Prove that if 2^n-2 is divisible by n, then 2^(2n-1)-2 is divisible by 2^n-1

Answer 1

Since 2^n - 2 is divisible by n, 2^n - 2 = kn for some integer k.

Since 2^n - 2 = 2(2^(n-1) - 1), it is even.

Then kn is also even.

Then 2^(n-1) - 1 = m = kn/2.

Then 2^(n-1) = m +1

Then squaring both sides, 2^(2n-2) = m^2 + 2m + 1

Then 2^(2n-2) - 1 = m^2 + 2m = m(m + 2).

Multiplying both sides by 2, 2^(2n-1) - 2 = 2m(m + 2) = kn(m + 2).

What we're proving appears to not be true.

Counterexample n = 3:

2^3-2=6 which is divisible by 3, but 2^5-2=30 which is not divisible by 2^3-1=8.

So, it is not true.