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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
50 points in plane
pohoatza   13
N 3 minutes ago by cursed_tangent1434
Source: JBMO 2007, Bulgaria, problem 3
Given are $50$ points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least $130$ scalene triangles with vertices of that color.
13 replies
pohoatza
Jun 28, 2007
cursed_tangent1434
3 minutes ago
D1024 : Can you do that?
Dattier   5
N 4 minutes ago by SimplisticFormulas
Source: les dattes à Dattier
Let $x_{n+1}=x_n^2+1$ and $x_0=1$.

Can you calculate $\sum\limits_{i=1}^{2^{2025}} x_i \mod 10^{30}$?
5 replies
Dattier
Apr 29, 2025
SimplisticFormulas
4 minutes ago
Hardest N7 in history
OronSH   25
N 6 minutes ago by sansgankrsngupta
Source: ISL 2023 N7
Let $a,b,c,d$ be positive integers satisfying \[\frac{ab}{a+b}+\frac{cd}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}.\]Determine all possible values of $a+b+c+d$.
25 replies
+1 w
OronSH
Jul 17, 2024
sansgankrsngupta
6 minutes ago
Friends Status are changing
lminsl   64
N an hour ago by SteppenWolfMath
Source: IMO 2019 Problem 3
A social network has $2019$ users, some pairs of whom are friends. Whenever user $A$ is friends with user $B$, user $B$ is also friends with user $A$. Events of the following kind may happen repeatedly, one at a time:
[list]
[*] Three users $A$, $B$, and $C$ such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends, change their friendship statuses such that $B$ and $C$ are now friends, but $A$ is no longer friends with $B$, and no longer friends with $C$. All other friendship statuses are unchanged.
[/list]
Initially, $1010$ users have $1009$ friends each, and $1009$ users have $1010$ friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.

Proposed by Adrian Beker, Croatia
64 replies
lminsl
Jul 16, 2019
SteppenWolfMath
an hour ago
hard problem
Cobedangiu   2
N an hour ago by Cobedangiu
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
2 replies
Cobedangiu
Yesterday at 4:24 PM
Cobedangiu
an hour ago
well-known NT
Tuleuchina   9
N 2 hours ago by Blackbeam999
Source: Kazakhstan mo 2019, P6, grade 9
Find all integer triples $(a,b,c)$ and natural $k$ such that $a^2+b^2+c^2=3k(ab+bc+ac)$
9 replies
Tuleuchina
Mar 20, 2019
Blackbeam999
2 hours ago
Inequality involving square root cube root and 8th root
bamboozled   0
2 hours ago
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the minimum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
0 replies
bamboozled
2 hours ago
0 replies
Old problem
kwin   2
N 2 hours ago by kwin
Let $a, b, c \ge 0$ and $ ab+bc+ca>0$. Prove that:
$$ \frac{1}{(a+b)^2} + \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2} + \frac{15}{(a+b+c)^2} \ge \frac{6}{ab+bc+ca}$$Is there any generalizations?
2 replies
kwin
Sunday at 1:12 PM
kwin
2 hours ago
functional equation
henderson   4
N 2 hours ago by megarnie
Source: unknown
Find all functions $f :\mathbb{R^+}\to\mathbb{R^+}$, satisfying the condition

$f(1+xf(y))=yf(x+y)$

for any positive reals $x$ and $y$.
4 replies
henderson
Oct 8, 2015
megarnie
2 hours ago
Parallelograms and concyclicity
Lukaluce   31
N 2 hours ago by Ihatecombin
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
31 replies
Lukaluce
Apr 14, 2025
Ihatecombin
2 hours ago
My Unsolved FE in R+
ZeltaQN2008   2
N 2 hours 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
3 hours ago
megarnie
2 hours ago
Something nice
KhuongTrang   32
N 3 hours 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
3 hours ago
Infimum of decreasing sequence b_n/n^2
a1267ab   35
N 3 hours 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
a1267ab
Dec 16, 2019
shendrew7
3 hours ago
IMO Genre Predictions
ohiorizzler1434   52
N 3 hours 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
3 hours ago
Classical factorial number theory
Orestis_Lignos   20
N Mar 24, 2025 by anudeep
Source: JBMO 2023 Problem 1
Find all pairs $(a,b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$.

Nikola Velov, North Macedonia
20 replies
Orestis_Lignos
Jun 26, 2023
anudeep
Mar 24, 2025
Classical factorial number theory
G H J
G H BBookmark kLocked kLocked NReply
Source: JBMO 2023 Problem 1
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Orestis_Lignos
558 posts
#1 • 4 Y
Y by GeoKing, Sedro, ItsBesi, lian_the_noob12
Find all pairs $(a,b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$.

Nikola Velov, North Macedonia
This post has been edited 2 times. Last edited by Orestis_Lignos, Jun 26, 2023, 4:02 PM
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Orestis_Lignos
558 posts
#2 • 4 Y
Y by GeoKing, ItsBesi, FeridMath, Yassperx
WLOG assume that $a \geq b$. We split into two cases.

Case 1: $a=b$. Then, $a!+a=5^k$ with $k \geq 1$. If $k=1$ then obiously $(a,b)=(5,5)$ works. If $k \geq 2$, then $(a-1)!+1 \equiv 1 \pmod 5$, and since $5^k=a!+a=a((a-1)!+1),$ we must have $(a-1)!+1=1,$ a contradiction.

Case 2: $a>b$. Then, $a=b+s$ with $s \geq 1$. Let $a!+b=5^k$ with $k \geq 1$. Therefore,

$b((b-1)! \cdot (b+1) \cdots (b+s)+1)=5^k$.

Thus, $b=5^\ell$ with $\ell \geq 0$, and if $b-1 \geq 5$ then $(b-1)! \cdot (b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction. Moreover, if $s \geq 5$, then $5 \mid (b+1) \cdots (b+5) \mid (b+1) \cdots (b+s),$ and so $(b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction.

Therefore, $b \leq 5$ and $s \leq 4$.

Taking all cases, we obtain $(b,s)=(1,3),$ that is $(a,b)=(4,1)$.

To sum up, we obtain the solutions $(a,b)=(5,5), (4,1)$ and $(1,4)$.
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ismayilzadei1387
219 posts
#3 • 3 Y
Y by Aoxz, FriIzi, Nuran2010
Orestis_Lignos wrote:
WLOG assume that $a \geq b$. We split into two cases.

Case 1: $a=b$. Then, $a!+a=5^k$ with $k \geq 1$. If $k=1$ then obiously $(a,b)=(5,5)$ works. If $k \geq 2$, then $(a-1)!+1 \equiv 1 \pmod 5$, and since $5^k=a!+a=a((a-1)!+1),$ we must have $(a-1)!+1=1,$ a contradiction.

Case 2: $a>b$. Then, $a=b+s$ with $s \geq 1$. Let $a!+b=5^k$ with $k \geq 1$. Therefore,

$b((b-1)! \cdot (b+1) \cdots (b+s)+1)=5^k$.

Thus, $b=5^\ell$ with $\ell \geq 0$, and if $b-1 \geq 5$ then $(b-1)! \cdot (b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction. Moreover, if $s \geq 5$, then $5 \mid (b+1) \cdots (b+5) \mid (b+1) \cdots (b+s),$ and so $(b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction.

Therefore, $b \leq 5$ and $s \leq 4$.

Taking all cases, we obtain $(b,s)=(1,3),$ that is $(a,b)=(4,1)$.

To sum up, we obtain the solutions $(a,b)=(5,5), (4,1)$ and $(1,4)$.

ohh Same solution :D
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Leartia
93 posts
#4
Y by
We claim that the only solutions are $(1,4),(4,1)$, and $(5,5)$.
Suppose that we have integers $a$ and $b$ such that $a!+b=5^r$ and $b!+a=5^t$ for some $r,t\geq 1$.
If $a\geq 5$, we have that $5\mid a!$.
Since $5\mid a!$ and $5\mid 5^r$, we have $5\mid b$.
This means that $b\geq 5$. By the same argument, we have $5\mid a$.
So $a=5x$ and $b=5y$ for some positive integers $x,y$.

Suppose that $x>1$. Then $v_5(a!)>1$.
Since $a!+b=5^r$, we have $v_5(b)=v_5(a!).$
However, as $v_5(b)>1$ we have that $v_5(b!)>v_5(b)$. Hence $v_5(b!+a)=\min\{v_5(b!),v_5(a)\}=v_5(a)$, but this is clearly a contradiction as $t>v_5(a)$.
We thus have that $x=1$. By symmetry we get that $y=1$ which gives us the pair $(a,b)=(5,5)$ which is a solution.
Now it remains to consider $a<5$.
This implies $b<5$ due to the reasoning above. We check the following cases:
\begin{itemize}
\item If $a=1$, then $1!+b\equiv 0\pmod 5$ so $b=4$. This pair is a solution.
\item If $a=2$, then $2!+b\equiv 0\pmod 5$ so $b=3$. This pair is not a solution.
\item If $a=3$, then $3!+b\equiv 0\pmod 5$ so $b=4$. This pair is not a solution.
\item If $a=4$, then $4!+b\equiv 0\pmod 5$ so $b=1$. This pair is a solution.
\end{itemize}
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steppewolf
351 posts
#5
Y by
The following solution was proposed together with the problem:

The condition is symmetric so we can assume that $b \leq a$.

The first case is when $a=b$. In this case, $a!+a=5^m$ for some positive integer $m$. We can rewrite this as $a \cdot \left ( (a-1)!+1 \right)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be $0$. If $k \geq 2$, then $(a-1)!+1 = 5^{l}$ for some $l \geq 1$, but $a-1=5^{k}-1 > 5$, so $5|(a-1)!$, which is not possible because $5|(a-1)!+1$. This means that $k=1$ and $a=5$. In this case, $5!+5 = 125$, which gives us the solution $(5,5)$.

Let us now assume that $1 \leq b<a$. Let us first assume that $b=1$. Then $a+1=5^x$ and $a!+1=5^y$ for integers $x,y \geq 1$. If $x \geq 2$, then $a=5^x-1 \geq 5^2-1>5$, so $5|a!$. However, $5|5^y=a!+1$, which leads to a contradiction. We conclude that $x=1$ and $a=4$. From here $a!+b=25$ and $b!+a=5$, so we get two more solutions: $(1,4)$ and $(4,1)$.

Now we focus on the case $1 < b < a$. Then we have $a!+b = 5^x$ for $x \geq 2$, so $b \cdot \left ( \frac{a!}{b}+1 \right ) = 5^x$, where $b|a!$ because $a>b$. Because $b|5^x$ and $b>1$, we have $b=5^z$ for $z \geq 1$. If $z \geq 2$, then $5<b<a$, so $5|a!$, which means that $\frac{a!}{b}+1$ cannot be a power of $5$. We conclude that $z=1$ and $b=5$. From here $5!+a$ is a power of $5$, so $5|a$, but $a>b=5$, which gives us $a \geq 10$. However, this would mean that $25|a!$, $5|b$ and $25 \nmid b$, which is not possible, because $a!+b=5^x$ and $25|5^x$.

We conclude that the only solutions are $(1,4)$, $(4,1)$ and $(5,5)$.
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gnoka
245 posts
#6
Y by
If $a<5$, it is obvious that $5$ does not divide $a$, so $5$ does not divide $b!$. This implies that $b<5$. By checking all possible values, we find that $(a,b)=(1,4)$ satisfies the condition.

If $a\geq 5$, it is obvious that $5$ divides $a!$, which means $5$ divides $b$. In turn, this implies that $5$ divides $b!$, and therefore $5$ divides $a$. When $a=b=5$, the condition is satisfied.

If $a>5$, we will prove by induction that for any $k\in \mathbb{N}^*$, $5^k$ divides $a$.

(i) From $a>5$ and $5|a$, we have $a\geq 10$.

(ii) If there exists a positive integer $m>1$ such that $5^m|a$, then $5^{m+1}|a!$. Since $a!+b$ is a power of $5$, we have $5^{m+1}|b$. Hence, $5^{m+1}|b!$. Furthermore, since $b!+a$ is a power of $5$, we have $5^{m+1}|a$.

Based on (i) and (ii), we can conclude that for any $k\in \mathbb{N}^*$, $5^k|a$. However, this is clearly impossible.

In conclusion, $(a,b)$ can only be $(1,4)$ or $(4,1)$ or $(5,5)$.
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dancho
38 posts
#7
Y by
WLOG $a\geq b$
We have that $a!+b=5^t$ and $b!+a=5^s$

Let's assume that $a\geq10$
Let $v_5(a!)=x>1$
From $a!+b=5^t$ we have that $v_5(b)=x\implies b\geq5^x$
Now note that $b!=1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot b$ which has $\frac{b}{5}$ numbers divisible by 5. Thus $v_5(b!)\geq\frac{b}{5}\geq\frac{5^x}{5}=5^{x-1}$
From $b!+a=5^s$ and $v_5(b!)\geq5^{x-1}$ we get that $v_5(a)\geq5^{x-1}$
$5^{x-1}>x$ for $x>1$ which is a contradiction with $v_5(a!)\geq v_5(a)$
Hence $a<10$

When we check the remaining possibilities we conclude that $(a,b)=(1,4)$ or $(4,1)$ or $(5,5)$
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Baffledbanna74
241 posts
#8
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mathmax12 wanted me to share his solution:
WLOG $a\ge b$, now we have $2$ cases:

If $a=b$:
So $a!+a=5^n$, where $n \ge 1.$ When $n=3$, we have that $a=b=5$, works. If $n \ge 3$ we have contradiction.

If $a>b$:
Let $a=b+x$, so $s \ge 1$, we also have $a!+b=5^k$, we have $b\le 5$, and $s \le 4$, otherwise contradiction. If we take all cases, we get that, $(4,1)$ works, so the solutions are $(a,b)=(5,5), (4,1) and (1,4).$
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Assassino9931
1320 posts
#9 • 2 Y
Y by PennyLane_31, Mr_GermanCano
After the competition one of the Bulgarian contestants proposed looking at triples $(a,b,p)$, where $p$ is prime, such that $a! + b$ and $b!+a$ are both powers of $p$. I just solved this now, here is a solution. (This would have been a great Problem 4 (or maybe 3 if one wants the problem set to be extreme) in my opinion, shame we did not think about it in time.)

Answer. $(a,b,p) = (2, 2, 2), (3, 3, 3), (5, 5, 5)$, $(1, 1, 2)$, $(2, 1, 3)$, $(1, 2, 3)$, $(4, 1, 5)$, $(1, 4, 5)$

Without loss of generality treat $a\geq b$. Suppose firstly that $a \geq 2p$. In
\[a! + b = b \cdot (1 \cdot 2 \cdots (b-1) \cdot b+1 \cdots (a-1) \cdot a + 1)\]both multipliers on the right must be powers of $p$. The second multiplier is greater than $1$, while $1 \cdot 2 \cdots (b-1) \cdot b+1 \cdots (a-1) \cdot a$ is divisible by $p$, since for $b\neq p$ we have $p$ as a multiplier, while for $b=p$ we have $2p$ as a multiplier, hence $(1 \cdot 2 \cdots (b-1) \cdot b+1 \cdots (a-1) \cdot a + 1$ is not divisible by $p$, which is impossible. Hence $b \leq a \leq 2p-1$ and since $b$ is a power of $p$, we must have $b=p$ or $b=1$.

Let $b=p$. Since $b!+a$ is a power of $p$, we must have that $a$ is divisible by $p$ and now $a\leq 2p-1$ implies $a=p$. In this case $b! + a = p((p-1)! + 1)$ and so it remains to see for which $p$ there is $k$ with $(p-1)! + 1 = p^k$. For $p=2$ we have $k=1$, for $p=3$ we have $k=1$, for $p=5$ we have $k=2$ and we now show that no $p\geq 7$ works. Divide by $p-1$ in $(p-1)! = p^k - 1$ to obtain
\[ (p-2)! = p^{k-1} + p^{k-2} + \cdots + p + 1. \]Consider the latter modulo $p-1$. For $p\geq 7$ the left-hand side contains the different multipliers $2$ and $\frac{p-1}{2}$ and so it is divisible by $p-1$. Each summand on the right gives remainder $1$ when divided by $p-1$, so the right-hand side is congruent to $k$. Hence $k$ must be divisible by $p-1$, but then $k\geq p-1$ and $(p-2)! > p^k \geq p^{p-1}$, which is impossible.

Let $b=1$. We want $a+1 = p^m$ for some $m\geq 1$, i.e. $a = p^m-1$, as well as $a! + 1 = p^n$ for some $n\geq 1$, i.e. $(p^m-1)! = p^n - 1$. If $m\geq 2$, then the left-hand side is divisible by $p$, while the right-hand side is not, contradiction. For $m=1$ we get $(p-1)! = p^n - 1$, thus as in the previous case we obtain only $p=2$ (with $a=1$), $p=3$ (with $a=2$) and $p=5$ (with $a=4$).
This post has been edited 2 times. Last edited by Assassino9931, Sep 13, 2023, 12:18 PM
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GrantStar
821 posts
#10 • 2 Y
Y by dhfurir, ventic
If $a=b$, then $a((a-1)!+1)$ is a power of $5$. In particular, $a$ and $(a-1)!+1$ are powers of $5$, and if $a\geq 6$ the latter is equivalent to $1$ mod $5$. Check $1,2,3,4,5$ to get $(5,5)$.
Otherwise WLOG $a>b$. First, if $b=1$ then $a!+1$ and $1+a$ are powers of $5$. In particular, $a< 5$ since otherwise $a!+1\equiv 1 \pmod 5$. Else, $b>1$ so $b \mid a!+b$, implying $5\mid b$. Moreover, \[5^x=a!+b=b\left(\frac{a!}{b}+1\right)\]Then, if $a\geq 10$, we must have $\frac{a!}{b}+1 \equiv 1 \pmod 5$ so therefore $b=5$. A finite case check yields no solutions.
Therefore, $(a,b)=(4,1),(1,4),(5,5)$
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Jishnu4414l
154 posts
#11
Y by
Here is my solution sketch...
Case 1:Let $a,b >5$
It is easy to see that both $a$ and $b$ must be powers of $5$. Now, WLOG, let $a \geq b$,
$a!+b= b(\frac{a!}{b}+1)$
We see that $\frac{a!}{b}+1$ is $1$ modulo $5$, so this choice of $(a,b)$ fails.
Case 2:Let $a>5$ and $b<5$
We see that $b$ must be odd. So $b=1$ or $b=3$. However, both these choices of $b$ fail as $a!+b \equiv b \pmod 5$.
Thus we can't choose such pairs of $(a,b)$ either.
Case 3: Let $a,b<5$. Note that $(a,b)=(5,5)$ works.
In the third case, if $a,b\geq 2$, it is easy to conclude that both $a$ and $b$ must be odd. However, $(3,3)$ does not work.
So both $a$ or $b$ must be $1$, Now checking the last few cases left we see that $(a,b)=(4,1);(1,4)$ work.
We thus have gotten three such pairs of $a$ and $b$.
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PennyLane_31
77 posts
#12
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Notice that if $x\in\mathbb{Z^*_+}$, then $x!$ ends in 1 (1!), 2 (2!), 6(3!), 4(4!), or 0 ($x\geq 5$)
Let
$\begin{cases}
    a!+b= 5^x\\
    b!+a= 5^y
    \end{cases}$
Where $x,y\in\mathbb{Z^*_+}$

If $a=1\implies b!+1= 5^y\implies b!\equiv 4 \mbox{ (mod 10)}$
So, $b= 4$ (if $b\geq 5\implies b!\equiv 0 \mbox{ (mod 10)})$
$\therefore \boxed{(a,b)= (1,4)}$ is solution ($1!+4= 5= 5^1$)

If $a=2\implies b!+2= 5^y\implies b!\equiv 3\mbox{ (mod 10)}$, which is impossible.

If $a=3\implies b!+3= 5^y\implies b!\equiv 2 \mbox{ (mod 10)}\implies b=2$. But, we also need that $a!+b$ is a power of five, so $3!+2= 8$ is a power of 5, contradiction!

If $a= 4$, we get the solution $\boxed{(a,b)= (4,1)}$

If $a=5\implies 5!+b= 5^x\implies b!\equiv 5 \mbox{ (mod 100)}$. Also, we know that $b!+5=5^y\implies b!\equiv 20 \mbox{ (mod 100)}$.
But, if $b\geq 10, b!\equiv 0 \mbox{ (mod 100)}$. So, we must have $b<10$, giving us the solution $\boxed{(a,b)= (5,5)}$

Noooow, consider WLOG $5<a<b$.

Let $b=a+b_0$, where $b_0\in\mathbb{Z^*_+}$ (with a quick check, you can verify that $a=b$ gives us only the solution a=b=5)
We know that:

$a!(a+1)(a+2)\dots(a+b_0)+a= 5^y\implies a\mid 5^y$.

So, $\boxed{\mbox{a is a power of 5}}.$
Call $a= 5^c$, $c\in\mathbb{Z^*_+}$

We have:
$b!+5^c= 5^y\implies b!= 5^y-5^c= 5^c(5^{y-c}-1)$
$\implies v_5(b!)=c$, but $b!$ has at least $(c+1)$ factors of 5 (actually much more than this), because he has at least the 5 and $a$(a has c factors 5). $\implies c= v_5(b!)\geq c+1$, contradiction!!

Therefore, 3 solutions:
$(a,b)= (1,4); (4,1); (5,5)$
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Aiden-1089
280 posts
#13 • 1 Y
Y by Siddharthmaybe
Clearly $v_5(a!)=v_5(b)$ and $v_5(b!)=v_5(a)$. Since $v_5(a!) \geq v_5(a)$ and $v_5(b!) \geq v_5(b)$, it follows that $v_5(a)=v_5(b)=v_5(a!)=v_5(b!)$.
Thus, $v_5((a-1)!)=v_5((b-1)!)=0 \implies a,b \leq 5$. Checking gives $(a,b)=(1,4),(4,1),(5,5)$ as the only solutions.
This post has been edited 1 time. Last edited by Aiden-1089, May 18, 2024, 5:17 AM
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megarnie
5606 posts
#14
Y by
Find all pairs $(a,b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$.

Claim: If $x + y$ is a power of $5$, then $\nu_5(x) = \nu_5(y)$.
Proof: Suppose otherwise and WLOG $\nu_5(x) > \nu_5(y)$. Then $\nu_5(x+y) = \nu_5(y)$, so $x + y = 5^{\nu_5(y)} < y$, absurd. $\square$

Hence $\nu_5(a!) = \nu_5(b)$ and $\nu_5(b!) = \nu_5(a)$.

Claim: For all $a > 5$, we have $\nu_5(a!) > \nu_5(a)$.
Proof: This is equivalent to there being a positive integer multiple of $5$ less than $a$. $\square$

Now for all positive integers $a$, we have $\nu_5(a!) \ge \nu_5(a)$ since $a \mid a!$. Thus, $\nu_5(b)  = \nu_5(a!) \ge \nu_5(a)$ and $\nu_5(a) = \nu_5(b!) \ge \nu_5(b)$, meaning $\nu_5(a) = \nu_5(b)$. Then $\nu_5(a!) = \nu_5(b) = \nu_5(a)$ and $\nu_5(b) = \nu_5(a) = \nu_5(b!)$, so $a \le 5$ and $b \le 5$.

Thus, $1 < a! + b \le 5! + 5 = 125$, so $a! + b \in \{5, 25, 125\}$.

If $a!  + b = 5$, then $a = 1, b = 4$, which works, or $a = 2, b  = 3$, which doesn't work since $3! + 2$ isn't a power of $5$.

If $a! + b = 25$, then $a  = 4, b = 1$, which works.

If $a! + b = 125$, then $a = b = 5$, which works.

Hence the answer is $(a,b) \in \{(1,4), (4,1), (5,5)\}$.
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Sedro
5845 posts
#15 • 1 Y
Y by sami1618
We claim the only working ordered pairs are $(a,b)=\boxed{(1,4),(4,1),(5,5)}$. It is easy to see that they work; we now prove they are the only ones.

It is not hard to verify that these are the only solutions when $a,b\le 5$, so henceforth, FTSOC, assume $a,b>5$. We must have $a!+b = 5^x$ and $a+b! = 5^y$, for some positive integers $x$ and $y$. Adding these two equations together, we have $a((a-1)!+1) + b((b-1)!+1) = 5^x+5^y$. Note that $v_5(a!), v_5(b!) \ge 1$, and thus $5\mid a$, $5\mid b$. Furthermore, notice that $v_5((a-1)!), v_5((b-1)!) \ge 1$, and hence $5\nmid (a-1)!+1$, $5\nmid (b-1)!+1$. It follows that $v_5(\text{LHS}) = \min\{v_5(a), v_5(b)\}$. We also have $v_5(\text{RHS}) = \min\{x,y\}$.

WLOG assume $\min\{v_5(a), v_5(b)\}=v_5(a)$. Now, suppose $\min\{x,y\}=x$. Then, $v_5(a)=x$, and $a\ge 5^x$. But then $a!+b \ge (5^x)!+b > 5^x$, contradiction. Thus, $\min\{x,y\}=y$, and $v_5(a)=y$. Then, $v_5(b)\ge y$, and $b\ge 5^y$. But then $a+b!\ge a+(5^y)! > 5^y$, absurd. Thus, there are no solutions when $a,b>5$, and we are done. $\blacksquare$
This post has been edited 1 time. Last edited by Sedro, May 29, 2024, 8:40 PM
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Orestis_Lignos
558 posts
#16
Y by
This was N1 at last year's shortlist.
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PaixiaoLover
123 posts
#18
Y by
$(1,4), (4,1), (5,5)$ work.

Case 1: $a=b$: then $a((a-1)!+1)=5^k$ for some k. clearly a neq 1, so $a$ is a power of 5. $a=5$ works, but if $a$ is 25 or more, $(a-1)!+1)$ isnt a power of 5 since its 1 mod 5. So $(5,5)$ is the only solution here.

Case 2: $a>b$. Consider b=1 (edge case since we noticed $(1,4)$ works. Then $a!+1=5^k, a+1=5^j.$ So $a=5^j-1$. If a=4, it works. if a=24 or more, $a! +1 $ is 1 mod 5 again. (same technique as before. we will use it again soon). So $(1,4) (4,1)$ works.

Now if $a>b>1$, then $a!+b=5^k, a+b!=5^j$. Then factor out b since $a>b$ to get $b(\frac{a!}{b}+1)=5^k$. So b is a power of 5. Then $a+b!=5^k$ gives $a = 0 \mod 5$. However, then $(\frac{a!}{b}+1)$ would 1 mod 5 (again) ((a would be the next multiple of 5 after b)). So no more solutions.
This post has been edited 2 times. Last edited by PaixiaoLover, Jan 4, 2025, 1:40 AM
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eg4334
637 posts
#19
Y by
The only pairs are $\boxed{(1, 4), (5, 5)}$ in some permutation. If $a=b$, then we have $a((a-1)!+1)$ to be a power of $5$, or more specifically $(a-1)!+1$ to be a power of $5$. This implies $a-1 < 5$ else it will be $1 \pmod{5}$, and a finite check gives the only solution of $(5, 5)$ as advertised above. Else, WLOG $a>b$. The key idea is that looking at $a!+b$, we can factor out a $b$ from $a!$ and hence $\frac{a!}{b}$ cannot be a multiple of $5$. However, when $a \geq 10$, we can easily see that $v_5(a!)$ is strictly greater than $v_5(b)$ for all $b < a$. This is most easily seen by noticing the gaps between powers of $5$. Therefore, there is a finite check that we need to do with $a \leq 9$, noting again that when $5 \leq a \leq 9$ then $b=5$ to eliminate the factor of $5$. This lands us at the other advertised solution above.
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Perelman1000000
110 posts
#20 • 1 Y
Y by MIC38
Case 1)$a=b\implies x=y$
$a!+a=5^x$
Case 1.1)$a<5$
$a=1\implies 1+1\neq 5^x$
$a=2\implies 2+2\neq 5^x$
$a=3\implies 6+3\neq 5^x$
$a=4\implies 24+4\neq 5^x$ so this case fails
Case 1.2)$a\geq 5$
$v_5(a!)\geq v_5(a)$
If $v_5(a!)=v_5(a)\implies a=5\implies a!+a=120+5=125=5^3\implies x=3$
If $v_5(a!)>v_5(a)$ ans also we know that $5|a\implies v_5(a!+a)=min(v_5(a!),v_5(a))=v_5(a)=v_5(5^x)=x\implies a\geq 5^x$ contradiction
Case 2) Wlog $a>b>5$
Since $a>b\implies v_5(a!)>v_5(b)\implies v_5(a!+b)=v_5(b)=x\implies b\geq 5^x$ contradiction
Case 3)$5>a>b$
By checking we get $4!+1=25=5^2$
$1!+4=5$ so $(a,b)=(4,1)$ if $b>a\implies (a,b)=(1,4)$
I think someone will ask why not $a>5>b$ because if $a>5$ $5|a!\implies 5|b\implies b>5$ since $a\neq b$(case 1)
$\boxed{ANSWER:(a,b)=(1,4),(4,1),(5,5)}$
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iStud
268 posts
#21
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W.L.O.G. $a\ge b$.

Case 1. If $a=b$.
Then we have $a((a-1)!+1)\equiv 0\pmod{5}$. It's easy to check that any $a\le 4$ doesn't works, so we must have $a\ge 5$. For $a=5$, the problem is satisfied. For $a>5$, we have $(a-1)!+1\equiv 1\pmod{5}$. But we have $5^t=a((a-1)!+1)$ for some $t\in\mathbb{N}$, so $(a-1)!+1=1$, a contradiction. So for this case, the only pair that satisfy is $(a,b)=(5,5)$

Case 2. If $a>b$.
Let $a=b+k$ for some $k\in\mathbb{N}$. If $a>b\ge 5$, we have $5\mid a,b$. In other side, we have $5^t=b((b+1)(b+2)\dots a+1)$ for some $t\in\mathbb{N}$, so we must have $(b+1)(b+2)\dots a\equiv -1\pmod{5}$.This is a contradiction since we must have $(b+1)(b+2)\dots a\equiv 0\pmod{5}$ (since that $5\mid a$). So we must have $a,b\le 4$, which from that we easily get $(a,b)=(4,1)$.

In the end, the only pairs that satisfy the problem are $(a,b)=(5,5),(4,1),(1,4)$. $\blacksquare$
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anudeep
170 posts
#22
Y by
We claim $(1,4), (4,1)$ and $(5,5)$ are the only solutions.
We may assume $a\ge b$ and that implies $b\lvert a!$. As $a!+b=b\left(\frac{a!}{b}+1\right)$ is given to be a power of $5$, $b$ is necessarily a power of $5$ and is clear that $\gcd(a!/b, 5)=1$. This forces $b=5^{\nu_5(a!)}$ and is a moment to celebrate as $b$ seems to get larger and larger surpassing $a$ given $a\ge 10$ which is against our assumption. Performing a finite check we get the required. $\square$
This post has been edited 2 times. Last edited by anudeep, Mar 24, 2025, 4:39 PM
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