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Old 2017-05-17, 13:28   #1
allasc
 
Aug 2010
SPb

3410 Posts
Default Property of pseudoprime numbers by base 2 AND / OR 3

Suppose that the number k is pseudoprime with respect to the base 2 and / or 3
That is, one of the conditions is fulfilled .. or both conditions are satisfied

2 ^ {k-1} mod  (k) = 1
or
3 ^ {k-1} mod  (k) = 1

k = {91, 121, 286, 341, 561, 645, 671, 703, 949, 1105, 1387, 1541, 1729, 1891, 1905, 2047, 2465, 2665, 2701......}

If instead of the degree (k-1) we take the following expression
a = 256k ^ 8-2048k ^ 7 + 6784k ^ 6-11904k ^ 5 + 11680k ^ 4-6112k ^ 3 + 1380k ^ 2-36k

The number k is pseudoprime for any whole base b>1, where (k,b) are relatively prime

b^a mod (k) = 1

Examples:
K = 121

a= 11006388017805370080

7 ^ {11006388017805370080}  mod (121) = 1
13 ^ {11006388017805370080}  mod (121) = 1
37 ^ {11006388017805370080}  mod (121) = 1
......

Last fiddled with by allasc on 2017-05-17 at 13:37
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Old 2017-05-17, 15:00   #2
allasc
 
Aug 2010
SPb

2×17 Posts
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next

a = 256k ^ 8-2048k ^ 7 + 6784k ^ 6-11904k ^ 5 + 11680k ^ 4-6112k ^ 3 + 1380k ^ 2-36k

This formula will help to decompose the remainder

t

if k any integer

2^{k-1} mod k = t

where t>1

find the power q according to the formula

q = 256(kt) ^ 8-2048(kt) ^ 7 + 6784(kt) ^ 6-11904(kt) ^ 5 + 11680(kt) ^ 4-6112(kt) ^ 3 + 1380(kt) ^ 2-36(kt)

And now the most interesting

b^{q} mod (t) = 1

where (t,b) are relatively prime and b>1

Examples
k=1121

2^{1120} \bmod 1121 = 833 = 7\cdot7\cdot17

933793=1121\cdot833

q = 147992994097091629145891052601604192667783899221632

2^{147992994097091629145891052601604192667783899221632} mod (833) = 1

2^{147992994097091629145891052601604192667783899221632/32} mod (833) = 50

49 and 51 Here is the answer :))))
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