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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Show that three lines concur
benjaminchew13   2
N a minute ago by benjaminchew13
Source: Revenge JOM 2025 P2
t $A B C$ be a triangle. $M$ is the midpoint of segment $B C$, and points $E$, $F$ are selected on sides $A B$, $A C$ respectively such that $E$, $F$, $M$ are collinear. The circumcircles $(A B C)$ and $(A E F)$ intersect at a point $P != A$. The circumcircle $(A P M)$ intersects line $B C$ again at a point $D != M$. Show that the lines $A D$, $E F$ and the tangent to $(A E F)$ at point $P$ concur.
2 replies
benjaminchew13
13 minutes ago
benjaminchew13
a minute ago
slightly easy NT fe
benjaminchew13   2
N 3 minutes ago by benjaminchew13
Source: Revenge JOM 2025 P1
Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $$f(a) + f(b) + f(c) | a^2 + af(b) + cf(a)$$for all $a, b, c\in\mathbb{N}$
2 replies
benjaminchew13
16 minutes ago
benjaminchew13
3 minutes ago
Cheesy's math casino
benjaminchew13   1
N 4 minutes ago by benjaminchew13
Source: Revenge JOM 2025 P4
There are $p$ people playing a game at Cheesy's math casino, where $p$ is an odd prime number. Let $n$ be a positive integer. A subset of length $s$ from the set of integers from $1$ to $n$ inclusive is randomly chosen, with an equal probability ($s <= n$ and is fixed). The winner of Cheesy's game is person $i$, if the sum of the chosen numbers are congruent to $i mod p$ for $0 <= i <= p - 1$.

For each $n$, find all values of $s$ such that no one will sue Cheesy for creating unfair games (i.e. all the winning outcomes are equally likely).
1 reply
1 viewing
benjaminchew13
10 minutes ago
benjaminchew13
4 minutes ago
2013 Japan MO Finals
parkjungmin   0
4 minutes ago
help me

we cad do it
0 replies
parkjungmin
4 minutes ago
0 replies
Square number
linkxink0603   4
N 4 hours ago by pooh123
Find m is positive interger such that m^4+3^m is square number
4 replies
linkxink0603
Yesterday at 11:20 AM
pooh123
4 hours ago
Inequalities
sqing   7
N 5 hours ago by sqing
Let $ a,b>0, a^2+ab+b^2 \geq 6  $. Prove that
$$a^4+ab+b^4\geq 10$$Let $ a,b>0, a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$a^4+ab+b^4  \leq 10$$Let $ a,b>0,  a^2+ab+b^2 \geq \frac{15}{2}  $. Prove that
$$ a^4-ab+b^4\geq 10$$Let $ a,b>0,  a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$-\frac{1}{8}\leq  a^4-ab+b^4\leq 10$$
7 replies
sqing
Thursday at 2:42 PM
sqing
5 hours ago
Compilation of functions problems
Saucepan_man02   2
N Today at 12:45 AM by Saucepan_man02
Could anyone post some handout/compilation of problems related to functions (difficulty similar to AIME/ARML/HMMT etc)?

Thanks..
2 replies
Saucepan_man02
May 7, 2025
Saucepan_man02
Today at 12:45 AM
How many triangles
Ecrin_eren   5
N Today at 12:10 AM by jasperE3


"Inside a triangle, 2025 points are placed, and each point is connected to the vertices of the smallest triangle that contains it. In the final state, how many small triangles are formed?"


5 replies
Ecrin_eren
May 2, 2025
jasperE3
Today at 12:10 AM
Triangle on a tetrahedron
vanstraelen   2
N Yesterday at 7:51 PM by ReticulatedPython

Given a regular tetrahedron $(A,BCD)$ with edges $l$.
Construct at the apex $A$ three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance $l$ from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
2 replies
vanstraelen
Yesterday at 2:43 PM
ReticulatedPython
Yesterday at 7:51 PM
shadow of a cylinder, shadow of a cone
vanstraelen   2
N Yesterday at 6:33 PM by vanstraelen

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
2 replies
vanstraelen
Yesterday at 3:08 PM
vanstraelen
Yesterday at 6:33 PM
2023 Official Mock NAIME #15 f(f(f(x))) = f(f(x))
parmenides51   3
N Yesterday at 5:13 PM by jasperE3
How many non-bijective functions $f$ exist that satisfy $f(f(f(x))) = f(f(x))$ for all real $x$ and the domain of f is strictly within the set of $\{1,2,3,5,6,7,9\}$, the range being $\{1,2,4,6,7,8,9\}$?

Even though this is an AIME problem, a proof is mandatory for full credit. Constants must be ignored as we dont want an infinite number of solutions.
3 replies
parmenides51
Dec 4, 2023
jasperE3
Yesterday at 5:13 PM
Geometry
AlexCenteno2007   3
N Yesterday at 4:18 PM by AlexCenteno2007
Let ABC be an acute triangle and let D, E and F be the feet of the altitudes from A, B and C respectively. The straight line EF and the circumcircle of ABC intersect at P such that F is between E and P, the straight lines BP and DF intersect at Q. Show that if ED = EP then CQ and DP are parallel.
3 replies
AlexCenteno2007
Apr 28, 2025
AlexCenteno2007
Yesterday at 4:18 PM
Cube Sphere
vanstraelen   4
N Yesterday at 2:37 PM by pieMax2713

Given the cube $\left(\begin{array}{ll} EFGH \\ ABCD \end{array}\right)$ with edge $6$ cm.
Find the volume of the sphere passing through $A,B,C,D$ and tangent to the plane $(EFGH)$.
4 replies
vanstraelen
Yesterday at 1:10 PM
pieMax2713
Yesterday at 2:37 PM
Combinatorics
AlexCenteno2007   0
Yesterday at 2:05 PM
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
0 replies
AlexCenteno2007
Yesterday at 2:05 PM
0 replies
Easy Number Theory
math_comb01   38
N 2 hours ago by lakshya2009
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
38 replies
math_comb01
Jan 21, 2024
lakshya2009
2 hours ago
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
1609 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
182 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
............
This post has been edited 1 time. Last edited by Samujjal101, Jan 21, 2024, 4:53 PM
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Samujjal101
2799 posts
#15
Y by
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
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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
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#21
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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
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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
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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
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#26
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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
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#27
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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
1005 posts
#29
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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
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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
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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
395 posts
#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
Y by
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
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#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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lakshya2009
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#43
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the second inmo problem that I solved
solution
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