Question

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

Answer 1

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