Property of pseudoprime numbers by base 2 AND / OR 3
 

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

[TEX]2 ^ {k-1} mod (k) = 1[/TEX]
or
[TEX]3 ^ {k-1} mod (k) = 1[/TEX]

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 ([B]k-1[/B]) we take the following expression
[TEX]a = 256k ^ 8-2048k ^ 7 + 6784k ^ 6-11904k ^ 5 + 11680k ^ 4-6112k ^ 3 + 1380k ^ 2-36k[/TEX]

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

[TEX]b^a mod (k) = 1[/TEX]

Examples:
[TEX]K = 121[/TEX]

[TEX]a= 11006388017805370080[/TEX]

[TEX]7 ^ {11006388017805370080} mod (121) = 1[/TEX]
[TEX]13 ^ {11006388017805370080} mod (121) = 1[/TEX]
[TEX]37 ^ {11006388017805370080} mod (121) = 1[/TEX]
......

next

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

This formula will help to decompose the remainder

[TEX]t[/TEX]

if [TEX]k[/TEX] any integer

[TEX]2^{k-1} mod k = t[/TEX]

where [TEX]t>1[/TEX]

find the power q according to the formula

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

And now the most interesting

[TEX]b^{q} mod (t) = 1[/TEX]

where [TEX](t,b)[/TEX] are relatively prime and [TEX]b>1[/TEX]

Examples
k=1121

[TEX]2^{1120} \bmod 1121 = 833 = 7\cdot7\cdot17[/TEX]

[TEX]933793=1121\cdot833[/TEX]

[TEX]q = 147992994097091629145891052601604192667783899221632[/TEX]

[TEX]2^{147992994097091629145891052601604192667783899221632} mod (833) = 1[/TEX]

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

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