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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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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
Inspired by Austria 2025
sqing   4
N 16 minutes ago by Tkn
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
4 replies
+2 w
sqing
Today at 2:01 AM
Tkn
16 minutes ago
Erasing the difference of two numbers
BR1F1SZ   1
N 28 minutes ago by BR1F1SZ
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
1 reply
BR1F1SZ
Yesterday at 9:48 PM
BR1F1SZ
28 minutes ago
3-var inequality
sqing   1
N 41 minutes ago by Natrium
Source: Own
Let $ a,b,c\geq 0 ,a+b+c =1. $ Prove that
$$\frac{ab}{2c+1} +\frac{bc}{2a+1} +\frac{ca}{2b+1}+\frac{27}{20} abc\leq \frac{1}{4} $$
1 reply
sqing
May 3, 2025
Natrium
41 minutes ago
Geo metry
TUAN2k8   0
an hour ago
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
0 replies
TUAN2k8
an hour ago
0 replies
mathemetics
Pangbowen   0
an hour ago
Let a,b,c≥0 and a+b+c=7. Prove that : a/b+b/c+c/a+abc≥ab+bc+ca-2
0 replies
Pangbowen
an hour ago
0 replies
Property of a function
Ritangshu   1
N an hour ago by Natrium
Let \( f(x, y) = xy \), where \( x \geq 0 \) and \( y \geq 0 \).
Prove that the function \( f \) satisfies the following property:

\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]
for all \( (x, y) \ne (x', y') \) and for all \( \lambda \in (0, 1) \).

1 reply
Ritangshu
May 3, 2025
Natrium
an hour ago
max value
Bet667   2
N 2 hours ago by Natrium
Let $a,b$ be a real numbers such that $a^2+ab+b^2\ge a^3+b^3.$Then find maximum value of $a+b$
2 replies
Bet667
2 hours ago
Natrium
2 hours ago
Geometry
gggzul   2
N 2 hours ago by gggzul
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
2 replies
gggzul
3 hours ago
gggzul
2 hours ago
thank you !
Piwbo   2
N 2 hours ago by Piwbo
Given positive integers $a,b$ such that $a$ is even , $b$ is odd and $ab(a+b)^{2023}$ is divisible by $a^{2024}+b^{2024}$ .Prove that there exists a prime number $p$ such that $a^{2024}+b^{2024}$ is divisible by $p^{2025}$
2 replies
Piwbo
2 hours ago
Piwbo
2 hours ago
Inequality involving square root cube root and 8th root
bamboozled   2
N 2 hours ago by bamboozled
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 maximum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
2 replies
bamboozled
Today at 4:46 AM
bamboozled
2 hours ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   12
N 2 hours ago by n-k-p
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
12 replies
parmenides51
Jul 25, 2018
n-k-p
2 hours ago
hard problem
Cobedangiu   5
N 3 hours ago by KhuongTrang
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
5 replies
+1 w
Cobedangiu
Yesterday at 4:24 PM
KhuongTrang
3 hours ago
Nordic 2025 P3
anirbanbz   9
N 3 hours ago by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
3 hours ago
Aime type Geo
ehuseyinyigit   1
N 4 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
4 hours ago
Easy Number Theory
math_comb01   37
N Apr 23, 2025 by John_Mgr
Source: INMO 2024/3
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
37 replies
math_comb01
Jan 21, 2024
John_Mgr
Apr 23, 2025
Easy Number Theory
G H J
G H BBookmark kLocked kLocked NReply
Source: INMO 2024/3
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math_comb01
662 posts
#1 • 3 Y
Y by GeoKing, Rounak_iitr, polynomialian
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
This post has been edited 3 times. Last edited by math_comb01, Jan 22, 2024, 8:22 AM
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mannshah1211
651 posts
#2 • 3 Y
Y by ATGY, GeoKing, megarnie
If $p$ divides one of them, it divides all of them. Assume that it divides none of them. Then, there exists an inverse for all of them in modulo $p$ and then it's some algmanip to get $b \equiv c \pmod{p}$ a contradiction
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ATGY
2502 posts
#3
Y by
mannshah1211 wrote:
If $p$ divides one of them, it divides all of them. Assume that it divides none of them. Then, there exists an inverse for all of them in modulo $p$ and then it's some algmanip to get $b \equiv c \pmod{p}$ a contradiction

Did it the same for the first part but used sum of cubes to arrive at the result (more lengthy)
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LuciferMichelson
18 posts
#4 • 1 Y
Y by Aliosman
isn't that this question toooo easy for INMO p3
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starchan
1608 posts
#5 • 1 Y
Y by kamatadu
LuciferMichelson wrote:
isn't that this question toooo easy for INMO p3

On what sample set are you basing your opinion on difficulty?
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anantmudgal09
1980 posts
#6 • 1 Y
Y by Raj_singh1432
Proposed by Navilarekallu Tejaswi
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lelouchvigeo
181 posts
#7 • 1 Y
Y by S_14159
If $p$ divides one of them, it divides all of them. Assume that it divides none of them. Then just do some things to get $b \equiv c \pmod{p}$.
Which gives a contradiction
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Pluto1708
1107 posts
#8 • 4 Y
Y by ATGY, GeoKing, kamatadu, Rounak_iitr
math_comb01 wrote:
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$.
Prove that $p$ divides each of $a,b,c$.
Clearly if $p$ divides any of $a,b,c$ we are done, so from now onwards we assume none of them are divisible by $p$.Define $x,y$ such that
$x = \dfrac{a}{b} \mod p \Longleftrightarrow x^{2023}\equiv -1\mod p$ and $y = \dfrac{b}{c}\mod p \Longleftrightarrow y^{2024}\equiv -1\mod p$.Then notice
\[\dfrac{1}{xy} = \dfrac{c}{a}\mod p\Longleftrightarrow (x\cdot y)^{2025}\equiv -1\mod p\]Claim : $y\equiv 1\mod p$ that is $b\equiv c \mod p$
Based on previous equations $$-1 = (x\cdot y)^{2023\cdot 2025} = x^{2023\cdot 2025}\cdot y^{2023\cdot 2025} = -y^{2024^2-1}=-y^{-1}\mod p\Longleftrightarrow y\equiv 1\mod p $$
Now plugging back in the original equation we get $1=y^{2024}=-1\mod p\implies p\mid 2$ a contradiction to the fact that $p$ was an odd-prime.$\blacksquare$
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HoRI_DA_GRe8
597 posts
#9
Y by
Yeah basically belows solution I got.
This post has been edited 1 time. Last edited by HoRI_DA_GRe8, Jan 22, 2024, 1:05 PM
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kamatadu
480 posts
#10 • 4 Y
Y by GeoKing, polynomialian, iamahana008, kkloveMinecraft
Firstly note that if even one of them is divisible by $p$, all of them are. So assume on the contrary that none of them are divisible by $p$. So assume none of them are.

Then we get that,
\begin{align*}
    \left(\dfrac{a}{b}\right)^{2023} &\equiv -1 \pmod{p}\\
    \left(\dfrac{b}{c}\right)^{2024} &\equiv -1 \pmod{p}\\
    \left(\dfrac{c}{a}\right)^{2025} &\equiv -1 \pmod{p}
.\end{align*}
We multiply all three of them to get $a^2 \equiv -bc \pmod{p}$.

Now note that we have $a^{2023} \equiv -b^{2023} \pmod{p}$ and $a^{2025} \equiv -c^{2025} \pmod{p}$. Multiplying these, we get,
\[ a^{2023 + 2025} \equiv b^{2023}c^{2025} \implies (a^2)^{2024} \equiv c^2 \cdot (bc)^{2023} \implies bc \cdot (bc)^{2023} \equiv c^2 \cdot (bc)^{2023} \implies b \equiv c \pmod{p}. \]
But then,
\[ p\mid b^{2024} + c^{2024} \equiv 2b^{2024} \implies p \mid 2 \]which is a contradiction and we are done. :yoda:
This post has been edited 1 time. Last edited by kamatadu, Jan 21, 2024, 2:34 PM
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megarnie
5606 posts
#11 • 2 Y
Y by Anchovy, HoRI_DA_GRe8
Solved with Anchovy.

I assume the problem means that $p$ divides all of $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$
If $p$ divides one of $a,b,c$, clearly $p$ divides all of $a,b,c$, so assume $p$ divides none of them.

Let $x = \frac ab, y = \frac bc, z = \frac ca$. The condition $x^{2023} \equiv -1\pmod p, y^{2024} \equiv 1\pmod p, z^{2025} \equiv -1\pmod p$ and we also have $xyz = 1$, so multiplying the three equations gives $(xyz)^{2023} \cdot y z^2 \equiv 1\pmod p$, so $yz^2 \equiv 1\pmod p$. This implies that $y\equiv -\frac{1}{z^2}$, so $(-1/z^2)^{2024} \equiv 1$, so $z^{4048} \equiv 1$. But $(z^{2025})^2 = z^{4050}\equiv 1$, so dividing the two gives $z^2 \equiv 1$, so $z\in \{-1,1\}$ mod p. Since $z^{2025} \equiv -1$, we have $z\equiv -1$. But then since $yz^2 \equiv 1$, $y\equiv 1\pmod p$, so $b\equiv c \pmod p$. Now, this means that $2b^{2024}$ is a multiple of $p$, contradiction since $p$ doesn't divide $b$ and $p > 2$.
This post has been edited 1 time. Last edited by megarnie, Jan 21, 2024, 2:42 PM
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SatisfiedMagma
458 posts
#13 • 1 Y
Y by ATGY
Nah bro, I ain't even gonna comment on this one.

Solution: Observe that $p$ divides either of $a,b,c$, then it divides everything and we're done. Henceforth, assume that $p \nmid a,b,c$. We will now work modulo $p$ with special powers of existence of inverses of $a,b,c$.

From $p \mid a^{2023}+b^{2023}$, we get that $b^{2024} \equiv -a^{2023}b \pmod{p}$. Swing this into $p \mid b^{2024}+c^{2024}$ to get $c^{2025} \equiv a^{2023}bc \pmod{p}$. At last, putting this into $p \mid a^{2025} + c^{2025}$, we get
\[a^{2025} + a^{2023}bc \equiv 0 \iff a^2 + bc \equiv 0 \pmod{p}.\]From here, we wish to eliminate $b$ completely modulo $p$. This can be achieved via substituting $b \equiv -a^2/c$. Putting this in $p \mid a^{2023}+b^{2023}$ we get
\[a^{2023} - \frac{a^{4046}}{c^{2023}} \equiv 0 \iff a^{2023} \equiv c^{2023} \pmod{p}.\]This yields
\[a^{2025} \equiv a^2c^{2023} \equiv -c^{2025} \pmod{p} \implies a^2 + c^2 \equiv 0 \pmod{p}.\]Finally, on combining this with $a^2 \equiv -bc$, we get
\[p \mid c(c-b) \iff c \equiv b \pmod{p}\]since $\gcd(c,p) = \gcd(b,p) = 1$. Upon putting $c \equiv b$ in say $p \mid b^{2024}+c^{2024}$, we get immediately get $2b \equiv 0 \pmod{p}$ which is a contradiction since $p$ is odd and we're done. $\blacksquare$
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Samujjal101
2799 posts
#14
Y by
............
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Samujjal101
2799 posts
#15
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Samujjal101 wrote:
Just see that if gcd(a,b)=1 then the given conditions are not possible. So, a|b which means b=ak for all integers k. Now p divides (a^2023 + k^2023.a^2023) so p either divides a^2023 or (1+ k^2023) => p divides a^2023 =>p divides a

..............
This post has been edited 1 time. Last edited by Samujjal101, Jan 21, 2024, 4:53 PM
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taptya17
29 posts
#16 • 1 Y
Y by Om245
If $p$ divides any one of them, it divides the other two as well.

Assume for the sake of contradiction that $p$ doesn't divide any of the three. Since $p$ is an odd prime, it has a primitive root, say $g$.
Let $a=g^x,b=g^y,c=g^z$.

Claim. $g^m+g^n=0(p)\implies m-n=k(2k)$ where $p=2k+1$.
Proof. $g^{m-n}=-1(p)\implies m-n=k(2k)\implies m-n=2kq+k=k(2q+1)$.
Using the claim and given information,
$$0=(x-y)+(y-z)+(z-x)=\frac{k(2q_1+1)}{2023}+\frac{k(2q_2+1)}{2024}+\frac{k(2q_3+1)}{2025}$$Cancelling $k$ and the denominator and taking mod 2 leads to a contradiction. Done!
This post has been edited 2 times. Last edited by taptya17, Jan 21, 2024, 4:01 PM
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djmathman
7938 posts
#17 • 2 Y
Y by starchan, Rounak_iitr
math_comb01 wrote:
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$.
This sentence seems to be missing something. Are all three integers divisible by $p$?
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mannshah1211
651 posts
#18 • 3 Y
Y by GeoKing, sanyalarnab, Erratum
djmathman wrote:
math_comb01 wrote:
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$.
This sentence seems to be missing something. Are all three integers divisible by $p$?

Yes. It's supposed to say that all three are divisible by $p$, but I suppose a lot of us got tunnel vision and didn't see the error because we already saw it in the test lol :rotfl:
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maths_arka
6 posts
#19 • 1 Y
Y by sanyalarnab
First see that if $p|a$ or $p|b$ or $p|c$ then $p$ divides all of them.

Now suppose that $p{\not|}a$ $p{\not|}b$ $p{\not|}c$.
Then the desired contradiction will come from the fact that $p$ is odd prime, i.e we would prove $p=2$.

Since $p{\not|}a$ $p{\not|}b$ $p{\not|}c$ we could take inverses.
Reducing everything $(\textrm{mod}\ p)$ we get
$$(ab^{-1})^{2023}\equiv-1 ({mod}\,p) \cdots(1)$$$$(bc^{-1})^{2024}\equiv-1 ({mod}\,p)\cdots(2) $$$$(ca^{-1})^{2025}\equiv-1 ({mod}\,p)\cdots(3) $$Multiplying equations $1$, $2$ and $3$ we get
$$a^{-2}bc\equiv-1 ({mod}\,p)\cdots(4)$$.
$$\Rightarrow (ab^{-1})^{-1}ca^{-1}\equiv-1 ({mod}\,p) \cdots(5)$$i.e
$$\Rightarrow ca^{-1}\equiv-(ab^{-1}) ({mod}\,p) \cdots(6)$$Now from $3$ we get
$$(ab^{-1})^{2025}\equiv1 ({mod}\,p) \cdots(7)$$and from 1 we get
$$(ab^{-1})^{4046}\equiv1 ({mod}\,p) \cdots(8)$$Let the order of $ab^{-1}$ mod $p$ be k.
Thus $k|gcd(4046,2025)$ i.e $k|1$.
Therefore we have
$$ab^{-1}\equiv1 ({mod}\,p)$$or
$$a\equiv b ({mod}\,p)$$Putting this information in equation $1$ we get
$$1\equiv -1 ({mod}\,p)$$or $p=2$
which is a contradiction.
$\blacksquare$ :-D
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mannshah1211
651 posts
#20
Y by
Well, I'll post my solution just for the sake of it, I guess.

If $p$ divides one of $a, b, c,$ then it divides all of them, so henceforth assume it doesn't divide any of them. Thus, there exists a valid inverse in modulo $p$ for each of $a, b, c.$ Then, we have $\left(\frac{a}{b}\right)^{2023} \equiv -1 \pmod{p}, \left(\frac{b}{c}\right)^{2024} \equiv -1\pmod{p}, \left(\frac{c}{a}\right)^{2025}\equiv -1\pmod{p},$ and thus, multiplying all of them together, we have $bc \equiv -a^2 \pmod{p}.$ Thus, $b^{1012}c^{1012} \equiv a^{2024} \pmod{p},$ which gives $ab^{1012}c^{1012} + c^{2025} \equiv 0 \pmod{p} \implies ab^{1012} \equiv -c^{1013} \pmod{p}.$ Thus, $a \equiv \frac{-c^{1013}}{b^{1012}} \pmod{p},$ which by putting in the first equation gives $\frac{b^{1013 \cdot 2023} - c^{1013 \cdot 2023}}{b^{1012 \cdot 2023}} \equiv 0 \pmod{p} \implies b^{1013 \cdot 2023} - c^{1013 \cdot 2023} \equiv 0 \pmod{p}.$ From the second equation, we get $b^{4048} - c^{4048} \equiv 0 \pmod{p},$ so $p \mid b^{\gcd(4048, 1013 \cdot 2023)} - c^{\gcd(4048, 1013 \cdot 2023)} = b - c,$ so $p \mid 2b^{2024},$ a contradiction.
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Safal
169 posts
#21
Y by
My attempt,
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This post has been edited 8 times. Last edited by Safal, Jan 23, 2024, 12:46 PM
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lifeismathematics
1188 posts
#22 • 1 Y
Y by Rounak_iitr
we work with $a^{r}+b^{r} \equiv 0 \pmod p , b^{r+1}+c^{r+1} \equiv 0 \pmod p , a^{r+2}+c^{r+2} \equiv 0 \pmod p$ for any $r \in \mathbb{Z}^{+}$ and it follows for $r=2023$.

$\mathsf{Claim 1:-}$ If $p$ divides any one of $a,b,c$ then $p|a,b,c$.

$\mathsf{Proof:-}$ W.L.O.G $p|a \implies p|a^{r}$ , now since $a^{r}+b^{r} \equiv 0 \pmod p \implies p|b^{r} \implies p|b$ and also $b^{r+1}+c^{r+1} \equiv 0 \pmod p \implies p|c^{r+1} \implies p|c$. $\square$

From $\mathsf{Claim 1}$ it suffices to disprove the fact that that $p \nmid a, p \nmid b , p\nmid c$ ,so FTSOC we assume that is the case.

$\mathsf{Claim 2:-}$ $p$ does not divide any of $a-b , b-c,c-a$.

$\mathsf{Proof:-}$ FTSOC $p|a-b$ , then $a \equiv b \pmod p \implies a^{r} \equiv b^{r} \pmod p \implies 2a^{r} \equiv 0 \pmod p$, now since $\gcd(p,2)=\gcd(p,a)=1$ , it gives a contradiction. $\rightarrow \leftarrow$.

Now ,
$\bullet a^{r}+b^{r} \equiv 0 \pmod p$

$\bullet b^{r+1}+c^{r+1} \equiv 0 \pmod p$

$\bullet a^{r+2}+c^{r+2} \equiv 0 \pmod p$


so $a^{r} \equiv -b^{r}   \pmod p$ and $a^{r+2} \equiv -c^{r+2} \pmod p \implies a^2c^{r+1} \equiv -bc^{r+2} \pmod p$ and since $\gcd(c,p)=1 \implies a^2 \equiv -bc \pmod p , b^2 \equiv -ca \pmod p , c^2 \equiv -ab \pmod p \qquad (\star)$

now substract the subsequent cogruences to get $a^2-b^2 \equiv c(a-b) \pmod p$ similarly symmetrically other in $b$ and $c$ , but from $\mathsf{Claim 2}$ we have $p \nmid a-b , b-c , c-a$ , which implies $a+b \equiv c \pmod p , b+c \equiv a \pmod p , c+a \equiv b \pmod p \qquad (\dagger)$

Now from $(\star)$ we have $(a+b)^2+(b+c)^2+(c+a)^2 \equiv 0 \pmod p$ and from $(\dagger)$ we get this is equivalent to $a^2+b^2+c^2 \equiv 0 \pmod p \implies ab+bc+ca \equiv 0 \pmod p \implies a+b+c \equiv 0 \pmod p \implies 2a,2b , 2c \equiv 0 \pmod p$ , but this is not possible as $\gcd(2,p)=\gcd(a,p)=\gcd(b,p)=\gcd(c,p)=1$.

Hence we get a contradiction $\rightarrow \leftarrow$.

Hence $p$ must divide $a,b,c$. $\blacksquare$

A cute NT for sure! :blush:
This post has been edited 2 times. Last edited by lifeismathematics, Jan 23, 2024, 9:15 AM
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Master_of_Aops
71 posts
#24
Y by
Isn’t this a one-liner:
For $a^x \equiv -b^x$, raise both sides to power $yz$ to get terms $a^{xyz}, b^{xyz}, c^{xyz} (mod p)$ and done
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Tintarn
9042 posts
#25
Y by
Master_of_Aops wrote:
... to get terms $a^{xyz}, b^{xyz}, c^{xyz} (mod p)$ and done
How exactly are you done from here?
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Master_of_Aops
71 posts
#26
Y by
You get $a^{xyz} \equiv b^{xyz}, b^{xyz} \equiv -c^{xyz},  a^{xyz} \equiv c^{xyz}(mod p)$ so all are divisible by $p$
This post has been edited 1 time. Last edited by Master_of_Aops, Feb 1, 2024, 5:49 PM
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idkk
118 posts
#27
Y by
if p doesnt divide any of them

$(\frac{a}{b})^{2023} \equiv -1(mod p)$

$(\frac{b}{c})^{2024} \equiv -1(mod p)$

$(\frac{a}{c})^{2025} \equiv -1(mod p)$

$x^{2023} \equiv -1(mod p)$
$y^{2024} \equiv -1(mod p)$
$x^{2025}y^{2025} \equiv -1(mod p)$

so $x^2y \equiv -1(mod p)$

so $y \equiv \frac{-1}{x^2} (mod p)$

so $x^{2025} \equiv 1 (mod p)$

also $x^{2*2023} \equiv 1(mod p)$

let $k$ be the order of $x$ mod p so

$k | gcd(2025,2*2023) \implies k|1$

$x \equiv 1(mod p)$

then $1 \equiv -1(mod p)$

not possible as p is odd.

so p divides one of them then the answer is easy to conclude
This post has been edited 1 time. Last edited by idkk, Feb 3, 2024, 4:23 AM
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Pyramix
419 posts
#28 • 1 Y
Y by GeoKing
We first prove that if $p$ divides any one of $a,b,c$, then it divides all the three numbers. Suppose $p\mid a$. Then, $p\mid a^{2023}$ but $p\mid a^{2023}+b^{2023}$, which means that $p\mid b^{2023}$, forcing $p\mid b$. Also, $p\mid c^{2025}+a^{2025}$ and $p\mid a^{2025},$ which means $p\mid c^{2025}$, forcing $p\mid c$.

Similarly, if $p\mid b$, then $p\mid b^{2023}$ and $p\mid a^{2023}+b^{2023}$, which means $p\mid a^{2023}$, forcing $p\mid a$. Since $p\mid a$, we also have that $p\mid c$ from the earlier proof.

Finally, if $p\mid c$, then $p\mid c^{2024}$ and $p\mid b^{2024}+c^{2024}$, which means $p\mid b^{2024}$, forcing $p\mid b$, which in turn forces $p\mid c$. So, if $p$ divides any one of $a,b,c$, then it divides all the three numbers.

Assume that $p\nmid a,b,c$. For integer $a$ and prime $p$, we define the order of $a\pmod{p}$ to be the smallest positive integer $k$ such that $a^k\equiv1\pmod{p}$.

We prove a Lemma.

$\textbf{Lemma.}$ If $k$ is order of $a$, $\pmod{p}$ and $a^n\equiv1\pmod{p}$, then $k\mid n$.

$\emph{Proof.}$ Suppose it was possible that $k\nmid n$. Note that if $n<k$, then it will contradict the minimality of $k$. So, $n>k$. Hence, there exists $q>0$ and $0<r<k$ such that $n=kq+r$. Now, $a^{k}\equiv1\pmod{p}$, which means $a^{kq}\equiv1\pmod{p}$, so $a^n\equiv a^{kq+r}\equiv a^r\equiv1\pmod{p}$. But since $0<r<k$ and $a^r\equiv1\pmod{p}$, the minimality of $k$ gives us a contradiction. Hence, $k\nmid n$ is impossible, and $r=0$ is forced. The proof is complete. $\blacksquare$

We have the equations \[\left(\frac{a}{b}\right)^{4046}\equiv1\pmod{p} \ \ \ \ (1)\]\[\left(\frac{b}{c}\right)^{4048}\equiv1\pmod{p} \ \ \ \ (2)\]\[\left(\frac{c}{a}\right)^{4050}\equiv1\pmod{p} \ \ \ \ (3)\]Multiplying these equations $(1),(2),(3),$ we obtain $a^4\equiv b^2c^2\pmod{p}$. So, $a^2\equiv\pm bc\pmod{p}$.

Note, the equation $(1)$ is $\left(\frac{a}{b}\right)^{4046}\equiv1\pmod{p}$. So, \[\left(\frac{\left(bc\right)^{2023}}{b^{4046}}\right)\equiv\pm1\pmod{p},\]which gives \[\left(\frac{b}{c}\right)^{4046}\equiv1\pmod{p} \ \ \ \ (4)\]It follows that if $k$ is the order of $\left(\frac{b}{c}\right)$, $\pmod{p}$, then from equation $(1)$, we get $k\mid4048$, and from equation $(4)$, we get $k\mid4046$, using the $\textbf{Lemma}$ in each case.

This forces $k\mid2$, so $k=1,2$. Hence, we have that $\left(\frac{b}{c}\right)^{2}\equiv1\pmod{p}$. So, $b^2\equiv c^2\pmod{p}$.

Taking to the power $1012$, we get $b^{2024}\equiv c^{2024}\pmod{p}$, and since $p\mid b^{2024}+c^{2024}$, we have $p\mid2b^{2024}$, which forces $p\mid b$, as $p$ is an odd prime. However, if $p\mid b$ then $p\mid a$, and $p\mid c$ are forced by our initial arguments.

The proof is complete. $\blacksquare$
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AshAuktober
1004 posts
#29
Y by
In contest solution:

Observe that if $p$ divides any one of $a, b, c$, then it divides all of them. Assume this to not be the case. Then we get the system of congruences $$a^{2023} \equiv -b^{2023} \pmod{p} \cdots(A)$$$$b^{2024} \equiv -c^{2024}  \pmod{p} \cdots(B)$$$$c^{2025} \equiv -a^{2025}  \pmod{p} \cdots(C)$$Multiplying $(A)$ and $(B)$, $$b \equiv c\left(\frac{c}{a}\right)^{2023}  \pmod{p}$$$$\implies b^{2025} \equiv c^{2025}\left(\frac{c}{a}\right)^{2023 \cdot 2025} \pmod{p}$$but from $(C)$, $$\left(\frac{c}{a}\right)^{2025} \equiv -1 \pmod{p}$$Substituting in, $$b^{2025} \equiv c^{2025}(-1)^{2023} \equiv -c^{2025} \pmod{p}$$Dividing $(B)$ from this new equation, $$b \equiv c \pmod{p}$$Therefore $$b^{2024} \equiv -b^{2024} \pmod{p}$$$$\implies 2b^{2024} \equiv 0 \pmod{p}$$$$\implies b^{2024} \equiv 0 \pmod{p}$$$$\implies p \mid b, $$a contradiction. $\square$
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Shreyasharma
682 posts
#30
Y by
We are given,
\begin{align*}
a^{2023} + b^{2023} \equiv b^{2024} + c^{2024} \equiv c^{2025} + a^{2025} \equiv 0 \pmod{p}
\end{align*}Now assume that $a, b, c \not\equiv 0 \pmod{p}$ as if one of the three is $0$ modulo $p$, all of them must be. Then we find that,
\begin{align*}
a^{2023}b \equiv c^{2024} \pmod{p}
\end{align*}However then from this we may conclude that,
\begin{align*}
a^{2023}bc \equiv -a^{2025} \pmod{p}
\end{align*}Dividing as $a \not\equiv 0$ modulo $p$, we find that,
\begin{align*}
-bc \equiv a^2 \pmod{p}
\end{align*}Now rewriting our conditions we have,
\begin{align*}
b^{1012} &\equiv c^{1011}a \pmod{p}\\
b^{2024} &\equiv -c^{2024} \pmod{p}\\
c^{1013} &\equiv -b^{1012}a \pmod{p}
\end{align*}Squaring the first equation we have,
\begin{align*}
b^{2024} &\equiv c^{2022}a^2 \pmod{p}\\
-c^{2024} &\equiv c^{2022}a^2 \pmod{p}\\
-c^2 &\equiv a^2 \pmod{p}\\
-c^2 &\equiv -bc \pmod{p}\\
b &\equiv c \pmod{p}
\end{align*}From the first equation we then find $a \equiv b \bmod p$. Thus we have,
\begin{align*}
a \equiv b \equiv c \pmod{p}
\end{align*}Thus we have from the original equations,
\begin{align*}
2a^{2023} \equiv 2a^{2024} \equiv 2a^{2025} \equiv 0 \pmod{p}
\end{align*}from which we easily see $a \equiv 0 \bmod{p}$ and hence are done.
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SilverBlaze_SY
66 posts
#31
Y by
The solution I originally did in the contest:
First, FTSOC, we assume that $p \nmid a$, $p \nmid b$, $p \nmid c$.
$$p \mid a^{2023}+b^{2023}...(i)$$$$p \mid b^{2024}+c^{2024}...(ii)$$$$p \mid a^{2025}+c^{2025}...(iii)$$From $(i)$, we get that: $p \mid a^{2025}+a^2b^{2023}$
Combining this with $(iii)$, $p \mid a^2b^{2023}-c^{2025}$

As $c^{2024} \equiv -b^{2024}$, $p \mid a^2b^{2023}+b^{2024}c \implies p \mid b^{2023}(a^2+bc)$
$\implies \boxed{p \mid a^2+bc}...(A)$
$\implies a^2 \equiv -bc \pmod{p}$

From $(iii)$, we have: $p \mid a^{2025}+c^{2025} \implies p \mid a.(bc)^{1012}+c^{2025}$
$\implies p \mid c^{1012}(ab^{1012}+c^{1013})$
$\implies p \mid a^2b^{2024}-c^{2026}$
Combining with $(ii)$, we get: $p \mid a^2b^{2024}+b^{2024}c^2 \implies p \mid b^{2024}(a^2+c^2)$
$\implies \boxed{p \mid a^2+c^2}...(B)$

Therefore, from $(A)$ and $(B)$, we have: $p \mid c(b-c) \implies b \equiv c \pmod{p}$
But then, in $(ii)$, $p\mid 2b^{2024} \implies p \mid 2$, which is a contradiction!

Then, $p$ must divide one of $a,b,c$. If $p$ divides any one of the numbers, $p$ must divide all the numbers individually, and we're done! :wow:
This post has been edited 1 time. Last edited by SilverBlaze_SY, Apr 27, 2024, 7:17 AM
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shafikbara48593762
17 posts
#32 • 3 Y
Y by E.Sultan, allaith.sh, BR1F1SZ
$$a^{2023} \equiv -b^{2023} \pmod{p} $$$$b^{2024} \equiv -c^{2024} \pmod{p} $$$$c^{2025} \equiv -a^{2025} \pmod{p} $$
Let $n$=2023×2024×2025
$$\implies a^{n} \equiv b^{n} \pmod{p} $$$$\implies b^{n} \equiv -c^{n} \pmod{p} $$$$\implies a^{n} \equiv c^{n} \pmod{p} $$And from this we get
$$c^{n} \equiv -c^{n} \pmod{p} $$and we are done
This post has been edited 2 times. Last edited by shafikbara48593762, May 9, 2024, 2:20 PM
Reason: Nicer
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E.Sultan
7 posts
#33
Y by
shafikbara48593762 wrote:
$$a^{2023} \equiv -b^{2023} \pmod{p} $$$$b^{2024} \equiv -c^{2024} \pmod{p} $$$$c^{2025} \equiv -a^{2025} \pmod{p} $$
Let $n$=2023×2024×2025
$$\implies a^{n} \equiv b^{n} \pmod{p} $$$$\implies b^{n} \equiv -c^{n} \pmod{p} $$$$\implies a^{n} \equiv c^{n} \pmod{p} $$And from this we get
$$c^{n} \equiv -c^{n} \pmod{p} $$and we are done

what a solution!!!
I was surprised by how organized and simple this solution was, it is a MASTERPIECE!
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shafikbara48593762
17 posts
#34
Y by
E.Sultan wrote:
shafikbara48593762 wrote:
$$a^{2023} \equiv -b^{2023} \pmod{p} $$$$b^{2024} \equiv -c^{2024} \pmod{p} $$$$c^{2025} \equiv -a^{2025} \pmod{p} $$
Let $n$=2023×2024×2025
$$\implies a^{n} \equiv b^{n} \pmod{p} $$$$\implies b^{n} \equiv -c^{n} \pmod{p} $$$$\implies a^{n} \equiv c^{n} \pmod{p} $$And from this we get
$$c^{n} \equiv -c^{n} \pmod{p} $$and we are done

what a solution!!!
I was surprised by how organized and simple this solution was, it is a MASTERPIECE!

Thx bro
Such an easy problem for INMO P3
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peppapig_
281 posts
#35
Y by
Note that if $p\mid a$, then
\[a^{2023}+b^{2023}\equiv 0 \mod p,\]\[\iff b^{2023} \equiv 0 \mod p,\]\[\iff p\mid b,\]and similarly we can find that $p\mid c$. In general, using a similar proof, it can be shown that if $p\mid abc$, then $p$ divides each of $a$, $b$, and $c$. FTSOC, assume that $p$ divides none of $a$, $b$, or $c$.

Now, notice that
\[a^{2023}+b^{2023}\equiv 0 \mod p,\]\[\iff a^{2023}\equiv -b^{2023} \mod p,\]\[\iff \left(\frac{a}{b}\right)^{2023}\equiv -1 \mod p,\]\[\iff ord_p\left(\frac{a}{b}\right) \mid 4046 \text{ and } 2\mid ord_p\left(\frac{a}{b}\right),\]since we had that $\left(\frac{a}{b}\right)^{2023}\equiv -1 \mod p$. Similarly, from the second equation, we get that
\[ord_p\left(\frac{b}{c}\right) \mid 4048 \text{ and } 16\mid ord_p\left(\frac{b}{c}\right).\]
Now, since $p$ is prime, it is well-known that it must have a primitive root. Let said primitive root be $g$ and define $m$ and $n$ to be the unique integers such that $\frac{a}{b}\equiv g^m\mod p$ and $\frac{b}{c}\equiv g^n\mod p$, where $1\leq m,n\leq p-1$. Now, notice that if
\[2\mid ord_p\left(\frac{a}{b}\right) \text{ and } ord_p\left(\frac{a}{b}\right) \mid 4046,\]we get that
\[\nu_2\left(ord_p\left(\frac{a}{b}\right)\right)=1,\]which in turn gives that
\[\nu_2(m)=\nu_2(p-1)-1,\]since the order of $g$ mod $p$ is $p-1$. Similarly, if
\[16\mid ord_p\left(\frac{b}{c}\right) \text{ and } ord_p\left(\frac{b}{c}\right) \mid 4048,\]then we have that
\[\nu_2(n)=\nu_2(p-1)-4.\]
Now, notice that
\[\frac{a}{c}\equiv \left(\frac{a}{b}\right)\left(\frac{b}{c}\right)\equiv g^{m+n} \mod p,\]which means that
\[\nu_2\left(ord_p\left(\frac{a}{c}\right)\right)=\nu_2(p-1)-\nu_2(m+n).\]However, since
\[\nu_2(p-1)-1=\nu_2(m)>\nu_2(n)=\nu_2(p-1)-4,\]we get that
\[\nu_2(m+n)=\nu_2(p-1)-4,\]which gives that
\[\nu_2\left(ord_p\left(\frac{a}{c}\right)\right)=\nu_2(p-1)-(\nu_2(p-1)-4)=4.\]
However, we had that
\[c^{2025}+a^{2025}\equiv 0 \mod p,\]\[\iff a^{2025}\equiv -c^{2025} \mod p,\]\[\iff \left(\frac{a}{c}\right)^{2025}\equiv -1 \mod p,\]\[\iff ord_p\left(\frac{a}{c}\right) \mid 4050,\]a contradiction, since $16$ does not divide $4050$. Therefore $p$ must divide each of $a$, $b$, and $c$, as desired, finishing the problem.
This post has been edited 3 times. Last edited by peppapig_, Jun 7, 2024, 8:18 PM
Reason: Typo corrections
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alexanderhamilton124
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#36
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If \( p \) divides one of \( a, b, c \), WLOG \( a \), we have \( p \mid b^{2023} \implies p \mid b \), \( p \mid c^{2025} \implies p \mid c \), so then \( p \) divides all of them.

Assume, for the sake of contradiction, \( p \nmid a, b, c \). We have the following modular congruences:
\[ \left(\frac{a}{b}\right)^{4046} \equiv 1 \pmod{p} \]\[ \left(\frac{b}{c}\right)^{4048} \equiv 1 \pmod{p} \]\[ \left(\frac{c}{a}\right)^{4050} \equiv 1 \pmod{p} \]Firstly, observe that \( p \mid a^{2023} + b^{2023} \implies p \mid a^{2025} + a^2b^{2023} \). Since \( p \mid a^{2025} + c^{2025} \), we have \( p \mid c^{2025} - a^2b^{2023} \). Further, \( p \mid b^{2024} + c^{2024} \implies p \mid c^{2025} + b^{2024}c \). From this, we conclude that:
\[ p \mid b^{2024}c + a^2b^{2023} = b^{2023}(bc + a^2) \implies a^2 \equiv -bc \pmod{p} \]since \( p \nmid b^{2023} \).

We have \( a^{4046} \equiv -(bc)^{2023} \), which, replacing in our first congruence, gives \( c^{2023} \equiv -b^{2023} \pmod{p} \implies c^{4046} \equiv b^{4046} \pmod{p} \). Replacing this in our second congruence, we have
\[ \frac{b^{4046}}{c^{4046}} \cdot \frac{b^2}{c^2} \equiv 1 \pmod{p} \implies b^2 \equiv c^2 \pmod{p} \implies b^{2024} \equiv c^{2024} \pmod{p} \]Therefore, \( b^{2024} + c^{2024} \equiv 2b^{2024} \equiv 0 \pmod{p} \). Since \( p \) is odd, we have \( p \mid b^{2024} \implies p \mid b \), which gives \( p \mid a, c \) as well, so we are done.

Edit: I would like to see a solution that doesn't use the even parity of 2024.
This post has been edited 1 time. Last edited by alexanderhamilton124, Aug 4, 2024, 5:18 PM
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Siddharthmaybe
106 posts
#39
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Raise everything to the power of 2023x2024x2025 mod p, then p divides a^2023x2024x2025 + or - c^2023x2024x2025 which gives p | a and the rest follows
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little-fermat
147 posts
#40
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I have discussed this question in my INMO 2024 video on my channel "little fermat". Here is the Video
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anudeep
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#41
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Showing that any one of $a,b,c$ is divisible by $p$ would suffice. Assume all of them are invertible on $\mathbb{Z}_p$. As $b^{2024}\equiv -ba^{2023}$ we may say,
$$c^{2024}+b^{2024}\equiv c^{2024}-ba^{2023}\quad\text{and}\quad c^{2025}\equiv cba^{2023}.$$We then have $bca^{2023}+a^{2025}\equiv 0$. Since $a$ is invertible we are left with the key result,
$$a^2\equiv -bc.$$Using the above fact, $a(-bc)^{1012}\equiv-c^{2025}$ which deduces to $ab^{1012}\equiv -c^{1013}$ and as $a^{2025}\equiv cb^{2024}$ we can easily show that $b^{1012}\equiv ac^{1011}$. Now we are left with,
$$ab^{1012}\equiv a^2c^{1011}\equiv -c^{1013}\quad\implies\quad a^2\equiv -c^2.$$But we had already seen $a^2\equiv -bc$ and is absurd, contradicting our assumption, hence none of them are invertible as desired.$\square$
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John_Mgr
68 posts
#42
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If any of the $a,b,c$ is divisible by $p$, them all of them are .
FOTSOC, assume that $p\nmid a, b, c$.
$a^{2023}\equiv -b^{2023}\pmod{p} $, $c^{2024}\equiv -b^{2024} \pmod{p} $ and $c^{2025}\equiv -a^{2025}\pmod{p} $
then, $c^{2025}\equiv c\cdot -b^{2024}\equiv -a^{2025}\equiv a^2\cdot b^{2023} \implies b^{2023}(c+ab)\equiv 0\pmod{p} $
$p\nmid b$. so $ab+c \equiv 0\pmod{p} $, $ab\equiv -c\pmod{p}  \rightarrow a^{2023}\cdot b^{2023}\equiv-c^{2023}\pmod{p} \implies (b^2)^{2023}\equiv c^{2023}\pmod{p} \implies b^2 \equiv c\pmod{p} $. Also $ab+c \equiv 0(modp)\implies ab+b^2 \equiv 0\pmod{p} \equiv b(a+b) (modp) $. so, $a\equiv -b\pmod{p} $
$a^{2025}+c^{2025} \equiv -b^{2025}+(b^2)^{2025}\equiv b^{2025}(-1+b^{2025})\equiv 0(modp) \implies b\equiv 1\pmod{p} $
then $b^{2024}+c^{2024}\equiv 0\pmod{p}  \implies b^{2024}+(b^2)^{2024}\equiv 2\equiv 0\pmod{p} $, Contradiction!!
So, $p$ divides and one of them which leads to $p$ divides of them i.e $p\mid a,b,c$.
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