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Old 2018-03-15, 10:13   #1
JeppeSN
 
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Cool 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
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Old 2018-03-15, 10:37   #2
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Quote:
Originally Posted by JeppeSN View Post
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
a and b must be coprime. Their powers can't be addiive inverse remainders of each other.Etc.
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Old 2018-03-15, 11:22   #3
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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 2018-03-15 at 11:22
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Old 2018-03-15, 11:33   #4
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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?)
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Old 2018-03-15, 11:35   #5
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Quote:
Originally Posted by JeppeSN View Post
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.
I would say 2 (a = b = 1).
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Old 2018-03-15, 12:12   #6
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Quote:
Originally Posted by axn View Post
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?)
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 New GFN factors (Wilfrid Keller).

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Old 2018-03-15, 12:19   #7
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Quote:
Originally Posted by Dr Sardonicus View Post
I would say 2 (a = b = 1).
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.

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Old 2018-03-15, 13:23   #8
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Quote:
Originally Posted by JeppeSN View Post
I should have said I was interested in odd primes only (and the number 1 is not prime).
There are some large "generalized Fermat" primes listed, e.g. here; 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...
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Old 2018-03-15, 15:33   #9
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Quote:
Originally Posted by JeppeSN View Post
I should have said I was interested in odd primes only (and the number 1 is not prime).
/JeppeSN
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?
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Old 2018-03-15, 16:10   #10
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Quote:
Originally Posted by VBCurtis View Post
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?
No. I wanted to forbid \(1 = 1^{16384} + 0^{16384}\) as another trivial "solution".

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 2018-03-15 at 16:15
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Old 2018-03-15, 16:46   #11
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Me:
Quote:
Originally Posted by JeppeSN View Post
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
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
The next line takes more time. The question is: Hasn't this been considered before?!

/JeppeSN
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