<s>[INST] Divisibility Of $(2^{32} +1)$ Here is my problem:
If $2^{32} +1 $ is completely divisible by a whole number. Which of the following numbers is completely divisible by that number :
(A)($2^{16}+1$)
(B)($2^{16}-1$)
(C)$7*2^{23}$  
(D)($2^{96}+1$)
I don't know where to start any help is appreciated. Thank you
 [/INST] Hint 1: We want to find an option that is divisible by $2^{32} + 1$. Indeed, this is necessary and sufficient: it is necessary since we can choose $2^{32} + 1$ as the factor that divides $2^{32} + 1$ and it is sufficient since if a number is divisible by $r = de$, then it is divisible by $d$ and $e$ as well. 
Hint 2: $a^3 + 1 = (a + 1)(a^2 - a + 1)$.

These hints tell us that (d) is the right answer, since $$2^{96} + 1 = (2^{32})^3 + 1 = (2^{32} + 1)(2^{64} - 2^{32} + 1),$$ so $2^{32}+1 \mid 2^{96} + 1$.
We can also rule out the others if we are not given that there are multiple answers: (a) (b) and (c) are all less than $2^{32} + 1$, so $2^{32} + 1$ cannot divide all of them evenly. We may even show that they share no common factors with $2^{32} + 1$ at all.
(a) and (b) are out since their product is $2^{32} - 1$. The only common divisor of $2^{32} - 1$ and $2^{32} + 1$ could be $2$ since their difference is $2$, but they're both odd so $\gcd (2^{32} + 1, 2^{32} - 1) = \gcd(2^{32} + 1, 2^{16} - 1) = \gcd(2^{32} + 1, 2^{16} + 1) = 1$.
(c) can be ruled out since its only divisors are $2$ and $7$. $2$ is not a divisor of $2^{32} + 1$ since it is odd, $7$ is not a divisor since $2^3 \equiv 1 \pmod 7$ and so $2^{32} + 1 \equiv 2^2 (2^3)^{10} +1 1 \equiv 4 + 1 \equiv 5 \pmod 7$.
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