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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
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
Thursday at 11:16 PM
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
Solution needed ASAP
UglyScientist   0
a minute ago
$ABC$ is acute triangle. $H$ is orthocenter, $M$ is midpoint, $L$ is the midpoint of smaller arc $BC$. Point $K$ is on $AH$ such that, $MK$ is perpendicular to $AL$. Prove that: $HMLK$ is paralelogram

0 replies
UglyScientist
a minute ago
0 replies
Product is a perfect square( very easy)
Nuran2010   0
2 minutes ago
Source: Azerbaijan Junior National Olympiad 2021 P1
At least how many numbers must be deleted from the product $1 \times 2 \times \dots \times 22 \times 23$ in order to make it a perfect square?
0 replies
Nuran2010
2 minutes ago
0 replies
Find the product
sqing   2
N 5 minutes ago by sqing
Source: Ecrin_eren
The roots of $ x^3 - 2x^2 - 11x + k=0 $ are $r_1, r_2,  r_3 $ and $ r_1+2 r_2+3 r_3= 0.$ Find the product of all possible values of $ k .$
2 replies
sqing
3 hours ago
sqing
5 minutes ago
Combinatorial Sum
P162008   1
N 23 minutes ago by Mathzeus1024
Evaluate $\sum_{n=0}^{\infty} \frac{2^n + 1}{(2n + 1) \binom{2n}{n}}$
1 reply
P162008
Apr 24, 2025
Mathzeus1024
23 minutes ago
2xy is perfect square and x^2 + y^2 is prime
parmenides51   3
N 39 minutes ago by justaguy_69
Source: Dutch NMO 2020 p4
Determine all pairs of integers $(x, y)$ such that $2xy$ is a perfect square and $x^2 + y^2$ is a prime number.
3 replies
parmenides51
Nov 23, 2020
justaguy_69
39 minutes ago
Mmo 9-10 graders P5
Bet667   10
N 39 minutes ago by User141208
Let $a,b,c,d$ be real numbers less than 2.Then prove that $\frac{a^3}{b^2+4}+\frac{b^3}{c^2+4}+\frac{c^3}{d^2+4}+\frac{d^3}{a^2+4}\le4$
10 replies
Bet667
Apr 3, 2025
User141208
39 minutes ago
IMO Genre Predictions
ohiorizzler1434   13
N 44 minutes ago by nunoarala
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
13 replies
ohiorizzler1434
Today at 6:51 AM
nunoarala
44 minutes ago
Number Theory Chain!
JetFire008   60
N an hour ago by whwlqkd
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
60 replies
JetFire008
Apr 7, 2025
whwlqkd
an hour ago
3 right-angled triangle area
NicoN9   1
N an hour ago by Mathzeus1024
Source: Japan Junior MO Preliminary 2020 P1
Right angled triangle $ABC$, and a square are drawn as shown below. Three numbers written below implies each of the area of shaded small right angled triangle. Find the value of $AB/AC$.

IMAGE
1 reply
NicoN9
Yesterday at 6:08 AM
Mathzeus1024
an hour ago
two sequences of positive integers and inequalities
rmtf1111   50
N an hour ago by math-olympiad-clown
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
50 replies
rmtf1111
Apr 10, 2019
math-olympiad-clown
an hour ago
inequalities
Tamako22   0
an hour ago
let $a,b,c> 1,\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1.$
prove that$$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \dfrac{2}{\sqrt{a}}+\dfrac{2}{\sqrt{b}}+\dfrac{2}{\sqrt{c}}$$
0 replies
Tamako22
an hour ago
0 replies
Problem 6
SlovEcience   2
N an hour ago by mashumaro
Given two points A and B on the unit circle. The tangents to the circle at A and B intersect at point P. Then:
\[ p = \frac{2ab}{a + b} \], \[ p, a, b \in \mathbb{C} \]
2 replies
SlovEcience
4 hours ago
mashumaro
an hour ago
A coincidence about triangles with common incenter
flower417477   3
N an hour ago by mashumaro
$\triangle ABC,\triangle ADE$ have the same incenter $I$.Prove that $BCDE$ is concyclic iff $BC,DE,AI$ is concurrent
3 replies
flower417477
Apr 30, 2025
mashumaro
an hour ago
this geo is scarier than the omega variant
AwesomeYRY   11
N an hour ago by LuminousWolverine
Source: TSTST 2021/6
Triangles $ABC$ and $DEF$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A,F,B,D,C,$ and $E$ occur in this order along $\Omega$. Let $\Delta_A$ be the triangle formed by lines $AB,AC,$ and $EF,$ and define triangles $\Delta_B, \Delta_C, \ldots, \Delta_F$ similarly. Furthermore, let $\Omega_A$ and $\omega_A$ be the circumcircle and incircle of triangle $\Delta_A$, respectively, and define circles $\Omega_B, \omega_B, \ldots, \Omega_F, \omega_F$ similarly.

(a) Prove that the two common external tangents to circles $\Omega_A$ and $\Omega_D$ and the two common external tangents to $\omega_A$ and $\omega_D$ are either concurrent or pairwise parallel.

(b) Suppose that these four lines meet at point $T_A$, and define points $T_B$ and $T_C$ similarly. Prove that points $T_A,T_B$, and $T_C$ are collinear.

Nikolai Beluhov
11 replies
AwesomeYRY
Dec 13, 2021
LuminousWolverine
an hour ago
IMO ShortList 1999, number theory problem 1
orl   62
N Apr 26, 2025 by Ilikeminecraft
Source: IMO ShortList 1999, number theory problem 1
Find all the pairs of positive integers $(x,p)$ such that p is a prime, $x \leq 2p$ and $x^{p-1}$ is a divisor of $ (p-1)^{x}+1$.
62 replies
orl
Nov 13, 2004
Ilikeminecraft
Apr 26, 2025
IMO ShortList 1999, number theory problem 1
G H J
Source: IMO ShortList 1999, number theory problem 1
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Cali.Math
128 posts
#51 • 1 Y
Y by cubres
We uploaded our solution https://calimath.org/pdf/IMO1999-4.pdf on youtube https://youtu.be/DtVQP4OXVCw.
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lelouchvigeo
181 posts
#52 • 1 Y
Y by cubres
If $x = 1$, then all $p$ work.
For $p=2$, then $x$ is $1$ or $2$. Assume now $p\ge 3$.
Let $x>1$, then let the samllest prime factor of x be q.
We have $(p-1)^{2x} \equiv 1 \pmod{q}$. This implies $ q-1 \mid 2x$. This forces $q=2.$
Now since $v_2(2x^{p-1}) > v_2(p-1)^x +1) $, the only case to be checked now is $x=p$
$p^{p-1} \mid (p-1)^p + 1 \implies p-1 \le \nu_p((p-1)^p + 1)   $ $ \implies p-1 \le 2 \implies p\le3$
Now checking for $p=3$. We get a solution as $(3,3)$
Therefore the solutions are $\boxed{(1, p), (2, 2), (3, 3)}$
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AlanLG
241 posts
#53 • 1 Y
Y by cubres
:coolspeak:
Solution without the bound of $x$
If $p=2$, then $x=2$, so for now assume $p\geq 3$
If $x=1$ all $p$ primes work, so also assume $x\geq 2$

Let $q$ be the smallest prime dividing $x$ then $(p-1)^{2x}\equiv 1\pmod q$ and as $(p-1)^{q-1}\equiv 1\pmod q$ then $\operatorname{Ord}_q(p-1)\mid 2\gcd\left(x,\frac{p-1}{2}\right)=2$ so $(p-1)^2\equiv 1\pmod q$ so $p=q$ or $p\equiv 2\pmod q$, the latter gives $(p-1)^x+1\equiv 2\pmod q$ impossible. If $p=q$ then, by Lifting the Exponent $$\nu_p((p-1)^x+1)=\nu_p(p)+v_p(x)\geq (p-1)\nu_p(x)$$so $1\geq (p-2)v_p(x)\geq (p-2)\cdot 1$ so $p=3$
which gives to find all integers $x$ such that $x^2\mid 2^x+1$ which by IMO 1990/3 only $x=3$ works so the solutions are $\boxed{(x,p)=(1,p),(2,2), (3,3)}$
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joshualiu315
2533 posts
#54 • 1 Y
Y by cubres
I was given no condition bounding $x$:


The solution set is $(x,p) \in \{(1,p), (2,2), (3,3)\}$.

Note that if $x=1$, all $p$ work. Also, if $p=2$, we must have $x \mid 2$, which gives the unique solution $(x,p)=(2,2)$. Assume $x>1$ and $p>2$ for the remainder of this solution.

Let $q$ be the smallest prime dividing $x$. We have

\[\operatorname{ord}_q(p-1) \mid \gcd(2x,q-1) = 2.\]
Hence,

\[(p-1)^2 \equiv 1 \pmod{q} \iff p(p-2) \equiv 0 \pmod{q}.\]
Hence, $p \equiv 0,2 \pmod{q}$. The latter gives

\[(p-1)^x+1 \equiv 2 \pmod{q},\]
implying that $q=2$. However, $(p-1)^x+1$ is odd, so this yields a contradiction.

Otherwise, we have $p \equiv 0 \pmod{q} \iff p=q$, so LTE gives

\[\nu_p((p-1)^x+1) = \nu_p(p)+\nu_p(x) \ge \nu_p(x^{p-1}) = (p-1) \nu_p(x)\]\[\implies 1+\nu_p(x) \ge (p-1) \nu_p(x) \iff (p-2) \nu_p(x) \le 1.\]
Since $\nu_p(x) \ge 1$, we must have $p=3$.

The problem is now reduced to finding all $x$ such that $x^2 \mid 2^x+1$.


Lemma The only value $x$ satisfying this is $x=3$.

Proof: Consider IMO 1990/3. $\square$


Hence, the unique solution here is $(x,p)=(3,3)$. This completes our solution set.
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shendrew7
794 posts
#55 • 1 Y
Y by cubres
The condition $x \leq 2p$ turns out to be unnecessary. Our solutions are $\boxed{(1,p),(2,2),(3,3)}$. We proceed first by tackling edge cases:
  • If $x=1$, $p$ can be any prime.
  • If $p=2$, $x$ is either 1 or 2.
  • Otherwise, $p$ is odd, forcing $x$ to be odd. Then we require
    \[(p-1) \cdot v_p(x) \leq v_p((p-1)^x+1) = v_p(x) + 1 \implies (p-2) \cdot v_p(x) \leq 1.\]If $p=3$, IMO 1990/3 tells us $x$ is either 1 or 3.
  • Otherwise, $\gcd(x,p)=1$. Suppose the least prime divisor of $x$ is $q>2$. Then
    \[\operatorname{ord}_q(p-1) \mid 2n, p-1 \text{ but not } n \implies \operatorname{ord}_q(p-1) = 2 \implies p=q,\]contradiction. $\blacksquare$
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SenorSloth
37 posts
#56 • 1 Y
Y by cubres
We claim that the only such pairs are $(2,2)$, $(3,3)$, and $(1,p)$ for any prime $p$.
(This proof assumes there is no bound on $x$.)

We start with the trivial case when $p=2$. We end up with $x\mid 2$, which clearly gives only $x=1$ and $x=2$ as solutions.

For odd primes, we know that $(p-1)^x+1$ is odd, and thus $x$ must be odd. For any $p$, $x=1$ clearly works since $1$ divides everything. Assuming $x\neq 1$, we let $q$ be the smallest prime divisor of $x$. We know that the order of $p-1\pmod{q}$ divides $\gcd(q-1,2x)$. Since $x$ only has prime factors at least $q$, it is clear $x$ and $q-1$ are relatively prime and thus $\gcd(q-1,2x)$ must be exactly $2$. The order cannot be $1$, because in that case $q\mid (p-1)^x-1$, not $(p-1)^x+1$, so the order must be $2$. This implies that $p-1\equiv -1\pmod{q}$, so $p\equiv 0\pmod{q}$ and thus $p=q$.

Now we use LTE to rule out all cases except $p=3$. We know that $\nu_p((p-1)-(-1))=1$, so $\nu_p((p-1)^x-(-1)^x)=1+\nu_p(x)$. We also have that $\nu_p(x^{p-1})=(p-1)\nu_p(x)$. Note that since $p=q$, we have $p\mid x$. This means that for $p>3$, $(p-1)\nu_p(x)>1+\nu_p(x)$, so there are no solutions. For $p=3$, if $\nu_3(x)=1$ then they have equal values, so this case has to be resolved separately.

We can check to see that $(3,3)$ is a solution. Now we have to rule out other solutions. Our previous bound showed that $\nu_3(x)=1$ was the only one possible. Let $r$ be the minimum other prime that is a factor of $x$. Then the order of $2\pmod{r}$ divides $\gcd(r-1,2x)$. Similar to before, other than a factor each of $2$ and $3$, $2x$ only contains prime factors greater than $r$, and thus the order must divide $6$. Furthermore, the order being $1$ or $3$ won't work, so it must be $2$ or $6$. However, $3$ is the only working prime for order $2$, while no primes exist for order $6$ since $2^6-1=63$ but $7$ has order $3$ since $2^3-1=7$. Thus, there can be no other prime factors, and we are done.
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de-Kirschbaum
198 posts
#57 • 1 Y
Y by cubres
Note that when $x=1$ every prime works so we will only consider the cases where $x \neq 1$. Now note that for any prime factor $q \mid x$ we have $(p-1)^x \equiv -1 \mod{q}, (p-1)^{2x} \equiv 1 \mod{q}$ which means $\delta_{q}(p-1)=\gcd(2x, q-1)$ which is either $2$ or $1$ since $q-1$ is corpime with $x$. If it is $1$, then $(p-1)^x \equiv 1 \equiv -1 \mod{q} \implies q=2 \implies p=2$ in which case we have $(p-1)^x+1=2$ which is divisible only by $1,2$ so $(x,p)=(2,2)$ is the new solution.

If $\delta_q(p-1)=2$ then $p-1 \equiv -1 \mod{q} \implies p \equiv 0 \mod{q} \implies p=q$. Then that means $q \leq x \leq 2q$ but if $x=2q$ then that means $(p-1)^x+1$ has a prime factor where the order is 1, but we dealt with that in the previous case already. Thus $x=q=p$. Then we have $p^{p-1} \mid (p-1)^p+1$. By LTE we get that $\nu_p((p-1)^p+1)=\nu_p(p)+\nu_p(p)=2$. Thus the division only holds when $p=2, 3$, but we dealt with the case when $p=2$ already so we check the solutions for $p=3$ which gives us a new solution $(x,p)=(3,3)$, and if $p \geq 5$ clearly there are no solutions as the division no longer holds.

Thus the solutions are $(x,p)=(1,p),(2,2),(3,3)$.
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BestAOPS
707 posts
#58 • 1 Y
Y by cubres
We claim the solutions are $(1,p)$ for all $p$, $(2,2)$, and $(3,3)$. They can be checked to work. Furthermore, if $p=2$, then $x$ must be a divisor of $1^x + 1 = 2$, so $(1,2)$ and $(2,2)$ are the only solutions. Thus, from now on, we can assume $x > 1$ and $p \geq 3$.

We know $x$ has a least prime factor; let that be $q$. Then, we have
\begin{align*}
    (p-1)^x \equiv -1 \pmod{q} \\
    (p-1)^{2x} \equiv 1 \pmod{q}.
\end{align*}Looking at the order of $(p-1)^2$, it must divide the GCD of $q-1$ and $x$. But $q$ being the least prime factor implies that GCD is $1$, so $(p-1)^2 \equiv 1 \pmod{q}$. This means $q \mid p(p-2)$.

However, $q \mid p-2$ is impossible. To see why, notice that $(p-1)^x + 1 \equiv 1 + 1 \pmod{p-2}$, so $(p-1)^x + 1 \equiv 2 \pmod{q}$. Since $q$ is odd, this means that $(p-1)^x + 1$ is not divisible by $q$, contradicting the fact that $x^{p-1}$ divides $(p-1)^x + 1$.

Thus, we must have $q = p$. Now, notice that since $x$ must be odd, we can use lifting the exponent:
\[ \nu_p((p-1)^x + 1) = 1 + \nu_p(x) \geq \nu_p(x^{p-1}) = (p-1)\nu_p(x). \]This implies
\[ \frac{1}{p-2} \geq \nu_p(x) \geq 1, \]so $p \leq 3$.

We only need to consider the case where $p = 3$. However, the result of IMO 1990/3 tells us that $(1,3)$ and $(3,3)$ are the only solutions in this case, so we are done.
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ezpotd
1262 posts
#59 • 1 Y
Y by cubres
The answer is only $(3,3), (2,2)$ and $(1,p)$ for all primes $p$. It is easy to verify these work.

Main idea is to consider the lowest prime divisor of $x$ (ignore $x =1$ as it always works, ignore $p < 4$ as it can be done manually), let it be $q$. Then take $\mathrm{ord}_q p - 1 \mid  q - 1, \mathrm{ord}_q p - 1 \mid 2x$, so $\mathrm{ord}_q p -1 = 1,2$. In the first case, we have $q = p - 2$, so for $p > 3$ we can check that $x = q$ by size, so we can eliminate this case by just showing $q^{p - 1} > p^{q } + 1$, by mod $p$ they are clearly never equal so we can just show $q^{p - 1} > p^{q }$ which forces no sol, since $(q + 1)^{p- 1} = (p - 1)^{q + 1}$, it suffices to show that $(\frac{q}{q + 1})^{p - 1} \ge \frac{1}{p - 1}$, rewrite as $(1 - \frac{1}{c})^{c} \ge \frac 1c$, which is obvious for $c > 4$ since the left is increasing in $c$ and the right is decreasing , in the latter we have $q = p$, so we can use LTE, check $p = 2$ manually then LTE on odd case gives $\nu_p$ on the right is $2$, on left is $p - 1$, so we must have $p \le 3$, giving the only solution as $p = 3$.
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pie854
243 posts
#60 • 1 Y
Y by cubres
Note that $(1,p)$ works for any prime $p$ and $(2,2)$ also works for any $x$. So let us assume $x>1$ and $p>2$.

We have $x^{p-1}\mid (p-1)^{2x}-1$. Let $q$ be a prime such that $q\mid x$. Let $\alpha=\text{ord}_q({p-1})\leq q-1$ then $q\mid (p-1)^\alpha-1$ and $\alpha\mid 2x$. So by LTE \begin{align*} v_q\left((p-1)^{2x}-1\right)\geq v_q(x^{p-1}) & \Rightarrow v_q\left((p-1)^\alpha-1\right)+v_q(2x/\alpha) \geq (p-1)v_q(x) \\ & \Rightarrow v_q\left((p-1)^\alpha-1\right)\geq (p-2)v_q(x) \\ & \Rightarrow q^{p-2} \mid (p-1)^\alpha-1 \\ & \Rightarrow q^{p-2} <(p-1)^\alpha\leq (p-1)^{q-1}.\end{align*}From this it's not very hard to show that $q\geq p$. So if $x=qk\leq 2p$ then either $k=2,q=p$ or $k=1$. If $x=2p$ then by mod 2 we get a contradiction. If $x=q$ then by FLT $$0\equiv (p-1)^q+1 \equiv (p-1)+1 \equiv p \pmod q,$$so $q=p$. Using LTE again $$p-1\leq v_p\left((p-1)^p+1\right)=v_p(p-1+1)+v_p(p)=2,$$so $p=3$. Checking we find that the other working pair is $(3,3)$.
This post has been edited 3 times. Last edited by pie854, Sep 28, 2024, 1:20 PM
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alexanderhamilton124
389 posts
#61 • 2 Y
Y by ubermensch, cubres
$x = 1$ and any prime work, and if $p = 2$, then $x = 2$, so assume $x \neq 1$, and $p$ is odd. Clearly, $(p - 1)^x + 1$ is odd, so $x$ must be odd as well. Consider the smallest prime divisor, $q$ (which is odd), of $x$, then $q \mid (p - 1)^x + 1 \implies (p - 1)^{2x} \equiv 1\mod{q} \implies \text{ord}_q(p - 1) \mid \gcd(2x, q - 1) = 2$. Note that $\text{ord}_q(p - 1) \neq 1 \implies \text{ord}_q(p - 1) = 2 \implies q \mid p - 1 + 1 = p \implies q = p$. So, $p \mid x$.

We have $(p - 1)v_p(x) \leq v_p((p - 1)^x + 1) = 1 + v_p(x)$, and since $v_p(x) = 1$, $p - 1 \leq 2 \implies p = 3$. Just checking gives only $x = 3$ works, so we are done.
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eibc
600 posts
#63 • 1 Y
Y by cubres
The only answers are $(x, p) = (2, 2), (3, 3)$, and $(1, p)$ for any prime $p$. It's easy to verify all of these solutions work, so now we show they are the only ones.

When $p = 2$, we have $x \mid 2$, so $x = 1$, or $x = 2$.

When $p = 3$, by IMO 1990/3, the only answers are $(x, p) = (1, 3)$ and $(3, 3)$.

When $p > 3$, if $x = 1$ then evidently $x^{p-1} \mid (p-1)^{x}+1$. If $x > 1$, suppose $q$ is the smallest prime factor of $x$. Note that $(p - 1)^x + 1$ is odd, so $q > 2$. Since $(p - 1)^x \equiv -1 \pmod p$, we find that $(p - 1)^{2x} \equiv 1 \pmod q$ so $\text{ord}_q(p - 1) \mid 2x$. Because $\text{ord}_q(p - 1) \le q - 1$, we must have $\text{ord}_q(p - 1) \in \{1, 2\}$.

If $\text{ord}_q(p - 1) = 1$ then $p - 1 \equiv 1 \pmod q$. Thus $0 \equiv (p - 1)^x + 1 \equiv 2 \pmod q$, so $q = 2$, which is impossible.

If $\text{ord}_q(p - 1) = 2$ then $p - 1 \equiv -1 \pmod q$, or $p \equiv 0 \pmod q$, so $p = q$. Then from LTE, we find that
$$(p - 1)\nu_p(x) = \nu_p(x^{p - 1}) \le \nu_p((p - 1)^x + 1) = \nu_p(p) + \nu_p(x) = 1 + \nu_p(x).$$However, because $p > 3$ and $\nu_p(x) > 1$, this is impossible. So, we are done.
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smileapple
1010 posts
#64 • 1 Y
Y by cubres
If $p\ge3$, note that $x$ must be odd. Then $\nu_p(x^{p-1})=(p-1)\nu_p(x)$ and $\nu_p((p-1)^x+1)=\nu_p(p)+\nu_p(x)=\nu_p(x)+1$. Hence $(p-2)\nu_p(x)\le1$, so that either $p=3$ or $\nu_p(x)=0$. In the first case, by 90IMO3 the only solutions are $(x,p)=(1,3)$ and $(x,p)=(3,3)$.

Otherwise, Let $q$ be the minimal prime divisor of $x$. We have $(p-1)^{2x}\equiv1\pmod q$ and $(p-1)^{(q-1)}\equiv1\pmod q$, so that $(p-1)^2\equiv(p-1)^{\gcd(2x,q-1)}\equiv1\pmod q$, so that $p\equiv2\pmod q$. But then $(p-1)^x+1\equiv2\pmod q$ and $x^{p-1}\equiv0\pmod q$, which is a contradiction as $p$ is odd. Thus $q$ cannot exist, so that $x=1$.

If $p=2$ then $x=1$ or $x=2$ both work.

Our solution set is thus $\boxed{\{(2,2),(3,3)\}\cup\{(1,p)\mid p\in\mathbb{P}\}}$.
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cursed_tangent1434
612 posts
#65
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We claim that the only pairs of solutions $(x,p)$ are $(1,p)$ for any prime $p$, $(2,2)$ and $(3,3)$. It is easy to check that all of these indeed work. We now show that they are the only ones.

First note that, if $p=2$ the resulting condition is $x \mid 2$ which is only possible if $x=1$ or $x=2$. If $p$ is odd, the right hand side is clearly odd as well, so $x$ must be odd. Further, if $x=1$ then it is clear that $p$ can be an arbitrary prime. Thus, in what follows we assume that $x>1$ is even and that $p$ is an odd prime. We first show the following claim.

Claim : For all pairs of solutions $(x,p)$ we must have $p \mid x$.

Proof : Let $q$ denote the smallest prime divisor of $x$. Then,
\[q\mid x^{p-1} \mid (p-1)^x +1 \]implies $(p-1)^{2x} \equiv 1 \pmod{q}$. Thus, $\text{ord}_{q}(p-1) \mid 2x$. Also, $\text{ord}_{q}(p-1) \mid q-1$. But then, if there exists an odd prime divisor $r \mid \text{ord}_q(p-1)$ then $r \mid 2x$ so $r\mid x$ and also $r \mid q-1$. Thus, $r \le q-1 <q$ which contradicts the minimality of $q$. Thus, $\text{ord}_{q}(p-1)$ is a perfect power of two. However, since $x$ is odd $\nu_2(2x)=1$ so $\text{ord}_{q}(p-1)=2$. This means, $(p-1)^2 \equiv 1 \pmod{q}$ and thus,
\[0 \equiv (p-1)^x +1 \equiv (p-1)+1 \equiv p \pmod{q}\]which is possible if and only if $p=q$ as desired.

Now note that by Lifting the Exponent Lemma we have,
\[(p-1)\nu_p(x) = \nu_p(x^{p-1}) \le \nu_p((p-1)^x+1) =\nu_p(p)+\nu_p(x)=\nu_p(x)+1 \]However, if $p>3$,
\[(p-1)\nu_p(x) > 2\nu_p(x) \ge \nu_p(x)+1\]which is a clear contradiction. Thus, we must have $p=3$ which reduces the desired divisibility to
\[x^2 \mid 2^x +1\]whose solutions are known to be $x=1$ and $x=3$ by IMO 1990/3 which finishes the proof.
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Ilikeminecraft
612 posts
#66
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Clearly, $x = 1$ is a solution to all $p.$ For $p = 2,$ the solution is $x = 1, 2,$ and for $p = 3,$ we have IMO 1990 P3, which gives us the solution $x = 3.$ Thus, we will ignore $x = 1$ and $p = 2, 3$ case.

Let $q$ be the smallest prime dividing $x.$ Thus, $(p - 1)^{2x}\equiv1\pmod q,$ while $(p - 1)^{q - 1} \equiv 1\pmod q.$ We know that it isn't 0 because otherwise, $q\not\mid (p - 1)^x + 1.$ Thus, we have that $\operatorname{ord}_q(p - 1) \mid \gcd(2x, q - 1) = 2.$ Thus, the order is 2, and hence $p^2 - 2p + 1 \equiv 1 \pmod q.$ Thus, either $p - 2 \equiv 0 \pmod q,$ or $p = q.$

We will handle the case where $p = q$ first. We have that $$\nu_p((p - 1)^x + 1) = \nu_p(p) + \nu_p(x)$$However, since $p - 1 \leq \nu_x((p - 1)^x + 1) \leq \nu_p((p - 1)^x + 1),$ we have that $p^{p - 2}\mid x.$ Thus, we are able to construct an infinite descent, which shows that this case has no solution.

Now, we have $p - 2\equiv 0\pmod q. $ However, by taking modulo $p - 2,$ we see that $(p - 1)^x + 1 \equiv 2 \pmod{p - 2}.$ Thus, $q = 2\implies p\equiv0\pmod2, p \geq 4$ which is a contradiction to the fact that $p$ is prime.

Thus, the answer is $\boxed{(1, p), (2, 2), (3, 3)}$ where $p$ is a prime.
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