<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] It is easy to see that $2^{2n}-1 \equiv 0 \mod 3$. This shows that for $n \geq 2$, $2^{2n}-1$ is never prime.
Also, since $2^3 \equiv 1 \mod 7$ it follows that $2^{3n} -1$ is always divisible by $7$.
Hence $2^{6n+2}-1, 2^{6n+3}-1, 2^{6n+4}-1$ are always composite. 
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