certain special types of posite. One such test is presented next. Theorem 11.3. If p and q = 2p +1 are primes, then either q |M, or q | Mp +2, but not both. Proof. With reference to Fermat's theorem, we know

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Chapter10: Sequences, Series, And Probability
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Problem 63RE
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Create two exercises related to the Theorem 11.3 and their solution
PECIAL FORM
229
hods for determining whether certain special types of
conditions 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 that
Mersenne 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 that
Theorem 11.3. If p and q = 2p+1 are primes, then either q | M, or q | Mp+ 2, but
not both.
Proof. With reference to Fermat's theorem, we know that
29-1 -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 both
g|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, then
a = 2p +1 = 47 is also a prime, so that we may consider the case of M23. The
question reduces to one of whether 47 | M23 Oor, to put it differently, whether 225 =
1 (mod 47). Now, we have
23 2°(2°)* = 2°(-15)*(mod 47)
But
(-15) = (225) = (-10)² = 6 (mod 47)
%3D
tke
Putting these two congruences together, we see that
223 = 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: What
out that Theorem 11.3 is of no help in testing the primality of
24-1/2 2"1 (mod q)
Transcribed Image Text:PECIAL FORM 229 hods for determining whether certain special types of conditions 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 that Mersenne 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 that Theorem 11.3. If p and q = 2p+1 are primes, then either q | M, or q | Mp+ 2, but not both. Proof. With reference to Fermat's theorem, we know that 29-1 -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 both g|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, then a = 2p +1 = 47 is also a prime, so that we may consider the case of M23. The question reduces to one of whether 47 | M23 Oor, to put it differently, whether 225 = 1 (mod 47). Now, we have 23 2°(2°)* = 2°(-15)*(mod 47) But (-15) = (225) = (-10)² = 6 (mod 47) %3D tke Putting these two congruences together, we see that 223 = 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: What out that Theorem 11.3 is of no help in testing the primality of 24-1/2 2"1 (mod q)
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