设 $a_0,a_1,a_2,\cdots,a_n$ 是正整数，且 $a_0>a_1>a_2>\cdots>a_n>1$，并满足\[\left(1-\dfrac 1{a_1}\right)+\left(1-\dfrac{1}{a_2}\right)+\cdots+\left(1-\dfrac{1}{a_n}\right)=2\left(1-\dfrac{1}{a_0}\right),\]试求出 $(a_0,a_1,a_2,\cdots,a_n)$ 的所有可能的解．
答案 $(24,4,3,2),(60,5,3,2)$．
解析 根据题意，有 $a_k\geqslant n+2-k$，于是\[\begin{split}\left(1-\dfrac 12\right)+\left(1-\dfrac 13\right)+\cdots+\left(1-\dfrac{1}{n+1}\right)&\leqslant \left(1-\dfrac 1{a_1}\right)+\left(1-\dfrac{1}{a_2}\right)+\cdots+\left(1-\dfrac{1}{a_n}\right)\\ &=2\left(1-\dfrac{1}{a_0}\right)\\ &<2,\end{split}\]因此 $n\leqslant 3$． [[case]]情形一[[/case]] $n=1$．此时\[1-\dfrac1{a_1}=2\left(1-\dfrac1{a_0}\right)\iff a_1=\dfrac{a_0}{2-a_0},\]无解． [[case]]情形二[[/case]] $n=2$．此时\[\left(1-\dfrac1{a_1}\right)+\left(1-\dfrac{1}{a_2}\right)=2\left(1-\dfrac1{a_0}\right)\iff \dfrac2{a_0}=\dfrac{1}{a_1}+\dfrac{1}{a_2},\]而 $\dfrac{1}{a_0}<\dfrac{1}{a_1}<\dfrac{1}{a_2}$，因此该方程无解． [[case]]情形三[[/case]] $n=3$．此时\[\left(1-\dfrac1{a_1}\right)+\left(1-\dfrac{1}{a_2}\right)+\left(1-\dfrac{1}{a_3}\right)=2\left(1-\dfrac1{a_0}\right)\iff 1+\dfrac{2}{a_0}=\dfrac{1}{a_1}+\dfrac1{a_2}+\dfrac{1}{a_3},\]于是 $a_3=2$，$a_2=3$，否则\[\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}<\dfrac{1}{2}+\dfrac{1}{4}+\dfrac 15<1,\]进而\[\dfrac{2}{a_0}+\dfrac16=\dfrac1{a_1}\iff a_0=\dfrac{12a_1}{6-a_1},\]其中 $a_1\geqslant 4$，因此 $(a_1,a_0)=(4,24),(5,60)$． 综上所述，$(a_0,a_1,a_2,\cdots,a_n)$ 的所有可能的解为 $(24,4,3,2),(60,5,3,2)$．