is it possible
is it possible ( 2 ^ n ) +1 and ( 2 ) ^ ( n + 1 ) + 1 can be prime

[QUOTE=drmurat;550032]is it possible ( 2 ^ n ) +1 and ( 2 ) ^ ( n + 1 ) + 1 can be prime[/QUOTE]
Yes, for n = 1. 
[QUOTE=rogue;550036]Yes, for n = 1.[/QUOTE]
lol and one more sample ? 
[QUOTE=drmurat;550037]lol
and one more sample ?[/QUOTE] n=0. 
[QUOTE=paulunderwood;550043]n=0.[/QUOTE]
yes but lets try n> 1 
For n > 1, N = 2^n +1 has to be a [URL="https://primes.utm.edu/top20/page.php?id=12"]generalized Fermat prime[/URL] with b=2 i.e 2^(2^a)+1, but 2^a+1 can never be a power of 2 :smile:

[QUOTE=drmurat;550032]is it possible ( 2 ^ n ) +1 and ( 2 ) ^ ( n + 1 ) + 1 can be prime[/QUOTE]
k*2^n+1 and k*2^(n+1)+1 can be both prime only for k divisible by 3, or one of these two numbers will be divisible by 3. 
[QUOTE=sweety439;550049]k*2^n+1 and k*2^(n+1)+1 can be both prime only for k divisible by 3, or one of these two numbers will be divisible by 3.[/QUOTE]
can you give sample 
yes it is impossible . one of rhem is diveded by 3 all the time

[URL="https://primes.utm.edu/glossary/page.php?sort=CunninghamChain"]Cunningham Chain of the second kind[/URL]
[QUOTE=PrimePages]A Cunningham chain of length k (of the first kind) is sequence of k primes, each which is twice the preceding one plus one. For example, {2, 5, 11, 23, 47} and {89, 179, 359, 719, 1439, 2879}. A Cunningham chain of length k (of the second kind) is a sequence of k primes, each which is twice the preceding one minus one. (For example, {2, 3, 5} and {1531, 3061, 6121, 12241, 24481}.) .[/QUOTE] OP is mentioning a special case of this (k=2, Fermat Primes), there are only finitely such primes as Paul mentioned. 
[QUOTE=carpetpool;550068][URL="https://primes.utm.edu/glossary/page.php?sort=CunninghamChain"]Cunningham Chain of the second kind[/URL]
OP is mentioning a special case of this (k=2, Fermat Primes), there are only finitely such primes as Paul mentioned.[/QUOTE] I asked it because if 2^(n) + 1 is prime and 2^(n+1) + 1 can be prime . it means (2^n) * (2^ (2n) + 1) *(2^ (2n+1) + 1) gives perfect number but it is impossible . one of them is devided by 3 
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