Find all positive integers n, such that $$\sum\limits_{k=1}^{n}(k+2)^n=(n+3)^n $$

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    $\begingroup$ Why don't you put the whole question in the title? $\endgroup$ – Jason DeVito Nov 1 '10 at 1:47

$k^n/(n+3)^n = \left(1-\frac{n+3-k}{n+3}\right)^n$, so using that $\log (1+x) \leq x$ for all $x$, \[ \frac{\sum_{k=3}^{n+2} k^n}{(n+3)^n} \leq \sum_{j=1}^n e^{-jn/(n+3)} \leq \frac{1}{e^{n/(n+3)}-1} \] This last term is decreasing, and for $n=7$ it is strictly less than $1$, so you are left with only $n \leq 6$ to check. It is probably possible to sharpen the inequalities so that there would be less to check.

In the end, you get $n=2$ or $3$.


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