<s>[INST] Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.

How can I show this equation has no solutions for $n>5$.

Thanks.
 [/INST] Let $n\ge5$ and suppose $$ \sum_{k=3}^{n+2}k^n<(n+3)^n.$$Then $$ \sum_{k=3}^{n+3} k^n <2(n+3)^n,$$ and we deduce \begin{align} \sum_{k=3}^{n+3} k^{n+1}<2(n+3)^{n+1}&<(n+4)^{n+1} \\ \left(1+\frac1{n+3}\right)^{n+1}\ge\left(\frac98\right)^6&>2, \end{align} where he have used the increasingness of $f(x)=\left(1+\frac1{x+3}\right)^{x+1}$ for $x>-3$: $f'(x)>0$ follows from $$ (x^2+6x+12)\log\left(1+\frac1{x+3}\right)>(x+3)^2((x+3)^{-1}-(x+3)^{-2})=x+2>x+1.$$
By induction, LHS < RHS for all $n\ge5$, and as you noted, it also holds for $n=4$.
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