Indian National Mathematical Olympiad

Set $k = 2023$. If one of $a, b, c$ is divisible by $p$, then all of them are. Indeed, for example, if $p \mid a$, then $p \mid a^k + b^k$ implies $p \mid b$, and then $p \mid b^{k+1} + c^{k+1}$ implies $p \mid c$. The other cases follow similarly.
So for the sake of contradiction assume none of $a, b, c$ is divisible by $p$. Then
$$a^{k(k+2)} \equiv (a^k)^{k+2} \equiv (-b^k)^{k+2} \equiv -b^{k(k+2)} \pmod{p}$$
and
$$a^{k(k+2)} \equiv (a^{k+2})^k = (-c^{k+2})^k \equiv -c^{k(k+2)} \pmod{p}.$$
So $b^{k(k+2)} \equiv c^{k(k+2)} \pmod{p}$. But then
$$c^{k(k+2)} \cdot c \equiv c^{(k+1)^2} \equiv (-b^{k+1})^{k+1} \equiv b^{(k+1)^2} \equiv b^{k(k+2)} \cdot b \equiv c^{k(k+2)} \cdot b \pmod{p}$$
which forces $b \equiv c \pmod{p}$. Thus
$$0 \equiv b^{k+1} + c^{k+1} = 2b^{k+1} \pmod{p}$$
implying $p \mid b$, a contradiction. Thus the proof is complete. □

As before, we may assume $p$ divides none of $a, b$, and $c$ and set $k = 2023$. Then
$$a^k \equiv -b^k \pmod{p}$$
$$b^{k+1} \equiv -c^{k+1} \pmod{p}$$
$$c^{k+2} \equiv -a^{k+2} \pmod{p}$$
and multiplying these three equations yields $a^k b^{k+1} c^{k+2} \equiv -b^k c^{k+1} a^{k+2} \pmod{p}$. By cancelling the factor $a^k b^k c^{k+1}$, we get $a^2 \equiv -bc \pmod{p}$. Now
$$p \mid a^k + b^k \implies a^{4k} \equiv b^{4k} \pmod{p} \implies c^{2k} \equiv b^{2k} \pmod{p}$$
so
$$p \mid b^{k+1} + c^{k+1} \implies b^{2(k+1)} \equiv c^{2(k+1)} \pmod{p} \implies b^2 \equiv c^2 \pmod{p}$$
so either $b \equiv c \pmod{p}$ or $b \equiv -c \pmod{p}$. In the latter case, $a^2 \equiv c^2 \pmod{p}$ so $a \equiv c \pmod{p}$ or $a \equiv -c \pmod{p}$. In any case, two out of $\{a, b, c\}$ are the same mod $p$, so one of the equations gives $p \mid 2x^y$ where $x \in \{a, b, c\}$ and $y \in \{k, k+1, k+2\}$, hence $p$ odd implies $p \mid x$ so $p \mid abc$, the desired contradiction. □

We have
$$a^{2023} \equiv -b^{2023} \pmod{p} \qquad (1)$$
$$b^{2024} \equiv -c^{2024} \pmod{p} \qquad (2)$$
$$c^{2025} \equiv -a^{2025} \pmod{p} \qquad (3)$$
Thus,
$$\begin{align*} a^{2023 \cdot 2024 \cdot 2025} &\equiv b^{2023 \cdot 2024 \cdot 2025} \pmod{p} & \text{by (1)} \\ &\equiv -c^{2023 \cdot 2024 \cdot 2025} \pmod{p} & \text{by (2)} \\ &\equiv -a^{2023 \cdot 2024 \cdot 2025} \pmod{p} & \text{by (3)} \end{align*}$$
Thus, $p \mid 2 \cdot a^{2023 \cdot 2024 \cdot 2025}$ and hence $p \mid a$ since $p$ is odd. Now, finish as before. □