\end{align*}
Comparing the coefficients and degrees on both sides, we have, $ q=n-1=2,\ k=1 $. Which gives us the only solution $ (m, n) = (5, 3) $.
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Aryan-23
558 posts
#17 • 4 Y
Y by AlastorMoody, parola, ImSh95, Mango247
The problem is a joke if someone knows a particular lemma. ;)
Here's my solution
This post has been edited 2 times. Last edited by Aryan-23, Jun 26, 2020, 7:51 AM
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dchenmathcounts
2443 posts
#18 • 1 Y
Y by ImSh95
Cool problem!

We require $a^n+a^2-1\mid a^m+a-1$, else the remainder (which is deg $n-1$ at most) will eventually be smaller than $a^n+a^2-1$ for large enough $a.$

Also note $n<2m$ otherwise $a^m+a>a^{2m}+a^2>a^n+a^2$ for $0<a<1$ and obviously one of the roots has this value. So then the remainder is $-a^{m-n+2}+a^{m-n}+a-1,$ so we want $a^n+a^2-1\mid a^{m-n+2}-a^{m-n}-a+1,$ which must be equal to $a^n+a^2-1$ or $(a^n+a^2-1)(a-1)$ as $m-n+2\leq m+1.$ For size reasons $m-n+2=m$ or $m-n+2=m+1$. The first case fails by comparison of $a$ coefficient, and the second case requires $n=3$, which gives us $(a^3+a^2-1)(a-1)=a^m+a-1,$ fixing $m=5.$

So just $(3,5)$ works.
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DottedCaculator
7348 posts
#19 • 2 Y
Y by megarnie, ImSh95
Solution
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i3435
1350 posts
#20 • 4 Y
Y by ImSh95, Mango247, Mango247, Mango247
Here is a weird reason why $m<2n$ which I can't find above, given that we already know $x^n+x^2-1$ has to divide $x^{m-n+1}+x^{m-n}-1$.

Both of these polynomials are monotonic when $x>0$, and one divides the other, so it is impossible for them to have different signs for some $x>0$. Consider $x=\frac{1}{2}^{\frac{1}{n}}$. $\left(\frac{1}{2}^{\frac{1}{n}}\right)^n+\left(\frac{1}{2}^{\frac{1}{n}}\right)^2-1=\frac{1}{2}+\frac{1}{2}^{\frac{2}{n}}-1>0$ since $\frac{2}{n}<1$. On the other hand, $\left(\frac{1}{2}^{\frac{1}{n}}\right)^{m-n+1}+\left(\frac{1}{2}^{\frac{1}{n}}\right)^{m-n}-1=\left(\frac{1}{2}^{\frac{1}{n}}+1\right)\left(\frac{1}{2}^{\frac{m-n}{n}}\right)-1$. The first part of the product is $<2$ and since $\frac{m-n}{n}\ge 1$, the second part of the product is $\le \frac{1}{2}$. Thus this is $<0$, giving the desired contradiction.
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MathLuis
1523 posts
#21 • 1 Y
Y by ImSh95
Let $f(x)=x^n+x^2-1$ and $g(x)=x^m+x-1$ be polynomials in $\mathbb Z[x]$. We claim that $(m,n)=(5,3)$ works.
Claim 1: $f \mid g$ holds always.
Proof: For any $x \in \mathbb C$ we do polynomial division with quotient and residue in $\mathbb Z[x]$
$$g(x)=f(x)q(x)+r(x) \implies \frac{g(x)}{f(x)}=q(x)+\frac{r(x)}{f(x)}$$From here note that $\text{deg} \; r<\text{deg} \; f$ so set $x$ to be a extremely large positive integer such that $-f(x)<r(x)<f(x)$ meaning that for infinite values $r$ is 0 so $r(x)=0$ holds always meaning that $f \mid g$ as desired.
Claim 2: $2n-1 \ge m$
Proof: From Claim 1 one has that all the roots of $f$ are also roots of $g$ now $f'(x)=nx^{n-1}+2x>0$ if $x>0$ hence $f$ is strictly increasing in $\mathbb R^+$, now $f(1)=1$ and $f\left(\frac{2}{3} \right)=\left(\frac{2}{3} \right)^n-\frac{5}{9}<0$ meaning that there exists a root of $f$ satisfying $\frac{2}{3}<z<1$.
Assume FTSOC that $m \ge 2n$ then by all said before and ineqs
$$(1-z^2)^2=z^{2n}\ge z^m=1-z \implies (1-z)(z^2+z+1+z) \ge 1 \implies 1-z^3+z-z^2 \ge 1 \implies 1 \ge z+z^2 \ge 3z-1 \implies \frac{2}{3} \ge z \; \text{contradicction!!}$$Contradiction comes from assuming $m \ge 2n$ hence $2n-1 \ge m$ as desired.
A nice finish: Lets now work in integers and divisions
$$a^n+a^2-1 \mid a^m+a-1 \implies a^n+a^2-1 \mid a^{m-n+2}-a^{m-n}+1-a=(a-1)(a^{m-n}(a+1)-1) \implies a^n+a^2-1 \mid a^{m-n}(a+1)-1$$From here as RHS has at most degree $n$ we see that both have to be equal and $m=2n-1$ hence $a^n+a^2-1=a^n+a^{n-1}-1$ which means $n=3$ and this means $m=5$.
Hence the pair $(m,n)=(5,3)$ is the unique pair that works, thus we are done :D
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SatisfiedMagma
458 posts
#22 • 1 Y
Y by ImSh95
Did it back like 2 years ago? Very cool problem. Solved with quirtt and Fafnir.

Solution: We will show that $(m,n)=(5,3)$ is the only solution. One can see that it works as
\[a^5+a-1=(a^2-a+1)(a^3+a^2-1).\]We will break the solution into two claims which will bound the values of $m$. We define \begin{align*} P(a) \overset{\text{def}}{=} a^m+a-1 \\ Q(a) \overset{\text{def}}{=} a^n+a^2-1 \end{align*}where $P,Q$ are polynomials. We will now prove a well-known but extremely useful lemma which will take the problem from Number Theory to Algebra!

Lemma: If two polynomials $f,g \in \mathbb{Z}[x]$ are such that $f(n) \mid g(n)$ for infinitely many integers $n$, then $f(x) \mid g(x)$.

Proof: Due to size reasons, it is easy to see $\deg(g) > \deg(f)$. By the division algorithm over $\mathbb{Q}[x]$, write $g = fq + r$ for $f,q \in \mathbb{Q}[x]$ and $\deg(r) < \deg(f)$. One can clear out the denominators by multiplying by suitable integer, so we assume that all the polynomials are in $\mathbb{Z}[x]$. Just pick huge $k$ such that $f(k) \mid g(k)$. It would actually mean that $f(k) \mid r(k)$. But since $\deg(r) < \deg(f)$, we must have $r \equiv 0$ as desired. $\square$
So, we can actually apply the lemma to reduce the problem down to just two polynomials dividing each other!

Claim: $2n-1 \le m$.

Proof: First of all notice that $\gcd(a+1,Q(a))=1$. The algebra for divisibility begins here \begin{align*} P(a)                   & \equiv 0 \pmod{Q(a)} \\ (a+1)P(a)              & \equiv 0 \pmod{Q(a)} \\ a^{m+1}+a^2-a+ a^m+a-1 & \equiv 0 \pmod{Q(a)} \\ a^{m+1}+a^m+a^2-1      & \equiv 0 \pmod{Q(a)} \\ a^{m+1}+a^m-a^n        & \equiv 0 \pmod{Q(a)} \\ a^{m-n+1}+a^{m-n}-1    & \equiv 0 \pmod{Q(a)} \\ \end{align*}Bringing this into divisibility condition, we get \begin{align} a^n+a^2-1 \mid a^{m-n+1}+a^{m-n}-1 \label{1} \end{align}Due the bound on the degree, we get $m-n+1 \ge n$ which on simplifying gives $2n-1 \le m$ as desired. $\square$
The next is claim is a little harder to come up with and prove.

Claim: $m < 2n$.

Proof: Assume for the sake of contradiction, that $m>2n$. By the Intermediate Value Theorem, there exists a real root of $Q$ which lies between 0 and 1. Let this root be $\alpha$. By the problem condition, $\alpha$ is also a root of $P$. Now $P(\alpha)$ and $Q(\alpha)$ gives the following \begin{align} \alpha^m + \alpha = 1 \\ \alpha^n + \alpha^2 = 1 \end{align}Now we compute $\alpha^m$ and $\alpha^{2n}$. \begin{align*} \alpha^m    & = 1- \alpha      \\ \alpha^{2n} & = (1-\alpha^2)^2 \\ \end{align*}By assumption, we must have $\alpha^m < \alpha^{2n}$. Plugging in the values, we get \begin{align*} 1-\alpha    & \le (1-\alpha^2)^2           \\ 1-\alpha    & \le (1-\alpha)^2(1+\alpha)^2 \\ 1           & \le (1-\alpha)(1+\alpha)^2   \\ \iff \alpha & \le \varphi \end{align*}where $\varphi$ is the golden ratio. We will now show that $\alpha > \varphi$ which would bring the desired contradiction. It is easy to see that $P$ is an increasing function in the interval $[0,1]$. Note that \begin{align*} P(\varphi)=\varphi^m+\varphi-1 & < \varphi^2+\varphi-1 = 0 = P(\alpha) \\ \implies P(\varphi)            & < P(\alpha) \end{align*}and we can conclude that $\alpha > \varphi$ and we are done with the claim. $\square$
Now, due to the bounds we just need to check the case for $m=2n-1$. Putting this in $(1)$ we get
\[a^n+a^2-1 \mid a^{n-1}-a^2.\]This forces $n=3$ and we are done. $\blacksquare$
This post has been edited 2 times. Last edited by SatisfiedMagma, Jun 30, 2023, 6:27 PM
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huashiliao2020
1292 posts
#23
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solved with hint to consider r, otherwise im very proud of myself :) :) :)

It's obvious that m>n; we want to find m,n s.t. $a^{m-n+2}-a^{m-n}-a+1\equiv a^m+a-1-a^{m-n}(a^n+a^2-1)\equiv0\pmod{a^n+a^2-1}$.
We claim that $m\le 2n$.
Assume FTSOC m>2n, and take 0<r<1 as the real number root of both polynomials. We have $$0=r^m+r-1\le r^{2n+1}+r-1=-(1-r)(r^4+r^3-r^2-r+1),r^4+r^3-r^2-r+1>r^2(r^2+r-1)>r^2(r^m+r-1)=0,1-r>0\implies 0\le RHS<0,$$impossible.

Case 1. m=2n. We have $a(a^3-2a+1)\equiv a^{n+2}-a^n-a+1-(a^2-1)(a^n+a^2-1)\equiv0\pmod{a^n+a^2-1}\implies n=\{1,2,3\}$, none of which work.
Case 2. m=2n-1. We have $a^{n-1}+a^3-1\equiv a^{n+1}-a^{n-1}-a+1-a(a^n+a^2-1)\equiv0\pmod{a^n+a^2-1}$, and checking the possible n, only $\boxed{n=3,m=5}$ works.
Case 3. m=2n-2 has similar computation with degrees on n, and none of n work. Case 4 has $m\le 2n-3$, none of which work by looking at deg LHS<deg RHS except for a few n, and checking those don't work. Having exhausted all cases, we conclude the proof here.

Remark. After enough of these problems, reducing the degree of polynomials by suitably subtracting multiples of the modulus becomes second nature; it's very natural to do so by euclidean algorithm, and once LHS|RHS with LHS deg<RHS deg except for a few numbers, the problem is solved.
This post has been edited 2 times. Last edited by huashiliao2020, Sep 4, 2023, 7:19 PM
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IAmTheHazard
5001 posts
#24
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The answer is $(m,n)=(5,3)$ only, which clearly works since $a^3+a^2-1 \mid a^5+a-1$ in the polynomial sense. We now prove this is the only one.

For fixed $(m,n)$, since $a^n+a^2-1$ is monic we can write
$$\frac{a^m+a-1}{a^n+a^2-1}=p(a)+\frac{q(a)}{a^n+a^2-1}$$for some $p,q \in \mathbb{Z}[x]$ and $\deg q<n$. Then by considering sufficiently large $a$ we require $q \equiv 0$, so we should have $a^n+a^2-1 \mid a^m+a-1$ in the polynomial sense. Note that both of these polynomials have exactly one positive real root, so they must be the same. Call this common root $r$; clearly $r<1$.

Claim: $m<2n$.
Proof: Suppose otherwise. We should then have
$$1=r^n+r^2=r^m+r<r^{2n}+r \implies r-r^2>r^n-r^{2mn}.$$On the other hand, since $r^m<r$ but $r^n=1-r^2>1-r$, this is impossible, since $x-x^2$ increases as $x$ gets closer to $1/2$. $\blacksquare$

Now let $m=n+k$ with $0 \leq k \leq n-1$ (since clearly $m \geq n$). We should have
$$a^n+a^2-1 \mid a^k(a^n+a^2-1)-(a^{n+k}+a-1)=a^{k+2}-a^k-a+1=(a-1)(a^{k+1}+a^k-1).$$Since $\gcd(a^n+a^2-1,a-1)=1$ (in the polynomial sense) it follows that $a^n+a^2-1 \mid a^{k+1}+a^k-1$. In general $a^{k+1}+a^k-1$ is a nonzero polynomial, so $k<n-1$ is impossible, hence $k=n-1$ and $a^n+a^2-1 \mid a^n+a^{n-1}-1 \implies a^n+a^2-1 \mid a^{n-1}-a^2$. The last polynomial has degree less than $n$, hence it must be zero, i.e. $n=3$. Thus we extract the claimed solution. $\blacksquare$

Remark: Good problem to test "polynomial" intuition? To me considering $r$ felt "intuitive" but it's not super motivated in itself. I guess you start thinking about the roots after divisibility, and these are mostly hard to pin down except $r$.
This post has been edited 1 time. Last edited by IAmTheHazard, Oct 4, 2023, 1:16 PM
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thdnder
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#25
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The only answer is $(m, n) = (5, 3)$.

Obviously $m > n$. It's not hard to see that $x^n + x^2 - 1 \mid x^m + x - 1$. Let $r$ be a real root of $x^n + x^2 - 1$. Then $r$ is real root of $x^m + x - 1$. Thus $r^n + r^2 = r^m + r$, which is equivalent to $r(1 - r) = r^n(1 - r^{m-n})$.

If $m > 2n$, then $r(1 - r) = r^n(1 - r^{m-n}) > r^n(1 - r^{n}) \ge r(1 - r)$, a contradiction. Thus $m \le 2n$.

Now note that $x^n + x^2 - 1 \mid (x + 1)(x^m + x - 1) = x^{m+1} + x^m + x^2 - 1$, thus $x^n + x^2 - 1 \mid x^{m-n}(x + 1) - 1$. Since $m - n + 1 \le n + 1$, thus $x^{m-n}(x + 1) - 1 = (x + 1)(x^n + x^2 - 1)$ or $x^{m-n}(x + 1) - 1 = x^n + x^2 - 1$. In the first case we have $x^{n+1} + x^n + x^3 + x^2 - x - 1 = x^{n} + x^2 - 1$, which is impossible. In the second case we have $x^n + x^2 - 1 = x^n + x^{n-1} - 1$, thus $n = 3$. Thus $m = 5$. Therefore $(m, n) = (5, 3)$ is only answer. $\blacksquare$
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awesomeming327.
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Note that these are both polynomials, so if for infinitely many $a$, $a^n+a^2-1\mid a^m+a-1$ then the polynomial $a^n+a^2-1$ divides the polynomial $a^m+a-1$. Note that $P(x) \mid Q(x)$ implies $\text{deg} P\le \text{deg} Q$. In particular, $m\ge n$.

Since $a^n + a^2 - 1\mid a^m+a-1$, which implies $a^n+a^2-1|(a+1)(a^m+a-1)-(a^n+a^2-1)$. This becomes $a^n+a^2-1\mid a^{m+1}+a^{m}-a^n$. Note that $\gcd(a^n+a^2-1,a)=\gcd(a^2-1,a)=1$ so $a^n+a^2-1\mid a^{m-n+1}+a^{m-n}-1$. This implies $m\ge 2n-1$. This also means $m-n\ge n-1\ge 2$.

By IVT and taking the derivative, there exists a unique real $0<r<1$ for which $r^n+r^2-1=0$. We must have $r^m+r-1=0$. Setting them equal, we get
\[r=\frac{r^n(1-r^{m-n})}{1-r}=\frac{(1-r^2)(1-r^{m-n})}{1-r}=(1+r)(1-r^{m-n})\]and this becomes $r^{m-n}(r+1)=1=r^n+r^2\ge r^n+r^{m-n}$ which implies $r^{m-n+1}\ge r^n\implies m-n+1\le n$ so $m\le 2n-1$. Note that equality is dependent on $m-n=2$, so we need $(m,n)=(5,3)$ which works.
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HamstPan38825
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#27
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Claim. [Main Claim] $m \leq 2n$.

Proof. Let $r$ be a root of $a^n + a^2 - 1 = 0$, such that $r^n + r^2 = r^m + r = 1$. Suppose for the sake of contradiction that $m > 2n$. Then,
\begin{align*}
r^m + r &< r^{2n} + r \\
&= (r^n - r^{n+2}) + r \\
&= 1 - 2r^2 + r^3 + r \\
0 &< r^4 + r - 2r^2 \\
&=r(r-1)(r^2+r-1).
\end{align*}On the other hand, $r^2 + r > 1$ and $r - 1 < 0$, so this is a clear contradiction. $\blacksquare$

Now, by Euclidean algorithm we have $$a^n + a^2 - 1 \mid (a-1)(a^{m-n+1}+a^{m-n} - 1).$$In particular, as $a^n + a^2 - 1 \equiv 1 \pmod {a-1}$, we have $$a^n + a^2 - 1 \mid a^{m-n+1} + a^{m-n} + 1.$$If $m \leq 2n-2$, then the RHS is clearly smaller and $a^{m-n+1}+a^{m-n} + 1 = 0$, which is clearly impossible.

If $m \in \{2n-1, 2n\}$, perform one more step to get $$a^n + a^2 - 1 \mid a^{m-n} + a^{m-2n+1}-a^{m+3-2n} - 1.$$As $m-2n+1 \leq 1$, the RHS is smaller than the LHS now, so it must uniquely equal zero. When $m = 2n-1$, we get $a^{n-1} = a^2$, giving the solution pair $(3, 5)$. When $m = 2n$, we get $a^n + a - a^3 - 1=0$, which does not work for any value of $n$. Thus, $(3, 5)$ is the only solution.
This post has been edited 1 time. Last edited by HamstPan38825, Dec 16, 2023, 11:37 PM
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shendrew7
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#28
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We claim our only solution is $\boxed{(5,3)}$. Our condition, $N/D \in \mathbb{Z}$, can be rewritten as
\[D \mid a^{m-n}\left(a^n+a^2-1\right) - \left(a^m+a-1\right) \implies D \mid a^{m-n+1}+a^{m-n}-1.\]
There are clearly no solutions for $m \leq 2n-2$, and investigating $m=2n-1,2n$ gives the desired solution. Now assume FTSOC $m>2n$, and consider a root $0<r<1$ of $D \mid N$. Then
\[\left(a^n+a^2\right)^2 - (a^m+a) = 0 \implies a\left(2a^{n+1}+a^3-1\right) = a^m-a^{2n}.\]
The RHS is negative under the assumption, but the LHS can be expressed as
\[a\left((a^{n+1})+a(a^n+a^2)\right)-(a^n+a^2) = a(1-a)(a-a^n) > 0. \quad \blacksquare\]
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bjump
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#29 • 1 Y
Y by megarnie
solution (arch carried me :skull:)
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atdaotlohbh
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#30
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Let $P(a)$ be the enumerator and $Q(a)$ be the denominator. Divide $P$ on $Q$:
$$P=QT+R$$where $deg R<def Q$ and division is done in $\mathbb{Q}$. Multiply this equation by the lcm of denominators so that all coeffiecients were integer. Now $Pc$ is divisible by $Q$ and $(Tc)Q$ are divisible by $Q$ in infinitely many points from the problem statement, thus $Rc$ is also divisible by $Q$ in infinitely many points. But if $R \neq 0$, then eventually $|Rc| < |Q|$, so $R$ must be equal to zero. Hence $P$ is divisible by $Q$
Now $a^{m+1}+a^{m}-a^n=(a+1)P-Q \vdots Q$, but $(Q,a)=1$, thus $a^{m-n+1}+a^{m-n}-1 \vdots Q$. Denote $m-n=t$
Obviously $t+1 \geq$deg $Q=n$. Note that $Q(0)=-1$ and $Q(1)=1$, thus there exists a number $x_0 \in (0,1)$ such that $Q(x_0)=0$. Then $x_0^{t+1}+x_0^{t}=1=x_0^n+x_0^2$. Note that if $t>2$, then $x_0^{t+1} \leq x_0^n$ and $x_0^t<x_0^2$, contradicting the statement. Thus $t \leq 2$. But $n \leq t+1 \leq 3$, thus $n=3, t=2$, and $m=t+n=5$
To show that it works, just note that $x^5+x-1=(x^3+x^2-1)(x^2-x+1)$
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OronSH
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#31
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By polynomial division if $x^n+x^2-1\nmid x^m+x-1$ then there is a remainder term which is eventually always smaller than the denominator, giving non integers. Thus $x^n+x^2-1\mid x^m+x-1$. Rearranging, $x^n+x^2-1\mid\tfrac1x((x^m+x-1)-(x^n+x^2-1))+(x^m+x-1)=x^{n-1}(x^{m-n+1}+x^{m-n}-1)$ so $m-n+1\ge n$. Now let $r$ be a root of $x^n+x^2-1$ with $0<r<1$ which exists by IVT. Then $r^m=1-r,r^n=1-r^2\implies r^{m-n}=\tfrac1{r+1}$ which is between $\tfrac 12$ and $1$. However $r^n\le\tfrac12$ as otherwise $r^n+r^2-1>\tfrac12+\tfrac12-1=0$. Thus $m-n<n$. Together we have $m=2n-1$, which implies $x^n+x^2-1\mid x^n+x^{n-1}-1$, so $n=3,m=5$ is the only solution.
This post has been edited 1 time. Last edited by OronSH, Feb 27, 2025, 6:07 PM
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Ilikeminecraft
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#32
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Problem is equivalent to finding $m, n$ such that $a^n + a^2 - 1 \mid a^m + a - 1.$

Let $f(x) = x^m + x - 1, g(x) = x^n + x^2 - 1.$ Let $0 < r < 1$ be a real root to $g(x)$. This exists by IVT. Hence, we must have that $f(r) = 0.$ I claim that $m < 2n.$

First, we have that $r^m + r = r^n + r^2 = 1.$ Rearranging, we have that $r(1 - r) = r^{n}(1 - r^{m - n}) = (1 - r^2)(1 - r^{m - n}).$ Hence, $r^{m - n + 1} + r^{m - n} = 1.$ Now, assume for the sake of contradiction, that $m \geq 2n.$ It follows that $r^{m - n + 1} + r^{m - n} < r^{n + 1} + r^{n} = r^{n + 1} + 1 - r < 1.$ Thus, we have that $m < 2n.$

We now simplify the problem a bit:
\begin{align*}
  a^n + a^2 - 1 \mid a^m + a - 1 & \implies a^n + a^2 - 1 \mid (a^m + a - 1)(a + 1) \\
  & \implies a^n + a^2 - 1 \mid a^{m + 1} + a^m - a^n \\
  & \implies a^n + a^2 - 1 \mid a^{m - n + 1} + a^{m - n} - 1 \\
\end{align*}Hence, since $n < m < 2n,$ we have that the degree of the LHS is $m - n + 1 < n + 1.$ Hence, it must be $n,$ and so $m = 2n - 1.$ Hence, the only solution is $(5, 3)$ only.
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