Texte intégral
Nous cherchons les entiersa > b>1 de somme minimale tels que 2009a ≡2009b (mod 104).Puisque 2009 est premier avec 10, alors c’est équivalent à 2009a−b≡1 (mod 104).Nous vérifions que 250 est l’ordre multiplicatif de 2009 modulo 104. Ainsia−b= 250k oùk >0 est entier eta+b = 250k+ 2b est minimal pour k=b= 1 et donc a= 251.
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