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JBTST II 2008, Problema 4

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

PostPosted: Mon Jun 16, 2008 11:25 pm    Post subject: JBTST II 2008, Problema 4 Reply with quote

Fie a, b numere reale nenule astfel incat [an+b] este numar intreg par, oricare ar fi n \in \mathbb{N}. Sa se arate ca numarul a este intreg par.
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Beniamin Bogosel
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Joined: 07 Mar 2008
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PostPosted: Tue Jun 17, 2008 12:27 pm    Post subject: Reply with quote

Daca notam  b_n=[an+b]-[a(n-1)+b]=a+\{an+b\}-\{a(n-1)+b\} \in (a-1,a+1) , interval care contine un singur numar par. De aici sirul  (b_n) este constant. Deci, putem obtine  [an+b]=cn+[b] , unde  c=b_n care e constant si par.

Mai putem scrie  [an+b]=an+b-\{an+b\}=cn+[b]\Rightarrow (a-c)n+\{b\}=\{an+b\} . Deoarece partea din dreapta este marginita de 0 si 1, iar partea din dreapta este o functie liniara in  n daca  a-c\neq 0 de la un moment dat sirul din stanga trece de limita (principiul lui Arhimede... ), ceea ce este o contradictie. Deci  a=c , unde  c este un numar intreg par. Astfel  a este un intreg par.

Mi se pare ca am mai vazut problema asta in cartea lui Armand Martinov, "Probleme alese pentru copii alesi"... Smile
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Filip Chindea
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Joined: 27 Sep 2007
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PostPosted: Wed Jun 18, 2008 9:51 pm    Post subject: Reply with quote

De fapt, abordarea este oarecum standard.

Scriind 2k_n \le an + b < 2k_n + 1, obtinem k_n - \frac{b}{2} \le \frac{a}{2} \cdot n < k_n + \frac{1 - b}{2}. Deci \left\{ \frac{a}{2} \cdot n \right\} este intr-o reuniune de (unul sau doua) intervale (deshise la dreapta) cu suma lungimilor 1/2, si din Kronecker obtinem a \in \mathbb{Q}.

De aici, pentru a_1 = a/2 = p/q, (p, q) = 1, q > 0 intregi, avem p, q \neq 0 si deci ps \equiv 1 \pmod{q} (cum p este inversabil) implica nsa_1 = n(ps)/q = t + n/q, \left\{\frac{a}{2} \cdot ns \right\} = \frac{n}{q}, \forall n, unde s, t \in \mathbb{Z}. Obtinem din nou o contradictie pentru q \ge 2. Deci a = 2p \in 2\mathbb{Z} \qed.

Obs. Rezultatul este mai "dificil" de atins decat pentru implicatia \lfloor an + b \rfloor \in k\mathbb{Z}, \forall n \Rightarrow a \in k\mathbb{Z} cu k \ge 3, in cazul de fata nemaifiind necesar faptul ca intervalele respective sunt deschise.
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