We use induction to prove that $ LHS<RHS $ $ \forall n \ge 5 $ ((which means $ n $ isn't a solution when $ n \ge 5 $ For $ n=5 $, it's easy to check (I just do it with wolframalpha ) Suppose it's true for $ n=5, 6, ..., k $. Now, we consider the case $ n=k+1 $ Then we have the following inequalities: $ LHS < (k+2){(k+3)}^{k}+{(k+3)}^{k+1} < 2{(k+3)}^{k+1} $ And also $ (\frac{k+4}{k+3})^{k+2} \ge 1+\frac{k+2}{k+3}+\frac{(k+2)(k+1)}{2(k+3)^2}>2 $ Thus, $ 3^{k+1}+4^{k+1}+...+{(k+2)}^{k+1}+{(k+3)}^{k+1} < {(k+4)}^{k+1} $ So, we finish this induction. Finally, we could easily check $ n=1,2,3,4 $ is a solution or not. Answer: $ n=3,4 $