20180316, 14:39  #23 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
23A3_{16} Posts 
Were these the droids you were looking for?
216^16384+109^16384 
20180316, 15:08  #24 
"Jeppe"
Jan 2016
Denmark
2^{5}×5 Posts 
Yes! And is it certain that 216^16384+109^16384 is minimal?
Did you search and find this now, or was it something that was known (to you) in advance? Of course, now I want then minimal odd prime a^(2^m)+b^(2^m) for all the next m values /JeppeSN 
20180316, 16:30  #25 
Sep 2002
Database er0rr
2^{2}·859 Posts 
For added confidence in Serge's result:
Code:
time ./pfgw64 k f0 od q"216^16384+109^16384"  ../../coding/gwnum/hybrid 
Testing (x + 1)^(n + 1) == 2 + 3 (mod n, x^2  3*x + 1)...
Likely prime!
real 1m45.793s
user 2m36.120s
sys 0m55.244s

20180316, 16:30  #26 
Feb 2017
Nowhere
3×7×13^{2} Posts 
If one wanted to check exponents 2^m for several values of m, one might want to compile a table of pairs (a,b) with a < b, b <= N (given upper bound), gcd(a, b) = 1, and a + b odd. This isn't entirely trivial. And I'm quite poor at this sort of thing.
Without the condition that a + b be odd, it would be equivalent to calculating the Farey sequence of order N. Perhaps the way to go is to calculate the Farey sequence, and then skip pairs where a and b are both odd. One idea I considered by rejected was, for even b, find the positive integers r < b with gcd(r, b) = 1, then fill in the entries (b, r + k*b) with r + k*b not exceeding N. Alas, you would have to make sure, for each r, that the multiplier k was relatively prime to r... Of course, viewing table computation as a preliminary calculation, it probably makes little difference how clumsily it is compiled, if N isn't very big. For large exponents, the value of b will be a good indicator of the size of a^(2^m) + b^(2^m) The idea of sieving out multiples of "small" primes p congruent to 1 (mod 2^(m+1)) makes sense. Each pair (a, b) sieved out as giving a multiple of p means you can skip a PRP test. I looked at the primes p congruent to 1 (mod 2^15) up to the limit 50 million. There are 174 such primes. Grasping at straws(*), I computed the product (1  2^14/p) over these primes, and got .507+ which would be great if it meant eliminating almost half the pairs (a, b) actually under consideration. (*) I reckoned that the relevant thing here was the 2^14 solutions to Mod(r,p)^(2^14) + 1 = Mod(0,p). Each pair (Mod(a,p), Mod(r, p)*Mod(a,p)) gives a multiple of p. This ignores the question of which such pairs might reduce to pairs of small coprime integers (a, b) of different parity. But hey  any PRP test you can skip cheaply is to the good, right? I did a quick check of the idea with the exponent 16. I looked at the smallest prime (p = 97) congruent to 1 (mod 32), and easily found from the root Mod(19,97) to x^16 + 1 = Mod(0,97) that 5^16 + 2^16 is a multiple of 97. Last fiddled with by Dr Sardonicus on 20180316 at 16:31 Reason: Fixing typos 
20180316, 16:52  #27  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·3,041 Posts 
Quote:
What puzzles me is why is this taking others so long to find the same? It only took overnight on 8 cores. The workflow is very simple: generate (a,b)s in 1 second with a trivial Pari script (opposite parity, gcd > 1, say, up to a<=400), then split into a few workfiles (hawever many cores you have) and run pfgw f"{65536}" N k l input on each chunk, periodically check, ..., done! For m=15, it is perhaps best to sieve, but for m=14, this f parameter takes care of the small factors just well enough. 

20180316, 17:15  #28 
Einyen
Dec 2003
Denmark
5×593 Posts 
It is just a question of resources. For this random question I only used 1 core for sieving and 1 core for pfgw, I had no need to use more, rest is running Prime95 on this machine, it is not like we had to find the answer within a certain time limit.
I only tried briefly using pfgw for trial factoring long ago, and my impression was it was horribly slow at it, so I have not tried again, I might try next time I need it. Edit: and yes I should have sorted the list so it ran pfgw on candidates by size, but I made this quick program last night before I went to bed, and I only got 45 hours of sleep before work. Last fiddled with by ATH on 20180316 at 17:17 
20180316, 17:41  #29 
Jun 2003
4706_{10} Posts 

20180316, 18:28  #30 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10001110100011_{2} Posts 
A typo, this commandline is already prepared for m=15. :)

20180316, 18:49  #31 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Technically if a and b are squares you find one for the next level up as well.
Last fiddled with by science_man_88 on 20180316 at 18:50 
20180316, 19:36  #32 
Aug 2006
13446_{8} Posts 

20180316, 20:00  #33 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,041 Posts 
Another droid crossed my path. 260^32768+179^32768
Not proven minimal yet, though. 
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