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