具体反应步骤：
$$ \text{C}_{15}\text{H}_{22}\text{O}_2 \xrightarrow{\text{O}_2,\ h\nu,\ \text{光敏剂}} \text{C}_{15}\text{H}_{24}\text{O}_2 \xrightarrow{\text{H}^+,\ \text{重排}} \text{C}_{15}\text{H}_{22}\text{O}_5 $$详细的三步合成法：
$$ \begin{aligned} \text{Step 1:}&\quad \text{青蒿酸} + \text{O}_2 \xrightarrow{h\nu,\ \text{Rose Bengal}} \text{氢过氧化物中间体} \\ \text{Step 2:}&\quad \text{氢过氧化物} \xrightarrow{\text{还原}} \text{二氢青蒿酸} \\ \text{Step 3:}&\quad \text{二氢青蒿酸} \xrightarrow{\text{H}^+,\ \text{O}_2,\ \text{光氧化}} \text{青蒿素}\ (\text{C}_{15}\text{H}_{22}\text{O}_5) \end{aligned} $$$$ \begin{aligned} c^d &\equiv (m^e)^d \pmod{n} \\ &\equiv m^{ed} \pmod{n} \\ &\equiv m^{1 + k\varphi(n)} \pmod{n} \quad (\text{因为 } ed \equiv 1 \pmod{\varphi(n)}) \\ &\equiv m \cdot (m^{\varphi(n)})^k \pmod{n} \\ &\equiv m \cdot 1^k \pmod{n} \quad (\text{由欧拉定理 } m^{\varphi(n)} \equiv 1 \pmod{n}) \\ &\equiv m \pmod{n} \end{aligned} $$