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