Numar compus
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BogdanCNFB
- Thales
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Numar compus
Fie \( n\geq 2,n\in\mathbb{N} \). Demonstrati ca numarul \( n^4+4^n \) nu poate fi niciodata prim.
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BogdanCNFB
- Thales
- Posts: 121
- Joined: Wed May 07, 2008 4:29 pm
- Location: Craiova
Daca n=par concluzia este evidenta.
Daca n=impar \( \Rightarrow n=2k+1;k\in \mathbb{N}^* \).
Atunci avem\( n^4+4^n=n^4+4^{2k+1}=n^4+4\cdot(2^k)^4=a^4+4\cdot b^4 \)(Identitatea Sophie Germain), unde am facut notatiile \( n=a,2^k=b \).
\( a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2=(a^2+2b^2)^2-(2ab)^2
=(a^2+2b^2+2ab)(a^2+2b^2-2ab) \). Ambii factori sunt supraunitari, deci numarul este compus.
Daca n=impar \( \Rightarrow n=2k+1;k\in \mathbb{N}^* \).
Atunci avem\( n^4+4^n=n^4+4^{2k+1}=n^4+4\cdot(2^k)^4=a^4+4\cdot b^4 \)(Identitatea Sophie Germain), unde am facut notatiile \( n=a,2^k=b \).
\( a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2=(a^2+2b^2)^2-(2ab)^2
=(a^2+2b^2+2ab)(a^2+2b^2-2ab) \). Ambii factori sunt supraunitari, deci numarul este compus.