Question

Which of the following numbers is completely divisible by $(2^{32}+1)$?

• A. $(2^{16}-1)$
• B. $(2^{96}+1)$
• C. $(2^{32}-1)$
• D. $(2^{16}+1)$

Answer 1

The correct option is B $(2^{96}+1)$

$\Rightarrow$ Let $2^{32}=x.$

$\Rightarrow$ Then, $(2^{32}+1)=(x+1)$

$\Rightarrow$ Let $(x+1)$ be completely divisible by the natural number $N$.

$\Rightarrow$ Then, $(2^{96}+1)=\left[\right(2^{32}{)}^3+1]=(x+1\left)\right(x^2-x+1)$, which is completely divisible by $N$, since $(x+1)$ is divisible by $N$.