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

Answered by witcher3 last updated on 27/Dec/23

$${\mathrm{p}}^{\mathrm{n}}-1=(\mathrm{p}-1)\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\mathrm{p}}^{\mathrm{k}}\right)$$$$\Rightarrow \mathrm{p}.\left({\mathrm{p}}^{\mathrm{n}}\right)+(\mathrm{p}-1).\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\mathrm{p}}^{1+\mathrm{k}}=\mathrm{p}$$$$\Rightarrow {\mathrm{p}}^{\mathrm{n}}(\mathrm{x}-\mathrm{p})+(\mathrm{p}-1)(\mathrm{y}+\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\mathrm{p}}^{1+\mathrm{k}})=0$$$$\Rightarrow {\mathrm{p}}^{\mathrm{n}}(\mathrm{x}-\mathrm{p})=(1-\mathrm{p})(\mathrm{y}+{\mathrm{c}}_{\mathrm{n}})$$$${\mathrm{p}}^{\mathrm{n}}\mathrm{and}1-\mathrm{p}\mathrm{are}\mathrm{coprime}$$$$\Rightarrow {\mathrm{p}}^{\mathrm{n}}\mid \mathrm{y}+{\mathrm{c}}_{\mathrm{n}}$$$$\mathrm{y}={\mathrm{kp}}^{\mathrm{n}}-\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\mathrm{p}}^{1+\mathrm{k}}$$$$\mathrm{x}=\mathrm{k}(1-\mathrm{p})+\mathrm{p}$$$$(\mathrm{x},\mathrm{y})=(\mathrm{k}(1-\mathrm{p})+\mathrm{p},{\mathrm{kp}}^{\mathrm{n}}-\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\mathrm{p}}^{\mathrm{k}+1});\mathrm{k}\in \mathbb{Z}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$$$$$$$

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