Aratati ca pentru orice n natural , numarul \( a=5^{5^{n+1}}+5^{5^n}+1 \) nu poate fi prim.
Concursul ,,Gh.Lazar'',2005
Numere compuse
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Marius Mainea
- Gauss
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
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Claudiu Mindrila
- Fermat
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Cu notatia de mai sus, avem:
\( a^{5}+a+1=a^{5}+a^{2}+a+1-a^{2}=a^{2}\left(a^{3}-1\right)+\left(a^{2}+a+1\right)=a^{2}\left(a-1\right)\left(a^{2}+a+1\right)+\left(a^{2}+a+1\right)=\left(a^{2}+a+1\right)\left(a^{3}-a^{2}+1\right) \) si cum \( a^{2}+a+1>1 \) si \( a^{3}-a^{2}+1>1 \) rezulta cerinta problemei.
\( a^{5}+a+1=a^{5}+a^{2}+a+1-a^{2}=a^{2}\left(a^{3}-1\right)+\left(a^{2}+a+1\right)=a^{2}\left(a-1\right)\left(a^{2}+a+1\right)+\left(a^{2}+a+1\right)=\left(a^{2}+a+1\right)\left(a^{3}-a^{2}+1\right) \) si cum \( a^{2}+a+1>1 \) si \( a^{3}-a^{2}+1>1 \) rezulta cerinta problemei.
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste