<s>[INST] 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.
 [/INST] You should be able to quickly see that $(a+b+c)$ must be a square. Since $999<1024$, we know that $(a+b+c)<16$ so we have ${a+b+c}\in\{1,4,9\}$, and thus $\sqrt{(a+b+c)^5}\in\{1,32,243\}$, the last of which produces a valid solution. Since you form $\overline{abc}$ as a number I'll assume $a$ cannot be zero.
</s>