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Concursul Matefbc editia a 4-a problema 3

 
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Sun Dec 06, 2009 9:44 pm    Post subject: Concursul Matefbc editia a 4-a problema 3 Reply with quote

Aratati ca daca x si y sunt numere naturale prime intre ele, atunci c.m.m.d.c.(x^2+y^2, xy)=1.
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Bogdan Stanoiu
Euclid


Joined: 28 Jul 2011
Posts: 13

PostPosted: Thu Jul 28, 2011 2:56 pm    Post subject: Re: Concursul Matefbc editia a 4-a problema 3 Reply with quote

Andi Brojbeanu wrote:
Aratati ca daca x si y sunt numere naturale prime intre ele, atunci c.m.m.d.c.(x^2+y^2, xy)=1.

Daca exista un factor prim p>2 astfel incat p divide x^2+y^2 si p divide xy rezulta ca pdivide x^2-2xy+y^2=(x-y)^2 si deci p divide x-y de unde rezulta ca
p divide pe x^2-y^2.
Deci p divide pe x^2+y^2 si p divide pe x^2-y^2 si facand suma si diferenta rezulta p divide 2x^2 si p divide 2y^2 si ca urmare rezulta p divide x si p divide y ceea ce contrazice faptul ca x si y sunt prime intre ele.
Daca 2 divide pe x^2+y^2 rezulta, din faptul ca x si y sunt prime intre ele ca x si y sunt ambele impare si deci xy este impar.
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