Question-163259

Question Number 163259 by SANOGO last updated on 05/Jan/22

Answered by TheSupreme last updated on 05/Jan/22

1) b^n −1≡0(b−1) (b−1)P_(n−1) (b)=b^n −1 P_(n−1) =Σ_(i=0) ^(n−1) b^i (b−1)Σ_(i=0) ^(n−1) b^i =Σ_(i=0) ^(n−1) b^(i+1) −b^i =b^n −1 2) a^k −1 is prime a^k −1≡1(2) a^k ≡0(2) a≡0(2) 2t=a 2^k t^k −1 is prime a^k −1 is always divided by a−1 so we have a−1=a^k −1 or a−1=1 ∀p<a^k −1 : a^k −1≡m(p) with m≠0

$$\left.\mathrm{1}\right)\: \\ $$$${b}^{{n}} −\mathrm{1}\equiv\mathrm{0}\left({b}−\mathrm{1}\right) \\ $$$$\left({b}−\mathrm{1}\right){P}_{{n}−\mathrm{1}} \left({b}\right)={b}^{{n}} −\mathrm{1} \\ $$$${P}_{{n}−\mathrm{1}} =\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{b}^{{i}} \\ $$$$\left({b}−\mathrm{1}\right)\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{b}^{{i}} =\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{b}^{{i}+\mathrm{1}} −{b}^{{i}} ={b}^{{n}} −\mathrm{1} \\ $$$$\left.\mathrm{2}\right) \\ $$$${a}^{{k}} −\mathrm{1}\:{is}\:{prime} \\ $$$${a}^{{k}} −\mathrm{1}\equiv\mathrm{1}\left(\mathrm{2}\right) \\ $$$${a}^{{k}} \equiv\mathrm{0}\left(\mathrm{2}\right) \\ $$$${a}\equiv\mathrm{0}\left(\mathrm{2}\right)\: \\ $$$$\mathrm{2}{t}={a} \\ $$$$\mathrm{2}^{{k}} {t}^{{k}} −\mathrm{1}\:{is}\:{prime} \\ $$$${a}^{{k}} −\mathrm{1}\:{is}\:{always}\:{divided}\:{by}\:{a}−\mathrm{1} \\ $$$${so}\:{we}\:{have}\:{a}−\mathrm{1}={a}^{{k}} −\mathrm{1}\:{or}\:{a}−\mathrm{1}=\mathrm{1} \\ $$$$ \\ $$$$\forall{p}<{a}^{{k}} −\mathrm{1}\::\:{a}^{{k}} −\mathrm{1}\equiv{m}\left({p}\right)\:{with}\:{m}\neq\mathrm{0} \\ $$

Commented by SANOGO last updated on 06/Jan/22

$${merci}\:{bien} \\ $$

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