Chinese Girls' Mathematical Olympiad

Proof: By contradiction, suppose there exist two different pairs of positive integer solutions $(x_1, y_1)$ and $(x_2, y_2)$. Since $x_1$ and $y_1$ are coprime, it follows that $p \nmid x_1y_1$. Similarly, $p \nmid x_2y_2$. Given
$$ax_1^2 \equiv -by_1^2 \pmod{p^r}, \quad ax_2^2 \equiv -by_2^2 \pmod{p^r},$$
it is known that $abx_1^2y_2^2 \equiv abx_2^2y_1^2 \pmod{p^r}$. Since $p \nmid ab$, we have $p^r \mid (x_1^2y_2^2 - x_2^2y_1^2)$.
Note that $x_1y_2 - x_2y_1$ and $x_1y_2 + x_2y_1$ cannot both be divisible by $p$, otherwise $p \mid 2x_1y_2$, which contradicts with $p$ being an odd prime and $p \nmid x_1y_1x_2y_2$. Hence, $p^r \mid x_1y_2 - x_2y_1$, or $p^r \mid x_1y_2 + x_2y_1$.
If $x_1y_2 - x_2y_1 = 0$, then $\frac{x_1}{x_2} = \frac{y_1}{y_2}$, combined with $ax_1^2 + by_1^2 = ax_2^2 + by_2^2$, implies $x_1 = x_2$, $y_1 = y_2$, contradicting with $(x_1, y_1) \neq (x_2, y_2)$. Therefore, $x_1y_2 - x_2y_1 \neq 0$.
If $p^r \mid x_1y_2 + x_2y_1$, then $x_1y_2 + x_2y_1 \ge p^r$.
If $p^r \mid x_1y_2 - x_2y_1$, then $x_1y_2 + x_2y_1 \ge |x_1y_2 - x_2y_1| \ge p^r$. Thus, in all cases,
$$x_1y_2 + x_2y_1 \ge p^r. \quad (*)$$
Using the condition $ab > m^2$ and $(*)$, we get
$$\begin{aligned} m^2p^{2r} &= (ax_1^2 + by_1^2)(ax_2^2 + by_2^2) \\ &= (ax_1x_2 - by_1y_2)^2 + ab(x_1y_2 + x_2y_1)^2 \\ &\ge ab(x_1y_2 + x_2y_1)^2 \\ &> m^2p^{2r}, \end{aligned}$$
a contradiction. Therefore, the assumption by contradiction is false, and the original statement holds. $\Box$