<s>[INST] If $2^{n+1} - 1$ is composite, is $2^{n+2} - 1$ prime? In 1556, Tartaglia claimed that the sums
1 + 2 + 4
1 + 2 + 4 + 8
1 + 2 + 4 + 8 + 16
are alternative prime and composite. Show that his conjecture is false.     
With a simple counter example, $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !
I tried:
The sum $\sum_{i=0}^n 2^i$ is equal to $2^{n+1} - 1$. I assumed that $2^{n+1} - 1$ is prime, then we must show that $2^{n+1} - 1 + 2^{n+1} = 2^{n+2} - 1$ is not composite. Or we assume $2^{n+1}$ is composite and we must show that $2^{n+2} - 1$ is not prime. 
But I have no clue how $2^{n+2} - 1$ relates to its previous prime/composite. Any hint?
 [/INST] If $2^{n}-1$ is prime, then $n$ must be odd, otherwise, we could factor
$2^{n}-1$ as $(2^{n/2}+1)(2^{n/2}-1).$
So at least every other number in the series $2^n-1,n=1,2,3...$
must be composite, by the conjugacy rule.
EDIT: Reread the question, but you can do the same for $a^3-b^3 = (a-b)(a^2+ab+b^2)$ so whenever $n$ is divisible by 3, you can do a factorization as above. In general, $a^n-b^n$ is divisible by $a-b,$
so if $n$ is not prime, $n=pq$, and we have
$$2^{n}-1 = (2^q)^p - 1^p = (2^q-1)Q$$
where $Q$ is an integer.
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