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: [URL="https://primes.utm.edu/primes/page.php?id=109871"]72^8192 + 43^8192[/URL] So what about [$]m \ge 14[/$]? /JeppeSN 
[QUOTE=JeppeSN;482389]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: [URL="https://primes.utm.edu/primes/page.php?id=109871"]72^8192 + 43^8192[/URL] So what about [$]m \ge 14[/$]? /JeppeSN[/QUOTE] a and b must be coprime. Their powers can't be addiive inverse remainders of each other.Etc. 
Not sure about first, but...
[url]http://www.primenumbers.net/prptop/searchform.php?form=x%5E16384%2By%5E16384&action=Search[/url] these are of the form a^16384+(a+1)^16384 
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?)

[QUOTE=JeppeSN;482389]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. [/QUOTE] I would say 2 (a = b = 1). 
[QUOTE=axn;482393]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?)[/QUOTE]
I would say the multiplier must be even, so it becomes (under a new convention where k is odd) [$]k\cdot 2^n + 1[/$] where [$]n \ge 15[/$]. See also [URL="http://www.prothsearch.com/GFNfacs.html"]New GFN factors (Wilfrid Keller)[/URL]. /JeppeSN 
[QUOTE=Dr Sardonicus;482395]I would say 2 (a = b = 1).[/QUOTE]
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 
[QUOTE=JeppeSN;482398]I should have said I was interested in odd primes only (and the number 1 is not prime).[/QUOTE]
There are some large "generalized Fermat" primes listed, e.g. [url=https://primes.utm.edu/top20/page.php?id=12]here[/url]; these have the forms 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... 
[QUOTE=JeppeSN;482398]I should have said I was interested in odd primes only (and the number 1 is not prime).
/JeppeSN[/QUOTE] Your OP never mentioned a and b should be prime. I don't see why "1 is not prime" is relevant. Do you mean to now require a>b>1? 
[QUOTE=VBCurtis;482416]Your OP never mentioned a and b should be prime. I don't see why "1 is not prime" is relevant. Do you mean to now require a>b>1?[/QUOTE]
No. I wanted to forbid [$]1 = 1^{16384} + 0^{16384}[/$] as another trivial "solution". I do not want [I]any[/I] 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 
Me: [QUOTE=JeppeSN;482389]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: [URL="https://primes.utm.edu/primes/page.php?id=109871"]72^8192 + 43^8192[/URL][/QUOTE]
To be explicit, this is what brute force finds: [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[/CODE] The next line takes more time. The question is: Hasn't this been considered before?! /JeppeSN 
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