See https://artofproblemsolving.com/community/q1h1809843p12051170 https://artofproblemsolving.com/community/c6h1950915p13466196

Basically, note that $U<6$ or else the LHS must be smaller than the RHS because $10^4<7^6$. Also note that $U>1$ or else the LHS must be greater than the RHS because $10^3>28$. If $U=5$, then $(P+E+R+5)^5$ must end in $5$, which implies $P+E+R+5$ is end in $5$. However, since $P>0$, then $P+E+R+5$ must be at least $15$, making the LHS smaller than the RHS because $10^4<15^5$ If $U=4$, then $(P+E+R+4)^4$ must end in a $4$. But it is easily verified that no integer raised to the fourth power ends in a four by using mod 10. If $U=3$, then $(P+E+R+3)^3$ must end in a $3$. It is easily verified that this implies $P+E+R+3$ ends in a $7$ by using mod 10. But $7^3<10^3$ and $27^3>10^4$, so $P+E+R+3$ must be $17$, so $\overline{PERU}=17^3$. We see this indeed does work because $(4+9+1+3)^3=\boxed{4913}$, so this is our desired number. If $U=2$, then $(P+E+R+2)^2$ must end in a $2$. But it is easily verified that no perfect square ends in a two by using mod 10. This completes our search $\ \blacksquare$