Wagstaff numbers have the potential still unexplored, and their definition is:
W = (2 ^ p +1) / 3 (1) In (1) there are several things to highlight: 1) W is first 2) 3 is first 3) 1 is coprime 4) p is prime Generalizing we are working, then, with expressions like: 2 ^ x = p1 + p2 * d ^ 2 or p1 = p2 * d - x (2) where p1, p2 are prime numbers and d is odd. In (2) p2, compared to (1), has the role of 3, p3, the role of the role of 1 W ex. Suppose at x = 1 to (2) leads to these conclusions: p1 = log2 (p2-d * x) (3) Then from (3) for a given p1 (because there is assigned the Mersenne number 2 ^ p1-1), if p2 is fixed (the denominator in (1)) and if x = 1, then by varying p3 we obtain the equality (3). In particular, with x = 1 the largest number of equalities is obtained for p2 = 3 Another way to exploit the (2) is, if x = 1: Mp = 2 ^ p1-1 = p2 * d-2 (4) Mp +2 = p2 * d (5) The (5) says that Mp +2 is a composite number (with Wagstaff we know that p2 = 3). Theorem A (Rosario Turco): Necessary and sufficient condition so that Mp = 2 ^ p-1 is a prime number is that all three conditions are true: 1) p is prime and of any shape (4k +1 or 4k +3) 2) If p = 4k +3 then S = 2p +1 mustn't be a prime number 3) Mp +2 = 3 * d with d prime number Ex: 2 ^ 3-1 = 7 Mersenne prime then 1) is true, the 2) also, for the 3) is: 7 +2 = 3 * 3 which Mp is prime 2 ^ p1 = p2 * d - x form 2^p1 = p2*d - x Interesting to see if x is not 1, so (2) becomes: Mp = 2 ^ p1 -1 = p2 * d - x - 1 (6) equivalent to: Mp + x +1 = p2 * d We will see that also serve two conditions: x +1 = p1-1 p2 = p1 or Mp+p1-1 = p1 * d (7) or Mp=p1*d-(p1-1) Mp=p1*d-phi(p1) With the generalization (7) we have the Theorem B (Rosario Turco): Necessary condition so that Mp = 2 ^ (p1-1) is a prime number is that all three conditions: 1) p1 is a prime number and any shape (4k +1 or 4k +3) 2) If p1 = 4k +3 then S = 2p1 + 1 must be non-prime 3) Mp + p1-1 = p1 * d with d odd but they aren't sufficient conditions. Now we will see why (7) is true: If in (6) say for example: x = 3 Mp + 4 = p2 * p3 Ex: 2 ^ 3-1 = 7 Mersenne prime 7 +4 = 11 is prime. it is impossible 11=p1*p3, because x +1 = 4 is not good 7 +2 = 3 * 3 OK!! 2 ^ 7-1 is prime. Note that in 3 * 3 is just the exponent p1 = 3 Mp = 2 ^ 3-1 Examples 2 ^ 5-1 = 31 31 + 4 = 35 = 5 * 7 = 35 then 31 is a Mersenne prime! Note that in 5 * 7 is precisely the exponent p1 = 5 Mp = 2 ^ 5-1 2 ^ 7-1 = 127 127 + 6 = 133 = 7 * 19 2 ^ 13-1 + 12 = 8191 +12 = 8203 = 13 * 631 Hence: x +1 = p1-1 Primality? Only with the Theorem A The (6), without generalization has another interest (primality), but the previous conditions are only necessary. If d is chosen as a big odd number (eg 127) you can check that (Mp + p1-1) / d = p1 (8) or (Mp+phi(p1))/d = p1 (9) Eg 2 ^ 13-1 = (8191 + 12) / 631 = 13 and d is a divisor of Mp+p1-1. So I must search a divisor d of Mp+p1-1 such that the (8) is true. Only Theorem A is useful for primality. Theorem B: (mp + p-1) / d = p As seen in previous Theorem B gives only necessary conditions, and is not effective for the primality of Mp, but only by p. We can prove this? yes. The Fermat's little theorem states that if p is prime then must divide 2 ^ (p-1) -1. We are in these conditions? Let's see. Mp + p - 1 = p * d (1) 2 ^ p - 1 + p -1 = p * d 2 ^ p = p * d-p + 2 I multiply and divide by 2 both sides: 2 / 2 * 2 ^ p = 2 / 2 (p-p * d) + 2 / 2 * 2 2 * 2 ^ (p-1) = 2 / 2 (p-p * d) + 2 / 2 * 2 I move left the term 2 / 2 * 2 = 2 2 * (2 ^ (p-1) -1) * d = p - p I divided by p and 2 (2 ^ (p-1) -1) / p = (d-1) / 2 (2) That we are in terms of Fermat's little theorem. For example: 2 ^ 37-1 with this technique we can only say that 37 is prime, because: 2 ^ 36-1/37 = 1857283155 n = (2 ^ p-1 + p-1) / p = 3714566311 (n-1) / 2 = 1857283155 The (2), which incorporates the significance of Fermat's little theorem, is obtained from (1) and hence Theorem B leads only to say that the exponent of Mp is prime. Why Theorem A is true? We saw in the previous blog that Mp + 2 = p2 * p2 in general with Mp = 2 ^ p1-1 Which is equivalent to the fact that: (2 ^ p1 +1) / p2 = p3 Now 2 ^ p1 +1 is a prime number if p1 = 2k; but if p1 is a prime number then 2 ^ p1 +1 is composite. So p1 = 4k +1 has got k odd. Ex: 2 ^ 3 +1 = 9 2 ^ 5 +1 = 33 2 ^ 7 +1 = 129 etc But for k odd p2 is a divisor of 2 ^ P1 +1 and P2 = 3; infact each (2 ^ p1 +1)% 9 = 3. If p2 = 3, p3 is a prime number? Yes, because it is also true that (2 ^ p1 +1) / p3 = p2 _______________________________________________ Prime mailing list Prime@hogranch.com http://hogranch.com/mailman/listinfo/prime
