Question-202462

Question Number 202462 by pticantor last updated on 27/Dec/23

Answered by witcher3 last updated on 27/Dec/23

p^n −1=(p−1)(Σ_(k=0) ^(n−1) p^k ) ⇒p.(p^n )+(p−1).Σ_(k=0) ^(n−1) p^(1+k) =p ⇒p^n (x−p)+(p−1)(y+Σ_(k=0) ^(n−1) p^(1+k) )=0 ⇒p^n (x−p)=(1−p)(y+c_n ) p^n and 1−p are coprime ⇒p^n ∣y+c_n y=kp^n −Σ_(k=0) ^(n−1) p^(1+k) x=k(1−p)+p (x,y)=(k(1−p)+p,kp^n −Σ_(k=0) ^(n−1) p^(k+1) );k∈Z Σp^(1+k) = { ((((p(1−p^n ))/(1−p)),p#1)),((=n,p=1)) :}

$$\mathrm{p}^{\mathrm{n}} −\mathrm{1}=\left(\mathrm{p}−\mathrm{1}\right)\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{k}} \right) \\ $$$$\Rightarrow\mathrm{p}.\left(\mathrm{p}^{\mathrm{n}} \right)+\left(\mathrm{p}−\mathrm{1}\right).\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{1}+\mathrm{k}} =\mathrm{p} \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{n}} \left(\mathrm{x}−\mathrm{p}\right)+\left(\mathrm{p}−\mathrm{1}\right)\left(\mathrm{y}+\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{1}+\mathrm{k}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{n}} \left(\mathrm{x}−\mathrm{p}\right)=\left(\mathrm{1}−\mathrm{p}\right)\left(\mathrm{y}+\mathrm{c}_{\mathrm{n}} \right) \\ $$$$\mathrm{p}^{\mathrm{n}} \:\mathrm{and}\:\mathrm{1}−\mathrm{p}\:\mathrm{are}\:\mathrm{coprime} \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{n}} \mid\mathrm{y}+\mathrm{c}_{\mathrm{n}} \\ $$$$\mathrm{y}=\mathrm{kp}^{\mathrm{n}} −\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{1}+\mathrm{k}} \\ $$$$\mathrm{x}=\mathrm{k}\left(\mathrm{1}−\mathrm{p}\right)+\mathrm{p} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{k}\left(\mathrm{1}−\mathrm{p}\right)+\mathrm{p},\mathrm{kp}^{\mathrm{n}} −\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{k}+\mathrm{1}} \right);\mathrm{k}\in\mathbb{Z} \\ $$$$\Sigma\mathrm{p}^{\mathrm{1}+\mathrm{k}} =\begin{cases}{\frac{\mathrm{p}\left(\mathrm{1}−\mathrm{p}^{\mathrm{n}} \right)}{\mathrm{1}−\mathrm{p}},\mathrm{p}#\mathrm{1}}\\{=\mathrm{n},\mathrm{p}=\mathrm{1}}\end{cases} \\ $$$$ \\ $$$$ \\ $$

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