Solve the equation $ (100 a+10b+c)^2 =(a+b+c)^5.$ Find a three-digits number  $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that 
$$
(a+b+c)^5 \leq 999^2  \implies a+b+c< \sqrt[5]{999^2}\leq 15
$$
and 
$$
(100 a+10b+c)^2<15^5 \implies 100 a+10b+c>\sqrt{15^5} \leq 871.
$$
Also 
$$
(100 a+10b+c)^2 =(a+b+c) \mod 2 
$$
implies $a+b=0 \mod 2$.
Similarly 
$$
(100 a+10b+c)^2 = c^2 = (a+b+c) \mod 5. 
$$
But it not enough to find the solution $243$.
No more ideas.

Let
$$x = \sqrt{a+b+c} = \sqrt[5]{\overline{abc}},\quad a\in \{1, 2\dots9\}, \quad b,c\in\{0,1\dots9\},$$
then 
$$\begin{cases}
\sqrt{1} <= x < \sqrt{27}\\
\sqrt[5]{100} < x < \sqrt[5]{999}\\
x\in\mathbb N,
\end{cases}$$
$$\begin{cases}2.5 < x < 3.99\\
x\in\mathbb N,
\end{cases}$$
$$x=3,$$
$$\begin{cases}
\overline{abc} = x^5 = 243\\
a+b+c = x^2 = 9,
\end{cases}$$
$$\boxed{a=2,\quad b=4,\quad b=3}.$$
