22nd 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 $p\nmid x_1{y}_1$. Similarly, $p\nmid x_2{y}_2$. Given $$a{x}_1^2\equiv -b{y}_1^2(\mathrm{mod}{p}^r),a{x}_2^2\equiv -b{y}_2^2(\mathrm{mod}{p}^r),$$ it is known that $ab{x}_1^2{y}_2^2\equiv ab{x}_2^2{y}_1^2(\mathrm{mod}{p}^r)$. Since $p\nmid ab$, we have $p^r\mid (x_1^2{y}_2^2-x_2^2{y}_1^2)$. Note that $x_1{y}_2-x_2{y}_1$ and $x_1{y}_2+x_2{y}_1$ cannot both be divisible by $p$, otherwise $p\mid 2x_1{y}_2$, which contradicts with $p$ being an odd prime and $p\nmid x_1{y}_1{x}_2{y}_2$. Hence, $p^r\mid x_1{y}_2-x_2{y}_1$, or $p^r\mid x_1{y}_2+x_2{y}_1$. If $x_1{y}_2-x_2{y}_1=0$, then $\frac{x_1}{x_2}=\frac{y_1}{y_2}$, combined with $a{x}_1^2+b{y}_1^2=a{x}_2^2+b{y}_2^2$, implies $x_1=x_2$, $y_1=y_2$, contradicting with $(x_1,y_1)\ne (x_2,y_2)$. Therefore, $x_1{y}_2-x_2{y}_1\ne 0$. If $p^r\mid x_1{y}_2+x_2{y}_1$, then $x_1{y}_2+x_2{y}_1\ge p^r$. If $p^r\mid x_1{y}_2-x_2{y}_1$, then $x_1{y}_2+x_2{y}_1\ge |x_1{y}_2-x_2{y}_1|\ge p^r$. Thus, in all cases, $$x_1{y}_2+x_2{y}_1\ge p^r.(\ast )$$ Using the condition $ab>m^2$ and $(\ast )$, we get $$\begin{array}{rl}{m}^2{p}^{2r}& =(a{x}_1^2+b{y}_1^2)(a{x}_2^2+b{y}_2^2)\\ & =(a{x}_1{x}_2-b{y}_1{y}_2{)}^2+ab(x_1{y}_2+x_2{y}_1{)}^2\\ & \ge ab(x_1{y}_2+x_2{y}_1{)}^2\\ & >m^2{p}^{2r},\end{array}$$ a contradiction. Therefore, the assumption by contradiction is false, and the original statement holds. $\square$