Given that (2^32 + 1) is divisible by some whole number N, which of the following is also divisible by N?

Given data

• Let a = 2^32. We are told N | (a + 1).

Concept / Approach

• If N divides (a + 1), then for any odd k: N | (a^k + 1) because a ≡ −1 (mod N) ⇒ a^k ≡ (−1)^k.

Step-by-step reasoning

a ≡ −1 (mod N)For k = 3 (odd): a^3 + 1 ≡ (−1)^3 + 1 = −1 + 1 = 0 (mod N)Thus N divides a^3 + 1 = (2^32)^3 + 1 = 2^96 + 1.

Why others need not work

• 2^64 + 1 = a^2 + 1 ≡ 1 + 1 = 2 (mod N), not generally 0 since N is odd (2^32 + 1 is odd).
• 2^16 + 1 and 2^48 + 1 are not of the form a^k ± 1 with k odd in terms of a = 2^32, hence no guaranteed divisibility by N.

Final Answer

The expression 2^96 + 1 is also divisible by N.