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JBTST IV 2008, Problema 3

 
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Laurian Filip
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Joined: 25 Nov 2007
Posts: 473
Location: Bucuresti

PostPosted: Tue Jun 17, 2008 12:06 am    Post subject: JBTST IV 2008, Problema 3 Reply with quote

Sa se determine perechile (m,n) de numere intregi n,m>1, cu proprietatea ca mn-1 divide n^3-1.
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Omer Cerrahoglu
Euclid


Joined: 17 Mar 2008
Posts: 34

PostPosted: Wed Jun 18, 2008 11:44 am    Post subject: Reply with quote

Avem ca  \exists d natural nenul a.i.  d(mn-1)=n^3-1 \Rightarrow 
dmn-d=n^3-1 \Rightarrow n(dm-n^2)=d-1. Pentru d=1 avem  mn-1=n^3-1 \Rightarrow m=n^2. In cele ce urmeaza presupunem ca  d\geq 2. Avem ca  n|d-1, deci  \exists p natural nenul a.i.  d-1=np . Deci  (np+1)m-n^2=p \Rightarrow n(pm-n)+m=p. Avem ca  d \leq \frac{n^3-1}{n-1} \Rightarrow d \leq n^2+n+1 . Deoarece  d=np+1 avem ca  np \leq n(n+1) \Rightarrow p \leq n+1. Daca pm-n \g 0 atunci  p \geq m+n. Deoarece  m \g 1 avem ca  p \g n+1 ceea ce este fals. Deci  pm-n \leq 0. Daca  pm-n=0 atunci  p=m deci  n=m^2. Daca  pm-n < 0 atunci trebuie sa avem m < n. Dar deoarece  n(pm-n)+m=p si  pm-n < 0 trebuie sa avem ca  m \geq n ceea ce este fals deoarece trebuie sa avem  m < n. Deci singurele solutii sunt  m=n^2 si  n=m^2.
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