Create two exercises related to the Theorem 11.3 and their solution PECIAL FORM229hods for determining whether certain special types ofconditions on q will ensure that q | M,? The answer is to be found in Theorem 11.4.In terms of the Legendre symbol, the latter condition becomes the requirement thatMersenne numbers are prime or composite. One such test is presented next.Theorem 11.4. If q = 2n + 1 is prime, then we have the following:(b) q| M,+2, provided that q 3 (mod 8) or q = 5 (mod 8).(a) q| M, provided that q 1 (mod 8) or q =7 (mod 8).Proof. To say that q | M, is equivalent to asserting thatTheorem 11.3. If p and q = 2p+1 are primes, then either q | M, or q | Mp+ 2, butnot both.Proof. With reference to Fermat's theorem, we know that29-1 -1= 0 (mod q)and, factoring the left-hand side, that(24-1)/2 - 1)(29-1)/2 + 1) = (2P – 1)(2P + 1)= 0 (mod q)What amounts to the same thing:Mp(Mp+2) = 0 (mod q)The stated conclusion now follows directly from Theorem 3.1. We cannot have bothg|Mp and q | Mp+2, for then q|2, which is impossible.A single application should suffice to illustrate Theorem 11.3: if p = 23, thena = 2p +1 = 47 is also a prime, so that we may consider the case of M23. Thequestion reduces to one of whether 47 | M23 Oor, to put it differently, whether 225 =1 (mod 47). Now, we have23 2°(2°)* = 2°(-15)*(mod 47)But(-15) = (225) = (-10)² = 6 (mod 47)%3DtkePutting these two congruences together, we see that223 = 23.6 = 48 = 1 (mod 47)whence M23 is composite.We might point" 29, say; in this instance, 59 M29, but instead 59| M29+2.he two possibilities q| M, or q| Mp+2, is it reasonable to ask: Whatout that Theorem 11.3 is of no help in testing the primality of24-1/2 2"1 (mod q)

Name: Create two exercises related to the Theorem 11.3 and their solution PE
Uploaded: 2024-03-20T07:07:26.000Z
Channel: Bartleby
Description: Create two exercises related to the Theorem 11.3 and their solution PECIAL FORM229hods for determining whether certain special types ofconditions on q will ensure that q | M,? The answer is to be found in Theorem 11.4.In terms of the Legendre symbol,