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JBTST I 2008, Problema 1

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

PostPosted: Sat May 03, 2008 9:07 pm    Post subject: JBTST I 2008, Problema 1 Reply with quote

Fie p un numar prim, p \neq 3, si a,b numere intregi astfel ca p|a+b si p^2|a^3+b^3. Sa se demonstreze ca p^2|a+b sau p^3|a^3+b^3.
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Ahiles
Euclid


Joined: 17 Apr 2008
Posts: 28

PostPosted: Sun May 04, 2008 7:32 pm    Post subject: Reply with quote

Avem p^2|(a+b)(a^2-ab+b^2). Fiindca p|a+b, avem p|\frac{a+b}{p}(a^2-ab+b^2). Fiindca p este prim, avem doua cazuri:
1) p|\frac{a+b}{p}, deci p^2|a+b;
2) p|a^2-ab+b^2 sau p|(a+b)^2-3ab, de unde p|ab. Fiindca p este prim p divide unul din numerele a,b. Fie p|a, dar p|a+b, deci p|a,b. Atunci p^3|a^3+b^3.
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