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Old 2020-07-08, 18:16   #874
sweety439
 
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Nov 2016

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Quote:
Originally Posted by sweety439 View Post
Of course,

* In the Riesel case if k and b are both r-th powers for an r>1

* In the Sierpinski case if k and b are both r-th powers for an odd r>1

* In the Sierpinski case if k is of the form 4*m^4, and b is 4th power

Then this k proven composite by full algebraic factors
Also, in the Sierpinski case if b = q^m, k = q^r, where q is not of the form t^s with odd s>1, and m and r have no common odd prime factor, and the exponent of highest power of 2 dividing r >= the exponent of highest power of 2 dividing m, and the equation 2^x = r (mod m) has no solution, then this k has no possible prime. (such k's are also excluded from the conjectures)

Examples:

b = q^7, k = q^r, where r = 3, 5, 6 (mod 7).
b = q^14, k = q^r, where r = 6, 10, 12 (mod 14).
b = q^15, k = q^r, where r = 7, 11, 13, 14 (mod 15).
b = q^17, k = q^r, where r = 3, 5, 6, 7, 10, 11, 12, 14 (mod 17).
b = q^21, k = q^r, where r = 5, 10, 13, 17, 19, 20 (mod 21)
b = q^23, k = q^r, where r = 5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22 (mod 23)
b = q^28, k = q^r, where r = 12, 20, 24 (mod 28)
b = q^30, k = q^r, where r = 14, 22, 26, 28 (mod 30)
b = q^31, k = q^r, where r = 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 (mod 31)
b = q^33, k = q^r, where r = 5, 7, 10, 13, 14, 19, 20, 23, 26, 28 (mod 33)
etc.

(these are all examples for m<=33)

Last fiddled with by sweety439 on 2020-07-10 at 06:18
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