<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] For $n \geq 6$ we have
$$\frac{(n+3)^n}{(n+2)^n} = \left( 1+\frac{1}{n+2} \right)^n = \left( 1+\frac{1}{n+2} \right)^{n+2} \cdot \left(\frac{n+2}{n+3}\right)^2$$
Both factors are monotonely increasing, hence this is at least its value for $n=6$, which is $\left(\frac{9}{8}\right)^6 \approx 2.072\dotsc > 2$.
Since $k \mapsto \frac{k+1}{k}$ is monotonely decreasing, we obtain
$$\frac{(k+1)^n}{k^n} \geq \frac{(n+3)^n}{(n+2)^n} > 2$$
for all $k = 1, \dotsc, n+2$, if $n \geq 6$. We can write this as
$$(k+1)^n-k^n > k^n$$
for all $k = 1, \dotsc, n+2$, if $n \geq 6$. This also holds for $k=0$.
In particular we have $$(n+3)^n = \sum_{k=0}^{n+2} ((k+1)^n-k^n) > \sum_{k=0}^{n+2} k^n = 1^n + 2^n + 3^n + \dotsb + (n+2)^n$$
for $n\geq 6$, an even stronger result.
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