n=p^n −pn n=?


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Question Number 6982 by FilupSmith last updated on 04/Aug/16

$n = p^{n} - p n$ $n = ?$
Commented by sou1618 last updated on 05/Aug/16

$i f n \in N . . . .$ $n = p (p^{n - 1} - n) \Rightarrow n = m_{1} p (m_{1} \in N)$ $m_{1} p = p (p^{n - 1} - m_{1} p)$ $m_{1} = (p^{m_{1} p - 1} - m_{1} p)$ $m_{1} = p (p^{m_{1} p - 2} - m_{1}) \Rightarrow m_{1} = m_{2} p (m_{2} \in N)$ $. . . . w h e n m_{x} p - x = 0 \Leftrightarrow m_{x} = \frac{x}{p}$ $m_{x} = p^{0} - m_{x} p$ $m_{x} (1 + p) = 1$ $\frac{x}{p} = \frac{1}{1 + p}$ $x = \frac{p}{1 + p}$ $m_{x} (1 + p) = 1$ $m_{x} = \frac{1}{1 + p}$ $n = m_{x} \times p^{x}$ $n = (\frac{1}{1 + p}) \times p^{\frac{p}{1 + p}}$ $I t m a y b e w r o n g!!!$
Commented by Yozzii last updated on 05/Aug/16

$n (1 + p) = p^{n}$ $n \in N \Rightarrow (1 + p) ∣ p^{\frac{p}{p + 1}} a n d p^{\frac{p}{p + 1}} \in N$
Commented by sou1618 last updated on 05/Aug/16

$o h . . . (\times_\times;)$
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