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

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

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

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
IMO 2014 Problem 1
Amir Hossein   132
N 11 minutes ago by maromex
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
132 replies
Amir Hossein
Jul 8, 2014
maromex
11 minutes ago
2-var inequality
sqing   0
17 minutes ago
Source: Own
Let $ a,b,c\geq 0    $. Prove that
$$ \frac{a }{a^2+2b^2+1}+ \frac{b }{b^2+2a^2+1}\leq \frac{1}{\sqrt{3}} $$$$   \frac{a }{2a^2+ b^2+2ab+1}+ \frac{b }{2b^2+ a^2+2ab+1}  \leq \frac{1}{\sqrt{5}} $$$$\frac{a }{a^2+2b^2+2ab+1}+ \frac{b }{b^2+2a^2+2ab+1}\leq \frac{1}{2} $$
0 replies
1 viewing
sqing
17 minutes ago
0 replies
IMO Genre Predictions
ohiorizzler1434   31
N 18 minutes ago by ohiorizzler1434
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
31 replies
ohiorizzler1434
Saturday at 6:51 AM
ohiorizzler1434
18 minutes ago
weird FE
tobiSALT   10
N 22 minutes ago by NicoN9
Source: Pan American Girls' Mathematical Olympiad 2024, P5
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(f(x+y) - f(x)) + f(x)f(y) = f(x^2) - f(x+y),$

for all real numbers $x, y$.
10 replies
tobiSALT
Nov 27, 2024
NicoN9
22 minutes ago
2015 solutions for quotient function!
raxu   49
N 31 minutes ago by blueprimes
Source: TSTST 2015 Problem 5
Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$.

Proposed by Iurie Boreico
49 replies
1 viewing
raxu
Jun 26, 2015
blueprimes
31 minutes ago
something...
SunnyEvan   0
44 minutes ago
Source: unknown
Try to prove : $$ \sum csc^{20} \frac{2^{i} \pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^{k} \pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
0 replies
SunnyEvan
44 minutes ago
0 replies
Cosine of polynomial is polynomial of cosine
yofro   1
N an hour ago by yofro
Source: 2025 HMIC #2
Find all polynomials $P$ with real coefficients for which there exists a polynomial $Q$ with real coefficients such that for all real $t$, $$\cos(P(t))=Q(\cos t).$$
1 reply
+1 w
yofro
an hour ago
yofro
an hour ago
Problem 1
randomusername   72
N an hour ago by blueprimes
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
72 replies
randomusername
Jul 10, 2015
blueprimes
an hour ago
Permutation with no two prefix sums dividing each other
Assassino9931   2
N 2 hours ago by Assassino9931
Source: Bulgaria Team Contest, March 2025, oVlad
Does there exist an infinite sequence of positive integers $a_1, a_2 \ldots$, such that every positive integer appears exactly once as a member of the sequence and $a_1 + a_2 + \cdots + a_i$ divides $a_1 + a_2 + \cdots + a_j$ if and only if $i=j$?
2 replies
Assassino9931
3 hours ago
Assassino9931
2 hours ago
Sum of Good Indices
raxu   25
N 4 hours ago by cj13609517288
Source: TSTST 2015 Problem 1
Let $a_1, a_2, \dots, a_n$ be a sequence of real numbers, and let $m$ be a fixed positive integer less than $n$. We say an index $k$ with $1\le k\le n$ is good if there exists some $\ell$ with $1\le \ell \le m$ such that $a_k+a_{k+1}+...+a_{k+\ell-1}\ge0$, where the indices are taken modulo $n$. Let $T$ be the set of all good indices. Prove that $\sum\limits_{k \in T}a_k \ge 0$.

Proposed by Mark Sellke
25 replies
raxu
Jun 26, 2015
cj13609517288
4 hours ago
f(x^2+y^2+z^2)=f(xy)+f(yz)+f(zx)
dangerousliri   7
N 4 hours ago by jasperE3
Source: FEOO, Shortlist A4
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}$ such that for any positive real numbers $x, y$ and $z$,
$$f(x^2+y^2+z^2)=f(xy)+f(yz)+f(zx)$$
Proposed by ThE-dArK-lOrD, and Papon Tan Lapate, Thailand
7 replies
dangerousliri
May 31, 2020
jasperE3
4 hours ago
foldina a rectangle paper 3 times
parmenides51   1
N 5 hours ago by TheBaiano
Source: 2023 May Olympiad L2 p4
Matías has a rectangular sheet of paper $ABCD$, with $AB<AD$.Initially, he folds the sheet along a straight line $AE$, where $E$ is a point on the side $DC$ , so that vertex $D$ is located on side $BC$, as shown in the figure. Then folds the sheet again along a straight line $AF$, where $F$ is a point on side $BC$, so that vertex $B$ lies on the line $AE$; and finally folds the sheet along the line $EF$. Matías observed that the vertices $B$ and $C$ were located on the same point of segment $AE$ after making the folds. Calculate the measure of the angle $\angle DAE$.
IMAGE
1 reply
parmenides51
Mar 24, 2024
TheBaiano
5 hours ago
Divisibility NT
reni_wee   0
5 hours ago
Source: Japan 1996, ONTCP
Let $m,n$ be relatively prime positive integers. Calculate $gcd(5^m+7^m, 5^n+7^n).$
0 replies
reni_wee
5 hours ago
0 replies
Modular arithmetic at mod n
electrovector   3
N 6 hours ago by Primeniyazidayi
Source: 2021 Turkey JBMO TST P6
Integers $a_1, a_2, \dots a_n$ are different at $\text{mod n}$. If $a_1, a_2-a_1, a_3-a_2, \dots a_n-a_{n-1}$ are also different at $\text{mod n}$, we call the ordered $n$-tuple $(a_1, a_2, \dots a_n)$ lucky. For which positive integers $n$, one can find a lucky $n$-tuple?
3 replies
electrovector
May 24, 2021
Primeniyazidayi
6 hours ago
Imo Shortlist Problem
Lopes   35
N Apr 24, 2025 by Maximilian113
Source: IMO Shortlist 2000, Problem N4
Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$.
35 replies
Lopes
Feb 27, 2005
Maximilian113
Apr 24, 2025
Imo Shortlist Problem
G H J
Source: IMO Shortlist 2000, Problem N4
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Lopes
42 posts
#1 • 6 Y
Y by itslumi, FaThEr-SqUiRrEl, Adventure10, Mango247, and 2 other users
Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$.
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Davron
484 posts
#2 • 5 Y
Y by mathisreal, FaThEr-SqUiRrEl, Adventure10, Mango247, and 1 other user
Answer: (a, m, n) = (1, m, n), (a, 1, n) or (2, 3, k) where k > 1.

It is obvious that (1, m, n) and (a, 1, n) are solutions, so assume a > 1 and m > 1. We show first that m must be odd.

Suppose there is an odd prime p dividing am + 1. Then p must also divide (a + 1)n and hence (a + 1). So a = -1 mod p, so am + 1 = (-1)m + 1 mod p. Hence m is odd. If there is no such prime p, then am + 1 = 2k for some k. But a and m are assumed to be at least 2, so am + 1 must be at least 5 and hence k ≥ 3. So am = -1 mod 4. But squares can only be 0 or 1 mod 4, so am cannot be a square, so m must be odd.

Suppose p divides m. Then (since m is odd) ap + 1 divides am + 1 and hence (a + 1)n. But p is odd, so a + 1 divides ap + 1 and (ap + 1)/(a + 1) = ap-1 - ap-2 + ap-3 - ... - p + 1 divides (a + 1)n-1. Put f(x) = xp-1 - xp-2 + ... - x + 1. Then f(-1) = p, so f(x) - p = (x + 1) g(x) for some polynomial g(x). Putting x = a, we get f(a) = (a + 1) g(a) + p. If a prime q divides f(a), then since f(a) divides (a + 1)n-1, q must divide (a + 1). Hence q divides p, so q = p. So f(a) = pk for some k ≥ 1. Since f(a) divides (a + 1)n-1, p must divide a + 1.

We show next that k = 1, so f(a) = p. If p2 divides (a + 1), then f(a) = (a + 1) g(a) + p implies that p2 does not divide f(a). If p2 does not divide (a + 1), then (a + 1) = ph (with h not divisible by p). So f(a) = (ap + 1)/(a + 1) = ( (1 + ph - 1)p)/ph = ( 1 - 1 + p2h + terms divisible by p3)/ph = p mod p2, so again p2 does not divide f(a). So in any case f(a) = p.

Now (a2 - a) ≥ 2, (a4 - a3) ≥ 8 and obviously (a2r - a2r-1) ≥ 2 for r ≥ 2. So f(a) ≥ 11 > 5 for p = 5, and hence by a trivial induction f(a) > p for p > 5. So the only possibility is p = 3. Then we have a2 - a + 1 = 3 and hence a = 2.

So far we have established that a = 2 and that m is a power of 3. But (29 + 1) = 513, which is not a power of 3 and so cannot divide (2 + 1)n. But (227 + 1), (281 + 1), ... are all divisible by (29 + 1) and hence are also not powers of 3. So m must be 3. Finally, we require that (am + 1) = 9 divides (2 + 1)n, so n ≥ 2.

davron ft. kalva
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trigg
36 posts
#3 • 3 Y
Y by FaThEr-SqUiRrEl, Adventure10, Mango247
I think this problem is not hard but it is nice :)
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ZetaX
7579 posts
#4 • 3 Y
Y by FaThEr-SqUiRrEl, Adventure10, Mango247
Like the similar ones just a corollary of Zsigmondy's theorem :P
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Scientist
11 posts
#5 • 2 Y
Y by FaThEr-SqUiRrEl, Adventure10
If a=1 then 〖 1〗^m+1|〖(1+1)〗^n this means for all m,n∈Z^+ and a=1 is solution. If a=2 and m=3 then 〖 2〗^3+1|3^n this means a=2 m=3 and for all n≥2 and n∈Z^+ is solution.If m=1 then for all a,n∈Z^+ is solution. If (a,m)≠(2,3) and a≠1 and m≠1 then from the Zsigmondy theorem ∃p∈prime p|a^m+1 but p∤a+1 this means contradiction.
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quantumbyte
547 posts
#6 • 6 Y
Y by Max D.R., nguyendangkhoa17112003, FaThEr-SqUiRrEl, Adventure10, Mango247, and 1 other user
Quick way to start the proof before some cases:
Solution outline
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sjaelee
485 posts
#7 • 3 Y
Y by FaThEr-SqUiRrEl, DCMaths, Adventure10
If $m$ is even, the any prime $p|a+1$ or $a\equiv -1\pmod{p}$ and $a^m+1\equiv 2\pmod{p}$. If $a+1$ is a power of $2$, then we have a contradiction as primes other than $2$ divide $a^m+1$ unless $a^m+1=2$, which means $a=1$ and discounts the case otherwise.

For $m$ odd, $a+1 | a^m+1$ and any prime divisor of $a^m+1$ divides $(a+1)^n$, implying all prime divisors of $a^m+1$ divides $a+1$. By LTE if a prime $p$ divides $a+1$, then $v_p(a^m+1)=v_p(a+1)+v_p(m)$. Therefore if all the prime divisors of $a+1$ are the prime divisors of $a^m+1$, and the $v_p$ of any prime dividing $a^m+1$ divides $m$, $\frac{a^m+1}{a+1} | m$.

If $m=1$ obviously all $a$ and $n$ work. If $a=1$ clearly all $m,n$ work. If $m>3$ the $\frac{a^m+1}{a+1}>m$ for $a\ge 2$. If $m=3$ the only solution is $a=2$ otherwise $\frac{a^m+1}{a+1}>m$.
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sayantanchakraborty
505 posts
#8 • 7 Y
Y by Sx763_, FaThEr-SqUiRrEl, ZHEKSHEN, NVA9205, Adventure10, Mango247, panche
It is easy to note that $(a,m,n)=(1,m,n)$ is a solution.Let $a \ge 2$.Also suppose $m \ge 2$ and $(a,m) \neq (2,3)$.Then by Zsigmondy's theorem there exists a primitive prime factor of $a^m+1$ which does not divide $(a+1)$ and hence doesn't divide $(a+1)^n$.For $m=1$ we get the trivial solution $(a,1,n)$.For $a=2,m=3$ the condition is true for $n \ge 2$ so $(2,3,n \ge 2)$ are also solutions.
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tenplusten
1000 posts
#9 • 2 Y
Y by FaThEr-SqUiRrEl, Adventure10
Too easy problem.
$(a,m,n)=(1,m,n),(2,3,n)$
It is obvious that $m$ is odd. İf $m$ is even:
Take $p|a^m+1$ then $p|a+1$ or $a\equiv -1 mod p$ $a^m+1\equiv (-1)^m+1 mod p$. Contradiction.
Claim: $a^m+1$ and $a+1$ have the same prime divisors.
Proof:Take an arbitrary $p|a^m+1$ then $p|(a+1)^n$ since $p$ is prime $p|a+1$.

By Zsigmondy's theorem there is $q|a^m+1$ but not $q|a+1$.Contradiction.

Comment
This post has been edited 2 times. Last edited by tenplusten, Mar 24, 2017, 5:10 AM
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ubermensch
820 posts
#10 • 3 Y
Y by FaThEr-SqUiRrEl, Adventure10, Mango247
Wait Zsigmondy was still so unpopular even as late as $2000$? (as this is not even a one-liner using Zsigmondy and its exceptions)...
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AlastorMoody
2125 posts
#12 • 1 Y
Y by FaThEr-SqUiRrEl
Solution
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R-sk
429 posts
#13 • 1 Y
Y by FaThEr-SqUiRrEl
Here since we can easily find the trivial solutions then by zsigmondy there exist a divisor of a^m+1 that does not divide a+1 and therefore does not divide (a+1) ^n and therefore there are no other solutions other than (1, m, n), (2, 3,n), (a, 1,n)
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MatBoy-123
396 posts
#14 • 1 Y
Y by FaThEr-SqUiRrEl
N5 is also an application of Zsimondy's..
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isaacmeng
113 posts
#15 • 1 Y
Y by FaThEr-SqUiRrEl
This can be trivialized by Zsigmondy's.

ISL 00N4. Find all $(a, m, n)\in\mathbb{N}^3$ such that $a^m+1\mid (a+1)^n$.

Solution. If $a=1$ then the divisibility condition obviously work. Assume now that $a\ge 2$.
If $m=1$ then the divisibility condition obviously work. Assume now that $a\ge 2$, $m\ge 2$.
If the two conditions $a=2$, $m=3$ are not both true then by Zsigmondy's there must exist a prime factor $p$ of $a^m+1$ which does not divide $a+1$ (and of course so does not $(a+1)^n$). Hence there is no solution in this case.
It suffices for us to consider the single case $a=2$, $m=3$, in which only all $n\ge 2$ satisfy the divisibility condition.
Therefore the solutions are only $(1, m, n), (a, 1, n), (2, 3, k)$ where $k\ge 2$.
This post has been edited 4 times. Last edited by isaacmeng, Apr 20, 2021, 1:30 PM
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hakN
429 posts
#16 • 4 Y
Y by FaThEr-SqUiRrEl, Mango247, Mango247, Mango247
If $a=1$ then we get $(1,m,n)$ as solutions for all $m,n\in \mathbb{N}$.
If $a=2$ then we get that $2^m + 1\mid 3^n$. If $m=1$ then we get $(2,1,1)$ as a solution. If $m\geq 2$, then $2^m + 1 = 3^k$ doesnt have any solution except $(m,k)=(3,2)$ becuase of Mihailescu's Theorem.

Let $a\geq 3$ and $m\geq 2$. By Zsigmondy, $a^m + 1$ has a prime divisor that doesnt divide $a+1$. Thus, if we take that divisor we clearly get no solution. If $m=1$ then we get $(a,1,n)$ as solutions.
So all solutions are $(1,m,n) , (a,1,n) , (2,3,n)$.
This post has been edited 2 times. Last edited by hakN, Apr 20, 2021, 6:43 PM
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oVlad
1742 posts
#17
Y by
Here's another way to solve this without Zsigmondy:

If $p\mid a^m+1$ then $p\mid (a+1)^n$ so $p\mid a+1$ so $$\nu_p(a^m+1)=\nu_p(a+1)+\nu_p(m)\leq \nu_p(a+1)\big(\nu_p(m)+1\big)\leq \nu_p(a+1)\big(\log_2 m+1\big)$$Hence, we gat that $a^m+1\mid (a+1)^{\log_2 m+1}$ so $(a+1)^{\log_2 m+1}\geq a^m+1.$ However, by Holder's Inequality, we get that \[2^{\log_2 m+1}\big(a^{\log_2 m+1}+1\big)\geq (a+1)^{\log_2 m+1}\geq a^m+1\]so by letting $\log_2 m=c$ we essentially got that $2^{c+1}\big(a^{c+1}+1\big)\geq a^{2^c}+1$ which, after some easy bounding, will give us $a=1, m=1$ or $a,m\leq 3.$
This post has been edited 2 times. Last edited by oVlad, Oct 21, 2021, 9:21 AM
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MathLuis
1523 posts
#18
Y by
Sorry for using Zsigmondy when almost everyone did :blush: , i just couldnt resist the free problem xD

Case 1: $m=1$
Then $a+1 \mid (a+1)^n$ which holds for any $a,n \in \mathbb N$ thus $(a,m,n)=(a,1,n)$
Case 2: $a=1$
Then $2 \mid 2^n$ which holds for any $m,n \in \mathbb N$ thus $(a,m,n)=(1,m,n)$
Case 3: $a,m>1$
Since $\gcd(a,1)=1$ we can use Zsigmondy assuming that $(a,m) \ne (2,3)$ so let $p$ such that $p \mid a^m+1$ and $p \not \; \mid a+1$ but since $p \mid a^m+1 \mid (a+1)^n$ we get a contraidiction.
Thus $a=2$ and $m=3$ and since $9 \mid 3^n$ holds for $n \ge 2$ the sol is $(a,m,n)=(2,3,n)$ where $n>2$

Thus we are done :blush:
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Mogmog8
1080 posts
#19 • 1 Y
Y by centslordm
By Zsigmondy, $a^m+1$ has a primitive prime factor if $(a,m)\neq(2,3);$ let the prime be $p.$ If $m>1$ and $a>1,$ $p\nmid a+1,$ so $a^m+1\nmid (a+1)^n.$ If $a=1$ or $m=1,$ notice that $1+1\mid 2^n$ and $a+1\mid(a+1)^n,$ respectively. If $(a,m)=(2,3)$ and $n>1,$ we have $2^3+1\mid(2+1)^n,$ so our solutions are $\boxed{(a,1,n),(1,m,n)},$ and $\boxed{(2,3,n)\text{ where }n>1}.$ $\square$
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hydo2332
435 posts
#20
Y by
We divide in two cases.

1) $m$ is even.
In this case, we have that if $p \mid a+1$, then $p \mid a^m + 1$, therefore $p \mid 2$ since $m$ is even. So we get $a^m + 1 = 2^x$ for some $x$, but notice that $4 \nmid a^m + 1$ so we conclude $x=1, a=1$. In this case we get the solution $(a,n,m) = (a,2m',n)$ for any $m',n$ positive integers.

2)$m$ is odd. In this case, we have that $a+1 \mid a^m + 1$. So we get that $\dfrac{a^m + 1}{a+1} \mid (a + 1)^{n-1}$. Notice that $\dfrac{a^m + 1}{a+1} = a^{m-1} - a^{m-2} + \dots + 1$ is odd since $m$ is odd and therefore it has no factors $2$.

So we can apply the LTE Lemma, which concludes that $\nu_p(\dfrac{a^m + 1}{a+1}) = \nu_p(m)$ for every prime $p$ that divides $\dfrac{a^m + 1}{a+1}$ (which are almost all the same primes that divide $a^m + 1$, the same as the ones that divide $a+1$). But this clearly implies that $\dfrac{a^m + 1}{a+1} \mid m$. Looking at the size of the left hand size, we get that either $a=1$, $m=3$ or $m=1$. If $a=1$, we get the solution $(a,n,m) = (1,n,2m'+1)$ for any $m',n$ positive integers. If $m=3$, we either have $a=2$ or $a=1$. If $a=1$ we are already done, and if $a=2$ we get that $n=2$ so we get the solution $(a,n,m) = (2,2,3)$. Finally, when $m=1$ we get the solution $(a,n,m) = (a,n,1)$ for any $a,n$ positive integers.

Finally, we get that the solutions are $(a,n,m) = (2,2,3), (1,n,m), (a,n,1)$.
This post has been edited 1 time. Last edited by hydo2332, Apr 1, 2022, 3:03 AM
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awesomeming327.
1712 posts
#21
Y by
oof^tm

Note that $(a,1,n)$, $(1,m,n)$ and $(2,3,n),n\ge 2$ are solutions. We claim that these are the only solutions. Suppose $a,m>1.$ Then, $n=1$ clearly doesn't work so $n>1$ as well. Also remove case $(2,3,n).$ By Zsigmondy's there exists a prime factor of $a^m+1$ that doesn't divide $a+1$ so it doesn't divide $(a+1)^n$ as well. Thus, we're done.
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megarnie
5604 posts
#22
Y by
The only solutions are $\boxed{(1,m,n), (a,1,n), (2,3,n+1)}$, where $a$, $m$, and $n$ are any positive integers.

The first one works because the LHS is $2$, the RHS is $2^n$, and $2\mid 2^n$.

The second one works because the LHS is $a+1$, the RHS is $(a+1)^n$, and $a+1\mid (a+1)^n$.

The third one works because the LHS is $9$, the RHS is $3^{n+1}$. Since $n+1\ge 2$, $9\mid 3^n$.


It remains to show that those are the only solutions.

If $a=1$, we get the solution $(1,m,n)$.

If $m=1$, we get the solution $(a,1,n)$.

Now suppose $a>1$ and $m>1$.

If $a=2$ and $m=3$, then we have $9\mid 3^n$, so $n\ge 2$, which gives the solution $(2,3,n+1)$.

Otherwise, by Zsigmondy's, there exists a prime divisor of $a^m+1$ that doesn't divide $a+1$, contradiction.
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HamstPan38825
8859 posts
#23 • 2 Y
Y by awesomeming327., Ritwin
I suppose here's a more instructive solution that isn't some variant of "trivial by Zsigmondy's."

The answers are $(1, m, n)$, $(2, 3, n)$, and $(a, 1, n)$, which work. Henceforth assume that $a, m > 1$ and $a \neq 2$. It suffices to show that $a^m+1$ contains a prime factor that does not divide $a+1$ for all choices of $a$ and $m$.

Observe that it suffices to show the case for $m$ a prime number $p$, as $$\nu_p(a^m + 1) = \nu_p(a^{m/k} + 1) + \nu_p\left(\frac mk\right) > 0$$for any $p \mid a^{m/k} + 1$ by choosing $\frac mk$ prime. In particular, we may choose $p \nmid a+1$ because the hypothesis holds for $\frac mk$ prime.

Now let $m$ be a prime number $p_1$. For any $p \neq p_1$ that divides $a+1$, $$\nu_p(a^{p_1} + 1) = \nu_p(a+1) + \nu_p(p_1) = \nu_p(a+1)$$by LTE, and furthermore if $p_1$ also divides $a+1$, $\nu_{p_1}(a^{p_1} + 1) = 2$ in this case as well. Suppose that $m$ has no prime factors that do not divide $n$; then, $$a^m+1 \leq (a+1)^2$$with equality holding when $a = p_1$ is prime and $m=p_1$. This forces $m=1, 2$ (and also the weird case $m=3, a=2$), from where we can check the cases manually. Otherwise, there must exist some $p$ that divides $a^m+1$ but not $n$, so the hypothesis is true and we are done.
This post has been edited 1 time. Last edited by HamstPan38825, Jul 20, 2022, 4:02 PM
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cj13609517288
1906 posts
#24
Y by
Solved with a hint from Ritwin.

For this to be possible, for all $p$ such that $p|a^m+1$, it must also have $p|(a+1)^n\Rightarrow p|a+1$. However, that is impossible by a variation of Zsigmondy's that says that $a^n+b^n$ has at least one primitive prime divisor with exceptions $2^3+1^3$ and trivial cases. The trivial cases are when $a=1$ and when $m=1$. Therefore, our answers are $(2,3,n)$ where $n\ge2$, $(1,m,n)$ for positive integers $m,n$, and $(a,1,n)$ for positive integers $a,n$.
This post has been edited 1 time. Last edited by cj13609517288, Aug 13, 2022, 5:24 PM
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Msn05
39 posts
#25 • 2 Y
Y by Ibrahim_K, teomihai
The answer is $(a,1,n), (1,m,n)$ and $(2,3,n)$ for $n \geq 2$.
Case 1: $m$ is even.
Notice we want to get $\gcd(a^m+1,(a+1)^n)=a^m+1$ but we have at most $\gcd(a^m+1,(a+1)^n)=2$ (When $a$ is odd). So, $a=1$ since $a \geq 2$ is impossible.
From here we get $(a,m,n)=(1,m,n)$ (Also works for all $m$).
Case 2: $m$ is odd.
$m=1$ obviously works. So from here $(a,m,n)=(a,1,n)$.
Now assume $m>1$. We want to get $\gcd(a^m+1,(a+1)^n)=a^m+1$ but we have $\gcd(a^m+1,(a+1)^n)=\gcd(m(a+1),(a+1)^n)$. After proving by induction we have $a^m+1 \geq am+m$ for all $a \geq 2$ and $m\geq 3$ so we need to solve $m(a+1)=a^m+1$ and from here we get $m \mid a^m+1$. So, $a$ and $m$ are coprime and thus we have $a^m \equiv -1 \pmod{m}$, $a^{2m} \equiv 1 \pmod{m}$ and $a^{\varphi(m)} \equiv 1 \pmod{m}$. So
from Zsigmondy's Theorem we get $a=2,m=3$. And thus we have $(a,m,n)=(2,3,n)$ for $n \geq 2$.
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teomihai
2961 posts
#26
Y by
Msn05 wrote:
The answer is $(a,1,n), (1,m,n)$ and $(2,3,n)$ for $n \geq 2$.
Case 1: $m$ is even.
Notice we want to get $\gcd(a^m+1,(a+1)^n)=a^m+1$ but we have at most $\gcd(a^m+1,(a+1)^n)=2$ (When $a$ is odd). So, $a=1$ since $a \geq 2$ is impossible.
From here we get $(a,m,n)=(1,m,n)$ (Also works for all $m$).
Case 2: $m$ is odd.
$m=1$ obviously works. So from here $(a,m,n)=(a,1,n)$.
Now assume $m>1$. We want to get $\gcd(a^m+1,(a+1)^n)=a^m+1$ but we have $\gcd(a^m+1,(a+1)^n)=\gcd(m(a+1),(a+1)^n)$. After proving by induction we have $a^m+1 \geq am+m$ for all $a \geq 2$ and $m\geq 3$ so we need to solve $m(a+1)=a^m+1$ and from here we get $m \mid a^m+1$. So, $a$ and $m$ are coprime and thus we have $a^m \equiv -1 \pmod{m}$, $a^{2m} \equiv 1 \pmod{m}$ and $a^{\varphi(m)} \equiv 1 \pmod{m}$. So
from Zsigmondy's Theorem we get $a=2,m=3$. And thus we have $(a,m,n)=(2,3,n)$ for $n \geq 2$.

BEAUTIFUL TREATMENT OF CASES!
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math_comb01
662 posts
#27
Y by
Overkill
Solution
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hectorleo123
344 posts
#28
Y by
Lopes wrote:
Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$.
If $a=1\Rightarrow (a,m,n)=(1,m,n)$ is a solution
If $a>1:$
If $m=1 \Rightarrow (a,m,n)=(a,1,n)$ is a solution
If $m>1:$
By Zsigmondy's Theorem:
$\exists$ prime $p/ p|a^m+1, p\nmid a+1$
$\Rightarrow p|(a+1)^n \Rightarrow p|a+1 (\Rightarrow \Leftarrow)$
But there is an exception$(m=3, a=2)$:
$\Rightarrow 9|3^n \Rightarrow n>1$
$\Rightarrow (a,m,n)=(2,3,n), \forall n>1$ is a solution
$\Rightarrow (a,m,n)=(1,m,n),(a,1,n), ((2,3,n),\forall n>1)$ are all solutions
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AleMM
17 posts
#29
Y by
Hello again! :welcomeani:
I know this is a direct application of Zsigmondy's theorem (with base cases exception, of course), but I'd like to present other (instructive) solution.
Any flaws or misstypes that you find you can tell me, I'd like to know. Any kind of constructive feedback is welcomed!


Solution:

Let $a \sim b$ denote that $a$ has the same prime divisors as $b$ and vice versa, that's the same as saying that for a prime $p$, $p|a \iff p|b$.

Lemma: $\boxed{a^m+1 \sim a+1}$
First note that if $a^m+1 \sim a+1$, then writing $a+1=p_1^{\alpha_1} \dots p_k^{\alpha_k}$ and $a^m+1=p_1^{\beta_1} \dots p_k^{\beta_k}$ we just take $n \geq \dfrac{\beta_i}{\alpha_i} \implies \alpha_i n \geq \beta_i \implies a^m+1|(a+1)^n$ (because every prime power in $(a+1)^n$ that is $\alpha_i n$ would be greater than $\beta_i$ due to how we defined $n$), so if $(a, m)$ satisfy $a^m+1 \sim a+1$ it tells us there are infinitely many suffuciently large $n$ for which $a^m+1|(a+1)^n$. So in fact $(a, m)$ such that $a^m+1 \sim a+1$ give us solutions, we just need to prove that all $(a,m)$ solutions are of that form. Let $p$ be any prime divisor of $a^m+1 \implies p|a^m+1|(a+1)^n \implies p|(a+1)^n \implies p|a+1$, so all prime divisors of $a^m+1$ appear in $a+1$ as well. Suppose $m$ is even, write it as $m=2m_0$ and take a prime divisor $p$ of $a^{2m_0}+1 \implies p|a^{2m_0}+1|(a+1)^n \implies p|a+1 \implies a \equiv -1 \pmod{p} \implies a^{2m_0} \equiv (-1)^{2m_0} \equiv -1 \pmod{p}$, but $(-1)^{2m_0}=1 \implies 1 \equiv -1 \pmod{p} \implies p|1-(-1)=2 \implies p=2$. It means that every prime divisor of $a^{2m_0}+1$ is 2, so it's a power of 2 $\implies a^{2m_0}+1 = 2^k$ for some $k \geq 1$. If $k = 1 \implies a=1$ and in fact $\boxed{(1,m,n)}$ works for all positive integers $m,n$. Now, suppose $k \geq 2 \implies 4|2^k \implies 4|a^{2m_0}+1 \implies (a^{m_0})^2 \equiv -1 \pmod{4}$, but -1 isn't a quadratic residue modulo 4! So $m$ is odd $\implies a^m+1=(a+1)(a^{m-1}-a^{m-2}+ \dots -a+1) \implies a+1|a^m+1$, so every prime divisor of $a+1$ must be a prime divisor of $a^m+1$ as well $\square$

Suppose $m > 1$ because $\boxed{(a,1,n)}$ works for $a,n$ positive integers. As we have $a^m+1 \sim a+1$, that is $p|a^m+1 \iff p|a+1$, but $p|a+1 \implies p|a^m+1$, so we will only use that $p|a^m+1 \implies p|a+1$. Choose $p$ to be a prime divisor of $(a^{m-1}-a^{m-2}+ \dots -a+1)$, but because $a^m+1=(a+1)(a^{m-1}-a^{m-2}+ \dots -a+1) \implies p|a^m+1 \implies p|a+1 \implies a \equiv -1 \pmod{p}$ $\implies$ $a^{m-1}-a^{m-2}+ \dots -a+1 \equiv (-1)^{m-1}-(-1)^{m-2}+ \dots -(-1)+1 \equiv \underbrace{1+1+ \dots +1+1}_{m \text{ times}} \pmod{p}$ because $m$ is odd so $m-1$ is even and $m-2$ is odd... But $a^m+1 \equiv 0 \pmod{p} \implies m \equiv 0 \pmod{p}$ for all prime divisors $p$ of $(a^{m-1}-a^{m-2}+ \dots -a+1)$. Which means that $p|(a^{m-1}-a^{m-2}+ \dots -a+1) \implies p|m$. But notice that $p|(a^{m-1}-a^{m-2}+ \dots -a+1) \implies p|a+1$ because $a^m+1 = (a+1)(a^{m-1}-a^{m-2}+ \dots -a+1) \sim a+1$, so by the LTE lemma we have that $V_p(a^{m-1}-a^{m-2}+ \dots -a+1)=V_p(\dfrac{a^m+1}{a+1})=(V_p(a+1)+V_p(m))-V_p(a+1)=V_p(m)$ because $p|a+1$, so every prime divisor $p$ of $(a^{m-1}-a^{m-2}+ \dots -a+1)=\dfrac{a^m+1}{a+1}$ is in $m$ and $V_p(\dfrac{a^m+1}{a+1})=V_p(m)$, thus $\dfrac{a^m+1}{a+1}|m$.

Now we just do case work because $\dfrac{a^m+1}{a+1}|m \implies \dfrac{a^m+1}{a+1} \leq m$ and $\dfrac{a^m+1}{a+1}=(a^{m-1}-a^{m-2}+ \dots -a+1)$ is growing as $a$ grows, so $\dfrac{a^m+1}{a+1} \geq \dfrac{2^m+1}{3} \implies 3m-1 \geq 2^m$, but $2^m > 3m-1$ for $m \geq 4$ and $m$ is odd, so $m=3$ ($m=1$ already seen) $\implies n \geq 2$ and in fact $\boxed{(2,3,n)}$ works for all integers $n \geq 2$. If $a=3 \implies 4m-1 \geq 3^m$ but for $m \geq 2$ we have that $3^m > 4m-1$ and if $a \geq 4$ we get $m \leq 1$, finishing cases. :D
This post has been edited 2 times. Last edited by AleMM, Jul 11, 2023, 6:35 PM
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SatisfiedMagma
458 posts
#30
Y by
Yeah sure, I agree Zsigmondy is good. But you know what's better? Mihăilescu's Theorem and bit of divisibility! The thing is, I was stuck on the last diophantine, but then someone told me this overkill theorem...

Solution: The answers are
  • $(a,m,n) = (1,m,n)$;
  • $(a,m,n) = (a,1,n), (2,3,n)$ for any positive integer $n$.
We will begin by showing that even values of $m$ mostly don't work.

Claim: For even $m$, the only solution is $(a,m,n) = (1,m,n)$ for any $m,n$.

Proof: Let $p$ be some prime divisor of $a^m+1$. Trivially, we get that $p \mid a + 1$. For an even $m$, this would imply that $p \mid (-1)^2 + 1 = 2$. Altogether, one gets that $a+1$ is a power of $2$ and so must be $a^m + 1$.

Write $a = 2^{\alpha} - 1$ for $\alpha \in \mathbb{Z}^+$. Then we have that
\[a^m + 1 = (2^\alpha - 1)^m + 1 = 2^\beta\]with $\beta \in \mathbb{Z}^+$ and $\beta > \alpha$. Reducing the above equation modulo $2^\alpha$, we get $\alpha = 1 \implies a = 2$. Plugging everything back one can see the claim holds true. $\square$
The rest of the solution will deal with the harder case when $m$ is an odd integer. For this, write $a = p^k\cdot l - 1$ for some positive integer $l$ such that $\nu_p(a+1) = k$.

Plugging this into the original relation, one would get
\[(p^kl - 1)^m + 1 \mid (p^kl)^n.\]By Lifting the Exponent Lemma, we can check that $(p^kl - 1)^m + 1$ always divide $p^k$. Since $p \nmid l$, we must have that that
\[(p^kl - 1)^m + 1 \mid l^n.\]Invoking the Binomial Theorem on the LHS, we would get that
\[l(z) + p^k \mid l^{n-1}\]for some some integer $z$. If $l \ne 1$ or $n \ne 1$, then pick some prime divisor $t$ of $l$ such that $t \mid lz + p^k$. But it would be immediately a contradiction. This means either of $l$ or $n$ must be unity.

The case for $n = 1$, can be easily checked out by hand. The final non-trivial case is for $l = 1$. As $m = 1$ clearly works, assume $m > 1$ henceforth. and the rest is easy. For this see that there exists some positive integer $e$ such that
\[(p^k-1)^m + 1 = p^e.\]You can immediately kill this with Mihăilescu's Theorem and suddenly, the solution is complete! $\blacksquare$
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huashiliao2020
1292 posts
#31
Y by
Note that $(a,1,n)$, $(1,m,n)$ and $(2,3,n)\forall n\ge2$ are solutions. We claim that these are the only solutions. Suppose $a,m>1.$ Then, $n=1$ clearly doesn't work so $n>1$ as well. Also remove case $(2,3,n).$

By Zsigmondy a prime dividing a^m+1 doesn't divide a+1 so it doesn't divide (a+1)^n. $\blacksquare$
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KOMKZ
50 posts
#32 • 1 Y
Y by Adil1
I will prove with another way lets take for some $m$ we have $a^m+1$ have all prime divisors $a+1$ and dont have another divisors so $a^m+1$ and $(a+1)^m$ have similar divisors and its will kill the problem by binomial theorem
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Eka01
204 posts
#33
Y by
Storage
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quantam13
112 posts
#34
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Storage: Trivial by zsigmondy
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HasnatFarooq
33 posts
#35
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$(a,1,n),(1,m,n)$ are trivial solutions. Suppose, $a,m>1$ then by Zsigmondy theorem there exist a prime $p$ such that $p \nmid a+1$ with the exception being $(a,m)=(2,3)$ which is another solution.
This post has been edited 1 time. Last edited by HasnatFarooq, Feb 25, 2025, 7:49 AM
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ray66
35 posts
#36
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Suppose that $a,m$>1. Then there are no solutions because there exists a prime $p$ that divides $a^m+1$ but not $a+1$ by Zsigmondy. The exception is $(2,3)$, and plugging this in gives $(2,3,n)$ as a solution. $a=1$ is also a solution, and $m=1$ is another. So the only solutions are $(1,m,n)$, $(m,1,n)$, and $(2,3,n)$
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Maximilian113
575 posts
#37
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If $a>2, m > 1$ by Zsigmondy's there is no solution. If $a > 2, m=1$ clearly $n$ can take any value so $(a, 1, n)$ works. If $a=2,$ by Zsigmondy's $m=3$ and then $n \geq 2$ suffices. So $(2, 3, k)$ where $k \geq 2$ works. Finally if $a=1$ we clearly have $(1, m, n)$ so the solutions (which clearly work) are $(a, 1, n), (2, 3, k), (1, m, n)$ with $k \geq 2.$
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