Problem 1 IMO 2009 (Day 1)

Soit n un entier positif et des entiers distincts dans l'ensemble .

Sachant que:

pour .

Prouvez que n ne divise pas

pour tout i €[1,k-1] , a_(i+1)a_i = a_i [n]

d'où a_i*a_(i-1).....*a1 = a1 [n] pour tout 1=< i =< k-1

Donc on aussi a_i *(a_(i-1).....*a1) = a_i * a1 [n]

Alors pour i = k on aura : a_k*a_1 = a1 [n]

Or si : n/a_k(a1-1) alors n/a_k(a1-1)-a1(a_k-1) = a_k-a1 contradiction

Car Ia_k-a1I < n

This is my solution, I don't no if it is true

First, we prove that n devide a1(ak-1)
then we prove that n don't devide ak(a1-1)

We have
n devide a1(a2-1) then there exist x1 that a1(a2-1)=x1.n ==>a1a2=x1.n+a1

multiplaying by (a3-1) then a1a2(a3-1)=x1.n.(a3-1)+a1(a3-1)

do a1(a3-1) is devided by n because a1a2(a3-1) is.

We repeat the same thing with a1(a3-1)

a1(a3-1) is devided by n then exist y such :
a1(a3-1)=y.n ==> a1a3=y.n+a1

multiplaying by (a4-1) .....(1)

a1a3(a4-1)=y.n(a4-1)+a1(a4-1)

a1a3(a4-1) is devided by n because a3(a4-1) is devided by n

then a1(a4-1) .........(2)


We continue, until we get that a1(ak-1) is devided by n

2) To prove that ak(a1-1) is not devided by n

we have ak(a1-1) = a1(ak-1) - (ak-a1)

a1(ak-1) s devided by n then :

ak(a1-1) is devided by n if (ak-a1) is devided by n

However ak-a1 < n then is not devided by n

Do ak(a1-1) is not devided by n..


Finally ,reply me if there is an error or a remark,please

Thank you

Dans le reste de cette preuve, on travaille uniquement des indices modulo k.

Démonstration par l'absurde : on suppose que a_k=a_k*a_1 (mod n).

En calculant de proche en proche avec toutes les relations a_i=a_i*a_(i+1) (mod n), on trouve facilement que : a_i=a_i*a_(i+1)=a_i*a_(i+1)*a_(i+2)=...=a_i*a_(i+1)*a_(i+2)*...*a_(i-1) (mod n).

Mais alors a_j=a_k (mod n) quels que soient j,k, puisque modulo n, ils sont tous égaux aux produits de tous les a_i.

Donc comme le seul multiple de n compris entre -n+1 et n-1 est 0, a_j-a_k=0 et donc a_j=a_k, ce qui est en contradiction avec les hypothèses lorsque j est différent de k.