Proof that there is no positive integer $n$ such that $n + 2 | 1^k + 2^k + \dots + n^k$ where $k$ is an odd integer.
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https://math.stackexchange.com/questions/427744/showing-that-1k2k-dots-nk-is-divisible-by-nn1-over-2 – lab bhattacharjee May 12 '19 at 16:21
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@InsideOut: $2+2=4$ does not divide $1^1+2^1=3$ – J. W. Tanner May 12 '19 at 16:35
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I read $k+2$.. I delete it – InsideOut May 12 '19 at 16:36
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Using modular arithmetic ($\text{mod }n+2$); $$\begin{align} 1^k+2^k+\dots+(n-1)^k+n^k &\equiv1^k+2^k+\dots+(-3)^k+(-2)^k\\ &\equiv\begin{cases}1+\left(\frac{n}2+1\right)^k&n\text{ even}\\1&n\text{ odd}\end{cases}\\ \end{align}$$

Peter Foreman
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