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Old 2018-03-15, 16:46   #11
JeppeSN
 
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"Jeppe"
Jan 2016
Denmark

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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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