Question

If $(2^{32} + 1)$ is exactly divisible by a certain number, which of the following is also definitely divisible by the same number?

• A. $(2^{16} + 1)$
• B. $(2^{16} - 1)$
• C. $7 \times 2^{13}$
• D. $(2^{96} + 1)$

Hint:

Use algebraic identities like $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ to factor high-power expressions and identify common divisors.

Answer 1

The Correct Option is D

Let us suppose $N = (2^{32} + 1)$.
Now observe: $2^{96} = (2^{32})^3$, so:
$2^{96} + 1 = (2^{32})^3 + 1 = a^3 + 1$ where $a = 2^{32}$
Recall the identity: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
So, $2^{96} + 1 = (2^{32} + 1)(2^{64} - 2^{32} + 1)$
Clearly, $(2^{32} + 1)$ divides $(2^{96} + 1)$
Therefore, if $(2^{32} + 1)$ is divisible by some number, then
$(2^{96} + 1)$ is also divisible by the same number.