Divisibility inference — If (2^32 + 1) is divisible by some positive integer m, which of the following is guaranteed also to be divisible by the same m?

Introduction / Context:
We are given that m | (2^32 + 1). Such numbers arise in the context of Fermat-type expressions. The key idea is to use modular arithmetic: if 2^32 ≡ −1 (mod m), then powers of 2 that are multiples of 32 behave predictably, allowing us to deduce divisibility of related expressions.

Given Data / Assumptions:

• m divides 2^32 + 1 → 2^32 ≡ −1 (mod m).
• We must test which option is also ≡ 0 (mod m).
• Basic exponent rules in modular arithmetic apply.

Concept / Approach:
If 2^32 ≡ −1 (mod m), then raise both sides to powers. In particular, (2^32)^3 = 2^96 ≡ (−1)^3 = −1 (mod m). Adding 1 gives 2^96 + 1 ≡ 0 (mod m), guaranteeing divisibility by m. Other options generally do not follow from the given congruence.

Step-by-Step Solution:
1) From m | (2^32 + 1), get 2^32 ≡ −1 (mod m).2) Cube both sides: (2^32)^3 = 2^96 ≡ (−1)^3 = −1 (mod m).3) Add 1: 2^96 + 1 ≡ 0 (mod m).4) Hence m | (2^96 + 1), which matches option “2^96 + 1.”

Verification / Alternative check:
Similarly, (2^32)^2 = 2^64 ≡ 1 (mod m), implying m | (2^64 − 1), not (2^64 + 1). This further confirms why option e is not guaranteed.

Why Other Options Are Wrong:
2^16 ± 1 and 7*2^13 have no direct necessity from 2^32 ≡ −1; 2^64 + 1 is not forced because 2^64 ≡ 1, making 2^64 + 1 ≡ 2 (mod m), not 0.

Common Pitfalls:
Assuming any multiple power automatically works; confusing (a ≡ −1) with (a ≡ 1); overlooking that squaring gives +1 while cubing keeps −1.

Final Answer:
2^96 + 1