Question

If \( \large 2^n -1 \) is prime, then n is a prime.

Answer 1

note that if \(n=2m\) is even then we most certainly have a difference of squares: $$2^{2m}-1=(2^m+1)(2^m-1)$$only in the case \(m=1\) i.e. \(n=2\) is this prime.

anyways, all other prime \(n\) are odd i.e. \(n=2m+1\)

Answer 2

if \(r | n\), then \(2^r-1 | 2^n-1\) implying that \(2^n-1\) is composite, so there will not be any \(r\) that divides \(n\)

Answer 3

I think he may want you to prove that \(2^r-1\big|2^n-1\)

Answer 4

I suppose if \(r\big|n\) then \(n=rs\) ergo \(2^n-1=(2^r)^s-1\)

Answer 5

using our knowledge of polynomials we have that:$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1})$$so we know \(2^n-1=(2^r-1)(\dots)\) Q.E.D.

Answer 6

\(2^2-1=3\\2^6-1=63\)

you calculated incorrectly