20180315, 10:13  #1 
"Jeppe"
Jan 2016
Denmark
2×3^{4} Posts 
Smallest prime of the form a^2^m + b^2^m, m>=14
I was creating a new OEIS entry as a kind of procrastination...
I wonder what the first (probable) prime of the form \(a^{16384}+b^{16384}\) is. Since this type of (extended) generalized Fermat numbers is mentioned in many sources (in the web), I would think someone had determined the answer? I had no lucky googling. For each \(m<14\), brute force will relatively early find a probable prime \(a^{2^m}+b^{2^m}\). The last of these is \(72^{8192} + 43^{8192}\) which can be found i Caldwell's database: 72^8192 + 43^8192 So what about \(m \ge 14\)? /JeppeSN 
20180315, 10:37  #2  
"Forget I exist"
Jul 2009
Dumbassville
8369_{10} Posts 
Quote:


20180315, 11:22  #3 
Jun 2003
4777_{10} Posts 
Not sure about first, but...
http://www.primenumbers.net/prptop/s...&action=Search these are of the form a^16384+(a+1)^16384 Last fiddled with by axn on 20180315 at 11:22 
20180315, 11:33  #4 
Jun 2003
1001010101001_{2} Posts 
what is the form of factors for these numbers? It will be k*2^14+1, but will k itself has other restrictions (like k is a multiple of 2 or 4 or something? other modular restrictions?)

20180315, 11:35  #5 
Feb 2017
Nowhere
3·5·11·23 Posts 

20180315, 12:12  #6  
"Jeppe"
Jan 2016
Denmark
A2_{16} Posts 
Quote:
/JeppeSN 

20180315, 12:19  #7 
"Jeppe"
Jan 2016
Denmark
2·3^{4} Posts 
I should have said I was interested in odd primes only (and the number 1 is not prime).
Therefore a and b cannot both be odd. And as said by science_man_88 above, additionally a and b must be coprime. These comments imply that a and b are distinct. Without loss of generality, a > b > 0. /JeppeSN 
20180315, 13:23  #8  
Feb 2017
Nowhere
3795_{10} Posts 
Quote:
a^(2^20) + 1, a^(2^19) + 1, and a^(2^18) + 1. Of course, they might not be the smallest prime a^(2^k) + b^(2^k) for their exponents. In looking at smaller exponents, it did occur to me to look at cases (a^(2^k) + b^(2^k))/2 with a*b odd. I noticed that (3^2 + 1)/2 and (3^4 + 1)/2 were primes, and, since I knew 2^32 + 1 isn't prime, I tried n = (3^32 + 1)/2. Pari's ispseudoprime(n) returned 1... 

20180315, 15:33  #9 
"Curtis"
Feb 2005
Riverside, CA
10572_{8} Posts 

20180315, 16:10  #10  
"Jeppe"
Jan 2016
Denmark
2×3^{4} Posts 
Quote:
I do not want any trivial solutions. I know \((a, b) = (67234, 1)\) gives an acceptable solution (\(b=1\) is allowed), but it is certainly not minimal (although we do not have the proof until someone gives an example (EDIT: axn's first post in this thread already gave examples)). Nitpicking is fine, but hopefully everyone sees I am just asking for the equivalent of \(72^{8192} + 43^{8192}\) with exponents \(16384\). /JeppeSN Last fiddled with by JeppeSN on 20180315 at 16:15 

20180315, 16:46  #11  
"Jeppe"
Jan 2016
Denmark
2×3^{4} Posts 
Me:
Quote:
Code:
m=0, 2^1 + 1^1 m=1, 2^2 + 1^2 m=2, 2^4 + 1^4 m=3, 2^8 + 1^8 m=4, 2^16 + 1^16 m=5, 9^32 + 8^32 m=6, 11^64 + 8^64 m=7, 27^128 + 20^128 m=8, 14^256 + 5^256 m=9, 13^512 + 2^512 m=10, 47^1024 + 26^1024 m=11, 22^2048 + 3^2048 m=12, 53^4096 + 2^4096 m=13, 72^8192 + 43^8192 /JeppeSN 

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