已知 $f\left( x \right)$ 为一元二次函数，且 $a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$ 为正项等比数列，求证：$f\left( a \right) = a$．
【难度】
【出处】
2012年北京大学保送生试题
【标注】
【答案】
略
【解析】
设 $f\left( x \right) = m{x^2} + nx + t\left( {m \ne 0} \right)$，数列 $a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$ 的公比为 $q\left( {q > 0} \right)$，
则 $f\left( a \right) = aq$，$f\left( {f\left( a \right)} \right) = f\left( {aq} \right) = a{q^2}$，$f\left( {f\left( {f\left( a \right)} \right)} \right) = f\left( {a{q^2}} \right) = a{q^3}$，所以$$\begin{cases} m{a^2} + na + t = aq,\\ m{(aq)^2} + naq + t = a{q^2},\\ m{(a{q^2})^2} + na{q^2} + t = a{q^3},\end{cases}$$进而有$$\begin{cases} m{a^2}\left( {1 - {q^2}} \right) + na\left( {1 - q} \right) = aq\left( {1 - q} \right),\\ m{a^2}{q^2}\left( {1 - {q^2}} \right) + naq\left( {1 - q} \right) = a{q^2}\left( {1 - q} \right).\end{cases}$$若 $q = 1$，则 $f\left( a \right) = a$；
若 $q \ne 1$，则 $m{a^2}\left( {1 + q} \right) + na = aq$ 与 $m{a^2}q\left( {1 + q} \right) + na = aq$ 矛盾．所以 $f\left( a \right) = a$．
则 $f\left( a \right) = aq$，$f\left( {f\left( a \right)} \right) = f\left( {aq} \right) = a{q^2}$，$f\left( {f\left( {f\left( a \right)} \right)} \right) = f\left( {a{q^2}} \right) = a{q^3}$，所以$$\begin{cases} m{a^2} + na + t = aq,\\ m{(aq)^2} + naq + t = a{q^2},\\ m{(a{q^2})^2} + na{q^2} + t = a{q^3},\end{cases}$$进而有$$\begin{cases} m{a^2}\left( {1 - {q^2}} \right) + na\left( {1 - q} \right) = aq\left( {1 - q} \right),\\ m{a^2}{q^2}\left( {1 - {q^2}} \right) + naq\left( {1 - q} \right) = a{q^2}\left( {1 - q} \right).\end{cases}$$若 $q = 1$，则 $f\left( a \right) = a$；
若 $q \ne 1$，则 $m{a^2}\left( {1 + q} \right) + na = aq$ 与 $m{a^2}q\left( {1 + q} \right) + na = aq$ 矛盾．所以 $f\left( a \right) = a$．
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