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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
d(2025^{a_i}-1) divides a_{n+1}
navi_09220114   2
N 23 minutes ago by mickeymouse7133
Source: TASIMO 2025 Day 2 Problem 5
Let $a_n$ be a strictly increasing sequence of positive integers such that for all positive integers $n\ge 1$
\[d(2025^{a_n}-1)|a_{n+1}.\]Show that for any positive real number $c$ there is a positive integers $N_c$ such that $a_n>n^c$ for all $n\geq N_c$.

Note. Here $d(m)$ denotes the number of positive divisors of the positive integer $m$.
2 replies
navi_09220114
Monday at 11:51 AM
mickeymouse7133
23 minutes ago
Funky function
TheUltimate123   22
N 42 minutes ago by jasperE3
Source: CJMO 2022/5 (https://aops.com/community/c594864h2791269p24548889)
Find all functions \(f:\mathbb R\to\mathbb R\) such that for all real numbers \(x\) and \(y\), \[f(f(xy)+y)=(x+1)f(y).\]
Proposed by novus677
22 replies
TheUltimate123
Mar 20, 2022
jasperE3
42 minutes ago
R+ FE f(f(xy)+y)=(x+1)f(y)
jasperE3   0
43 minutes ago
Source: p24734470
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that for all positive real numbers $x$ and $y$:
$$f(f(xy)+y)=(x+1)f(y).$$edit oops sorry I misinterpreted that the original poster had solved it, solvable tho
0 replies
jasperE3
43 minutes ago
0 replies
Inequality with x+y+z=1.
FrancoGiosefAG   1
N an hour ago by Blackbeam999
Let $x,y,z$ be positive real numbers such that $x+y+z=1$. Show that
\[ \frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leq 0. \]
1 reply
FrancoGiosefAG
4 hours ago
Blackbeam999
an hour ago
Find all numbers
Rushil   10
N an hour ago by SomeonecoolLovesMaths
Source: Indian RMO 1994 Problem 3
Find all 6-digit numbers $a_1a_2a_3a_4a_5a_6$ formed by using the digits $1,2,3,4,5,6$ once each such that the number $a_1a_2a_2\ldots a_k$ is divisible by $k$ for $1 \leq k \leq 6$.
10 replies
Rushil
Oct 25, 2005
SomeonecoolLovesMaths
an hour ago
Some number theory
EeEeRUT   3
N an hour ago by MathLuis
Source: Thailand MO 2025 P9
Let $p$ be an odd prime and $S = \{1,2,3,\dots, p\}$
Assume that $U: S \rightarrow S$ is a bijection and $B$ is an integer such that $$B\cdot U(U(a)) - a \: \text{ is a multiple of} \: p \: \text{for all} \: a \in S$$Show that $B^{\frac{p-1}{2}} -1$ is a multiple of $p$.
3 replies
EeEeRUT
May 14, 2025
MathLuis
an hour ago
Gcd
Rushil   5
N 2 hours ago by SomeonecoolLovesMaths
Source: Indian RMO 1994 problem 5
Let $A$ be a set of $16$ positive integers with the property that the product of any two distinct members of $A$ will not exceed 1994. Show that there are numbers $a$ and $b$ in the set $A$ such that the gcd of $a$ and $b$ is greater than 1.
5 replies
Rushil
Oct 25, 2005
SomeonecoolLovesMaths
2 hours ago
Solve the system
Rushil   20
N 2 hours ago by SomeonecoolLovesMaths
Source: 0
Solve the system of equations for real $x$ and $y$: \begin{eqnarray*} 5x \left( 1 + \frac{1}{x^2 + y^2}\right) &=& 12 \\ 5y \left( 1 - \frac{1}{x^2+y^2} \right) &=& 4 . \end{eqnarray*}
20 replies
Rushil
Oct 25, 2005
SomeonecoolLovesMaths
2 hours ago
Angles made with the median
BBNoDollar   1
N 2 hours ago by Ianis
Determine the measures of the angles of triangle \(ABC\), knowing that the median \(BM\) makes an angle of \(30^\circ\) with side \(AB\) and an angle of \(15^\circ\) with side \(BC\).
1 reply
BBNoDollar
4 hours ago
Ianis
2 hours ago
Find all rationals s.t..
Rushil   12
N 2 hours ago by SomeonecoolLovesMaths
Source: Indian RMO 1994 Problem 7
Find the number of rationals $\frac{m}{n}$ such that

(i) $0 < \frac{m}{n} < 1$;

(ii) $m$ and $n$ are relatively prime;

(iii) $mn = 25!$.
12 replies
Rushil
Oct 25, 2005
SomeonecoolLovesMaths
2 hours ago
An inequality
Rushil   11
N 2 hours ago by SomeonecoolLovesMaths
Source: Indian RMO 1994 Problem 8
If $a,b,c$ are positive real numbers such that $a+b+c = 1$, prove that \[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]
11 replies
Rushil
Oct 25, 2005
SomeonecoolLovesMaths
2 hours ago
Minimum moves to reach 25
lkason   0
3 hours ago
Source: Final of the XXI Polish Championship in Mathematical and Logical Games
Mateusz plays a game of erasing-writing on a large board. The board is initially empty.

In each move, he can either:
-- Write two numbers equal to $1$ on the board.
-- Erase two numbers equal to $n$ and write instead the numbers $n-1$ and $n+1$.

What is the minimal number of moves Mateusz needs to make for the number 25 to appear on the board?

Note: Numbers on the board retain their values; their digits cannot be combined or split.

Spoiler, answer:
Click to reveal hidden text
0 replies
lkason
3 hours ago
0 replies
Reals to reals FE
a_507_bc   7
N 3 hours ago by jasperE3
Source: IMOC 2023 A2
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$, such that $$f(f(x)+y)(x-f(y)) = f(x)^2-f(y^2).$$
7 replies
a_507_bc
Sep 9, 2023
jasperE3
3 hours ago
Generic Real-valued FE
lucas3617   3
N 4 hours ago by jasperE3
$f: \mathbb{R} -> \mathbb{R}$, find all functions where $f(2x+f(2y-x))+f(-x)+f(y)=2f(x)+f(y-2x)+f(2y)$ for all $x$,$y \in \mathbb{R}$
3 replies
lucas3617
Apr 25, 2025
jasperE3
4 hours 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
183 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
2534 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
799 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
201 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
1279 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
400 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
636 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
658 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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