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posons-1-2-3-n-a-n-b-n-3-montre-que-pgcd-a-n-b-n-1-




Question Number 87877 by Cmr 237 last updated on 06/Apr/20
posons   (1+2(√3))^n =a_n +b_n (√3)  montre que pgcd(a_n ;b_n )=1
$${posons}\: \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\boldsymbol{{n}}} =\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} +\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} \sqrt{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{montre}}\:\boldsymbol{\mathrm{que}}\:\boldsymbol{\mathrm{pgcd}}\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ;\boldsymbol{{b}}_{\boldsymbol{{n}}} \right)=\mathrm{1} \\ $$

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