We have $V+U+Q+A+R=1+2+3+4+5=15$ thus the equation is $$V^{{{U^Q}^A}^R}(V-U-Q+A+R)=(3^2.5^2)$$ We have $1 \leq V-U-Q+A+R \leq 9$ thus $V-U-Q+A+R \in \{1,3,5,9\}$ If $V-U-Q+A+R=1$ then $V^{{{U^Q}^A}^R}=3^2.5^2$ which it a contradiction because the the right hand side is a power on an integer less that $5$. If $V-U-Q+A+R=3$ then $V^{{{U^Q}^A}^R}=3.5^2$ which it a contradiction because the the right hand side is a power on an integer less that $5$. If $V-U-Q+A+R=5$ then $V^{{{U^Q}^A}^R}=3^2.5$ which it a contradiction because the the right hand side is a power on an integer less that $5$. We have $$V^{{{U^Q}^A}^R}=25$$ so $V=5$, $U=2$, $Q=1$, $A$ and $R$ are in $\{3,4\}$.

IstekOlympiadTeam wrote: All letters in the word $VUQAR$ are different and chosen from {1,2,3,4,5}. Find all solutions to the equation \[(V+U+Q+A+R)^2 : (V-U-Q+A+R)=V^{{{U^Q}^A}^R}\] $V + U + Q + A + R = \sum_{i=1}^{5} = 15$ $\implies$ $V^{U^{Q^{A^{R}}}} = 3^2*5^2*(V + A + R - U - Q)$ $\implies$ $V\in{3,5}$ and $U\in{1,2}$. If $U = 2$ $\implies$ $V=5$ and $Q=1$,$A\in{3,4}$ and $R = 7 - A$. If $U = 1$ $\implies$ if $Q = 4$ and $5V = 5^2*3^2$,contradiction $1 \leq V \leq 5$,if $Q = 2$ we obtain $V = 5^2$,contradiction. The solutions are $(V,U,Q,A,R)\in{(5,2,1,3,4),(5,2,1,4,3)}$.

Hi! Here's my solution.

We must have \[\frac{15^2}{15-2(U+Q)}\]as a positive integer, therefore we have $15-2(U+Q)|15^2 \implies (U,Q)=(3,4),(1,2),(2,3),(1,4),(1,5),(2,4)$ up to permutations. However, they give the desired expression as $15^2, 25, 45, 75$ which clearly only $25$ works as the others are not powers of a number in the set $\{1,2,3,4,5\}$. Hence, we must have $V=5, U=2, Q=1, (A,R)=(3,4) \ \textrm{or} \ (4,3)$.

@tchebytchev, I think you could reach easier to the desired result by finding whitch letter is equal to 1, from the RHS.... After that everything is clear and you can simply find the two solutions: (V, U, Q, A, R) = (5, 2, 1, 4, 3), (5, 2, 1, 3, 4)