ONM 2010 Iasi Problema 4

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Andi Brojbeanu
Bernoulli
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Location: Targoviste (Dambovita)

ONM 2010 Iasi Problema 4

Post by Andi Brojbeanu »

Fie \( a, b, c, d \) numere naturale nenule si \( p=a+b+c+d \). Stiind ca \( p \) este un numar prim, aratati ca \( p \) nu divide \( ab-cd \).
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Andi Brojbeanu
Bernoulli
Posts: 294
Joined: Sun Mar 22, 2009 6:31 pm
Location: Targoviste (Dambovita)

Post by Andi Brojbeanu »

\( p|a+b+c+d\Rightarrow p|a(a+b+c+d)=a^2+ab+ac+ad\Rightarrow p|a^2+ab+ac+ad-ab+cd=a^2+ac+ad+cd=(a+d)(a+c) \).
Cum p este prim, rezulta ca \( p|a+c \) sau \( p|a+d \), deci \( a+b+c+d<a+c \) sau \( a+b+c+d<a+d \), imposibil deoarece \( a, b, c, d \) sunt numere naturale nenule.
Andi Brojbeanu
profesor, Liceul Teoretic "Lucian Blaga", Cluj-Napoca
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