Another unproven conjecture is that there are an infinitude of primes that are $1$ less than a power of $2$. If $p=2^{k-1}$ is prime, show that $k$ is an odd integer, except when $k=2$.

Hint: $3|4^{n-1} \:\: \forall \:\: n \geq 1$

Can someone explain how to go about this proof step by step, I'm really confused. Thank you so much!

  • $\begingroup$ If $k$ is even, $2^k$ is a power of four. $\endgroup$ – user61527 Mar 14 '14 at 0:43

Your claim is: If p = 2^k - 1 is a prime number, show that k is an odd integer.

The proof is by contradiction,and it goes like this: assume otherwise that k is not an odd

number. That means k is even. Because it is even, k = 2n for some positive integer n. Thus

substituting 2n for k in the expression of p you get: p = 2^(2n) - 1. But 2^(2n) = 4^n. So

p = 4^n - 1. But you already knew that 3 divides 4^n - 1 for all naturals n. This means 3

divides p, and since p is a prime p must be 3. So 2^k - 1 = p = 3 ==> k = 2. Since k is not

equal to 2, p can't be 3. But then p is a composite number because 1, 3, and p are three

distinct factors of p, contradicting the assumption that it is a prime number. So k must

be odd.


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