We know $qb - pa = 1$ for some positive integers $a, b$ with $1 \le b \le p, 1 \le a \le q$; this is straightforward Bezout and noticing that if $(a, b)$ is solution, then so is $(a - q, b - p)$. If $a \le \frac{q}{2}$ then $b \le \frac{p}{2}$, and the largest prime divisor of $qb$ is $q$, and that of $pa$ is $p$ since $p < \frac{q}{2}$. If $a > \frac{q}{2}$ then $(q - a, p - b)$ satisfies $px - qy = 1$; and repeat same argument, since $q - a < \frac{q}{2}, p - b < \frac{p}{2}$.