$p$ is an odd prime, and $a,b,c$ are integers.
If $p$ divides each of
$$a^{2023}+b^{2023},\ b^{2024}+c^{2024}, \ c^{2025}+a^{2025}$$prove that $p$ divides $a,b,c$.
This question appeared in INMO 2024 (the contest is over now).
This is how I did it:
$$a^{2023} \equiv -b^{2023} \pmod p$$
$$b^{2024} \equiv -c^{2024} \pmod p$$
$$c^{2025} \equiv -a^{2025} \pmod p$$
Assume, for the sake of contradiction, $p$ does not divide all of them. $\implies$ $p$ does not divide any of them (it can not divide exactly two or one of them).
So, it makes sense to divide and multiply congruences by $a,b,c$.
Multiplying the above three and cancelling terms, we have:
$$-bc \equiv a^2 \pmod p$$
Now, we put this in the congruences:
$$-b^{2023} \equiv a^{2023} \equiv a(a^{2022}) \equiv -ab^{1011}c^{1011} \pmod p$$
$$\implies ab^{1012} \equiv a^2c^{1011} \pmod p$$
Putting $a^2 \equiv -bc \pmod p$ in the other congruence:
$$-c^{2025} \equiv a^{2025} \equiv a(a^{2024}) \equiv a(-bc)^{1012} \equiv ab^{1012}c^{1012} \equiv a^2c^{1011}c^{1012} \pmod p$$
$$\implies c^2 \equiv -a^2 \equiv bc \pmod p$$
$$\implies b \equiv c \pmod p$$
$$b^{2024} \equiv c^{2024} \equiv -b^{2024} \pmod p$$
Since $p$ is odd, this implies $p$ divides $b$, a contradiction.
However, this seems too easy to be true. Is there another way to present the proof that clarifies the correctness and key idea?
Best Answer
Clearer: we have $i,\color{#c00}j,k = 2023,\color{#c00}{2024},2025$ with $\color{#c00}j$ even, $\color{#90f}{i,k}$ odd, so we get a $\rm\color{darkorange}{contradiction}$ by raising the congruences to a common power $\,m = ijk\,$ then taking their product, i.e.
$$\begin{align} [a^i \equiv -b^i]^{jk}\ \Rightarrow\ \color{#0af}{a^m}&\equiv\ \color{#0a0}{b^m}\ \ \ \ \ {\rm by}\ \ \color{#c00}jk\ \ \rm even\\[.2em] [b^j \equiv -c^j]^{ki}\ \Rightarrow\ \color{#0af}{b^m}&\equiv \color{#0a0}{-c^m}\ \ \ {\rm by}\ \ \color{#90f}{ki}\ \ \rm odd\\[.2em] [c^k \equiv -a^k]^{ij}\ \Rightarrow\ \color{#0af}{c^m}&\equiv\ \color{#0a0}{a^m}\ \ \ \ \ {\rm by}\ \ i\color{#c00}j\ \ \rm even_{\phantom{|}} \\[.2em] \hline {\rm multiplying\ rows}\ \Rightarrow\ \color{#0af}{(abc)^m}&\equiv \color{#0a0}{-(abc)^{m^{\phantom|}}} \color{darkorange}{\Rightarrow\!\Leftarrow} \end{align}\qquad\qquad$$
Remark $ $ This boils down to a multiplicative form of a basic ($p$-adic) nonintegrality test for a sum of fractions, i.e. for $\,x,y,z = a/b,\,b/c,\,c/a\,$ the above proof is equivalent to the first line below. The 2nd line is an additive analog, which is rewritten in fraction language in the 3rd line.
$\begin{align} -1 = x^i = y^j = z^k,\:\!\ 1\:\! = x\:\!\cdot\:\! y\:\!\cdot\:\! z \,\Rightarrow\ &1^m\! = (x\cdot y\cdot z)^m\! =\! (-1)^s\text{ contra $\rm\color{#90f}{odd}$ } s \!=\! \!\color{#90f}{ki}\!+\!\color{#c00}j(k\!+\!i)\\[.4em] 1/2 = ix = jy = kz,\ n=x\!+\!y\!+\!z \, \Rightarrow\ &\!mn = m(x\!+\!y\!+\!z)\! = s/2\:\! \text{ contra odd } s,\ mn\in\Bbb Z\\ {\rm i.e.}\ \ \ x\!+\!y\!+\!z\, =\ \ &\!\! \frac{1}{2i} \!+\! \frac{1}{2\color{#c00}j}\!+\!\frac{1}{2k} \not\in \Bbb Z\,\text{ if exactly $\rm\color{#c00}{one}$ of $\,i,\color{#c00}j,k\,$ is even} \end{align}$
The prior line generalizes from $\,2\,$ to any prime $\,p,\,$ i.e. a sum of fractions is nonintegral if the highest power of $\,p\,$ in any denominator occurs in exactly $\rm\color{#c00}{one}$ denominator, and the above multiplicative form generalizes using this $ $ (replace $\,{-}1\,$ by any element of order $\,p,\,$ e.g. use $\,\zeta_p\,$ (a primitive $\:\!p$'th root of $1).\,$
This basic fact about fractions is ubiquitous in number theory but - curiously - it seems it is not widely taught in elementary courses (witness: this simple proof with that key idea got $> 300$ upvotes). That pedagogical gap will be remedied when one studies advanced number theory, where this method is a trivial prototypical example of local-global methods (valuation theory and $p$-adics). Further, group theory rigorously explains the analogy between the additive and multiplicative forms (see also posts on denominator and order ideals). After mastering these generalizations and abstractions, it is much easier to recognize such reformulations - whether their genesis is nature or a sneaky contest problem composer attempting to obfuscate such basic results to complicate solutions.