Question

Let 2^(k)+1 be a prime number. Prove that then k=0 or k=2^(n) for some n>=0 .​

Answer 1

Step-by-step explanation:

The factorization

a2k+1+b2k+1=(a+b)(a2k−a2k−1b+a2k−2b2−⋯+⋯−ab2k−1+b2k)

shows that (a+b)∣(a2k+1+b2k+1) whenever k∈Z≥0 .

In particular, (a+1)∣(a2k+1+1) .

So if d∣n and d is odd ( and positive ) , then (2n/d+1)∣(2n/d)d+1=2n+1 . Therefore, if 2n+1 is prime, then 2n/d+1=1 or 2n+1 . The first case is impossible, so the second case must hold. But then d=1 , and the only odd divisor of n is 1 . Thus n must be a power of 2 . ■