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.

$(a+b+c)$ must be a square, so it can only be $1,4,9,16$ or $25$.
By your inequalities we have $a+b+c=1,4$ or $9$.
notice that $(100a+10b+c)^2\geq 100^2$.On the other hand $1^5$ and $4^5$ are too small.
We conclude that $a+b+c=9$.
So now we must have $(100a+10b+c)=\sqrt{9^5}=3^5=243$
