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Question Number 148157 by aliibrahim1 last updated on 25/Jul/21

Answered by puissant last updated on 26/Jul/21

$$\mathrm{pgcd}(\mathrm{a};\mathrm{b})=\mathrm{pgcd}(\mathrm{a}-\mathrm{b};\mathrm{b})$$$$\Rightarrow \mathrm{pgcd}(2^{\mathrm{a}}-1;2^{\mathrm{b}}-1)=\mathrm{pgcd}(2^{\mathrm{a}}-2^{\mathrm{b}};2^{\mathrm{b}}-1)$$$$\mathrm{or}{2}^{\mathrm{a}}-2^{\mathrm{b}}=2^{\mathrm{b}}(2^{\mathrm{a}-\mathrm{b}}-1)$$$$\Rightarrow \mathrm{pgcd}(2^{\mathrm{a}}-1;2^{\mathrm{b}}-1)=\mathrm{pgcd}(2^{\mathrm{b}}(2^{\mathrm{a}-\mathrm{b}}-1);2^{\mathrm{b}}-1)$$$$=\mathrm{pgcd}(2^{\mathrm{a}-\mathrm{b}}-1;2^{\mathrm{b}}-1)$$$$\mathrm{because}{2}^{\mathrm{b}}\mathrm{et}{2}^{\mathrm{b}}-1\mathrm{first}\mathrm{among}\mathrm{them}..$$$$\mathrm{let}\mathrm{a}=\mathrm{bq}+\mathrm{r}$$$$\Rightarrow \mathrm{pgcd}(2^{\mathrm{a}}-1;2^{\mathrm{b}}-1)=\mathrm{pgcd}(2^{\mathrm{a}-\mathrm{bq}}-1;2^{\mathrm{b}}-1)$$$$=\mathrm{pgcd}(2^{\mathrm{r}}-1;2^{\mathrm{b}}-1)$$$$\mathrm{so}\mathrm{pgcd}(\mathrm{a}-\mathrm{bq};\mathrm{b})=\mathrm{pgcd}(\mathrm{r};\mathrm{b})$$$$\mathrm{pgcd}(\mathrm{a};\mathrm{b})=\mathrm{pgcd}(\mathrm{b};\mathrm{r})=.....=\mathrm{pgcd}({\mathrm{r}}^{\mathrm{n}};0)$$$$\Rightarrow \mathrm{pgcd}(2^{\mathrm{a}}-1;2^{\mathrm{b}}-1)=(2^{\mathrm{b}}-1;2^{\mathrm{r}}-1)$$$$=.......=\mathrm{pgcd}(2^{{\mathrm{r}}_{\mathrm{n}}}-1;2^0-1)=2^{{\mathrm{r}}_{\mathrm{n}}}$$$$$$$$\mathrm{so}\mathrm{pgcd}(2^{\mathrm{a}}-1;2^{\mathrm{b}}-1)=2^{\mathrm{pgcd}(\mathrm{a};\mathrm{b})}-1$$$$$$$$\mathrm{puissant}....$$

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