determinati n,p numere prime astfel incat 2^{p} ( 2^{n+p}-p)=2008

Răspuns :

[tex]2^{n+p}-p=\frac{2008}{2^p}[/tex]
[tex]2^{n+p}-p\in\mathbb{N}\Rightarrow2^p\mid2008[/tex]
[tex]2^3\mid2008,~dar~2^4\nmid2008[/tex]
[tex]p<4[/tex]
[tex]p~este~prim\Rightarrow p\in\{2,3\}[/tex]
[tex]p=2\Rightarrow2^{n+2}-2=\frac{2008}{4}[/tex]
[tex]2^{n+2}-2=502[/tex]
[tex]2^{n+2}=504[/tex]
[tex]n+2\not\in\mathbb{N}[/tex]
[tex]n\not\in\mathbb{N}[/tex]
[tex]p=3\Rightarrow2^{n+3}-3=\frac{2008}{8}[/tex]
[tex]2^{n+3}-3=251[/tex]
[tex]2^{n+3}=254[/tex]
[tex]n+3\not\in\mathbb{N}[/tex]
[tex]n\not\in\mathbb{N}[/tex]
Deci nu este scris corect sau ceva, oricum as face n nu este natural