Property of pseudoprime numbers by base 2 AND / OR 3

allasc · 2017-05-17, 13:28

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:
$K = 121$

$a= 11006388017805370080$

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

allasc · 2017-05-17, 15:00

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 :))))

2017-05-17, 13:28	#1
allasc Aug 2010 SPb 100010₂ Posts	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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2017-05-17, 15:00	#2
allasc Aug 2010 SPb 34₁₀ Posts	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 :))))