E139 - Les indices s’indicent ak

E139 - Les indices s’indicent a_k = 12 + (k−1)r

aa_k =a12+(k−1)r = 12 + (11 + (k−1)r)r= (k−1)r²+ 11r+ 12 2000 =a_aak =a12+(11+(k−1)r)r = 12 + (11 + (11 + (k−1)r)r)r

(k−1)r³+ 11r²+ 11r= 1988

r((k−1)r²+ 11r+ 11) = 2²×7×71 L’unique solution pour k et r entiers estk = 5 et r= 7

Ainsia2000 = 12 + 1999×7 = 14005