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

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)

isn't that this question toooo easy for INMO p3

LuciferMichelson wrote: isn't that this question toooo easy for INMO p3 On what sample set are you basing your opinion on difficulty?

Proposed by Navilarekallu Tejaswi

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

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.$\blacksquare$

Yeah basically belows solution I got.

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.

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 $p > 2$.

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. $\blacksquare$

............

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 ..............

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!

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 $p$?

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

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. $\blacksquare$

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.

My attempt,

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!

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

Master_of_Aops wrote: ... to get terms $a^{xyz}, b^{xyz}, c^{xyz} (mod p)$ and done How exactly are you done from here?

You get $a^{xyz} \equiv b^{xyz}, b^{xyz} \equiv -c^{xyz}, a^{xyz} \equiv c^{xyz}(mod p)$ so all are divisible by $p$

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

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. $\blacksquare$

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. $\square$

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.

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!

$$a^{2023} \equiv -b^{2023} \pmod{p} $$$$b^{2024} \equiv -c^{2024} \pmod{p} $$$$c^{2025} \equiv -a^{2025} \pmod{p} $$ 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

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!

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

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.

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.

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

I have discussed this question in my INMO 2024 video on my channel "little fermat". Here is the Video