# Prime Numbers: (2^n - 1) and (2^n + 1)

by smithg86
Tags: numbers, prime
 P: 60 1. The problem statement, all variables and given/known data I was able to prove both of these statements after getting some help from another website, but I am trying to find another way to prove them. Can you guys check my work and help me find another way to prove these, if possible? Thanks. Part A: Show that if 2^n - 1 is prime, then n must be prime. Part B: Show that if 2^n + 1 is prime, where n $$\geq$$ 1, then n must be of the form 2^k for some positive integer k. 2. Relevant equations (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + ... + x + 1) 3. The attempt at a solution Part A: Write the contrapositive, n is not prime (a.k.a. n is composite) ==> 2^n - 1 is composite Assume n is composite. Let n = p*q, where neither p nor q are 1. Then, 2^n - 1 = (2^p)^q - 1 = (2^p - 1)*((2^p)^(q-1) + (2^p)^(q-2) + ... + (2^p) + 1) Note that 2^p - 1 > 1. Also, ((2^p)^(q-1) + (2^p)^(q-2) + ... + (2^p) + 1) > 1. So we have factored 2^n - 1, thus it is not prime. We have proved the contrapositive, so the original statement is true. -------- Part B: Note that if n is of the form 2^k, then n's prime factorization is only composed of 2's. Thus, the contrapositive of the original statement is as follows: n = b*(2^k), where b is a positive odd number ==> 2^n + 1 is composite. Let n = b*(2^k). Then, 2^n + 1 = 2^(b*(2^k)) + 1 = ((2^(2^k))^b + 1 = (2^(2^k) + 1)*{[(2^(2^k)]^(k-1) + [2^(2^k)]^(k-2) + ... + [2^(2^k)] + 1} Observe that [2^(2^k) + 1)] > 1 and {[(2^(2^k)]^(k-1) + [2^(2^k)]^(k-2) + ... + [2^(2^k)] + 1} > 1. We have factored 2^n + 1, so it is composite. This proves the contrapositive of the original statement, so the original statement is true. ------- Is there another way to prove either one of these statements?
P: 291
 Is there another way to prove either one of these statements?
Why would you want to? This is both a short and elementary solution. That is often considered to be the nicest proof.
P: 60
 Quote by Kummer Why would you want to? This is both a short and elementary solution. That is often considered to be the nicest proof.
I thought it would provide more insight as to why the statements are true.

P: 218
Prime Numbers: (2^n - 1) and (2^n + 1)

 Quote by smithg86 [b] Part B: Note that if n is of the form 2^k, then n's prime factorization is only composed of 2's. Thus, the contrapositive of the original statement is as follows: n = b*(2^k), where b is a positive odd number ==> 2^n + 1 is composite. Let n = b*(2^k). Then, 2^n + 1 = 2^(b*(2^k)) + 1 = ((2^(2^k))^b + 1 = (2^(2^k) + 1)*{[(2^(2^k)]^(k-1) + [2^(2^k)]^(k-2) + ... + [2^(2^k)] + 1} Observe that [2^(2^k) + 1)] > 1 and {[(2^(2^k)]^(k-1) + [2^(2^k)]^(k-2) + ... + [2^(2^k)] + 1} > 1. We have factored 2^n + 1, so it is composite. This proves the contrapositive of the original statement, so the original statement is true.

Should it state

$$[(2^{2^{k}})^{b}+1]=(2^{2^{k}} + 1)([2^{2^{k}}]^{b-1} + [2^{2^{k}}]^{b-2} + ... + [2^{2^{k}}] + 1)$$?

Can someone prove the general case of this expansion via Binomial Theorem for me?
P: 450
 (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + ... + x + 1)
Where does this factorization come from? I just need a link or something. Thanks.
 P: 1 the factorizations come fromm doing polynomial long division of $x^n-1$ and $x^n+1$ with $x^p-1$ and $x^p+1$ if $n=pq$ $x^n-1 = (x^p-1)(x^{p(q-1)}+x^{p(q-2)}+\cdots +x^p+1)$ The other equation should read: $x^n+1 = (x^p+1)(x^{p(q-1)}-x^{p(q-2)}+\cdots -x^p+1 )$ With alternating signs. Again this comes from polynomial long division, taking $x^n+1$ and dividing by $x^p+1$
 HW Helper P: 1,965 For the first part I would start with setting out 2 = There is nothing more elementary in math, but I have found someone at least got stuck in thinking of anything that 2 = After that you do have to use the binomial theorem which was found easier.

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