show that 2^p+1 is a factor of n

P) _ p 02 - 1)/2 when 6=+ 1 e- 1

Show that n = 1 or 2. 8. Let m n be positive integers. Show that 4mn − m Show that if p = 2n + 1

and would have been sufficient to prove that 2P - 2 has a factor p for any a prime number p and 0 < r < n

2 янв. 2013 г. Its running time to find a factor p is exp[2 ln(p) lnln(p)]1/2. If n = pq with p and q both near n1/2 then this is L(1) = exp[ln(n) lnln(n)] ...

And we can show this if p-1 has a factor F exceeding p² with the property II(n−1 p - 1) <. F(a)' p❘n so that F(a) <nloglogn/loga. Hence it is correct ...

of these points is a 2-division point modulo p for some prime factor p of n

6 нояб. 2008 г. It hopes that some prime factor p of N has smooth p−1. It picks b0 ... Bostan shows that multipoint evaluation of a polynomial of degree < n ...

+ 1. There is a prime factor p Qn. Suppose p ≤ n; then p n! = n(n−1)(n−2)2 

associated to a p-paperfolding sequence. It is known that the number of factors of length n of a 2-paperfolding sequence (i.e. its complexity function) is 

24 нояб. 2016 г. ... factor v. The sub-Gaussian property implies that Z − EZ has a sub ... t) ≤ e−(n−1)t2/2 . In other words as soon as µ(A) ≥ 1/2

where q is some prime divisor of R4 depending on p. Proof Let r = r(p) be an order of apparition of p such that rl(N2 + 1); then.

If p is a prime and n is an integer such that p

For example we show PIoo

The graph suggests that the function has three zeros one of which is x = 2. It's easy to show that f(2) = 0

12 fév. 2006 a natural number n i.e.

Let n be an even positive integer and let p(x) be an n-degree polynomial such that 2. Factor p(x) + 1. 3. Prove that the sum is the root of a monic ...

(i) n is squarefree and (ii) for every prime p dividing n also (p ? 1)

We have 2n= p k+1 = (p+1)(pk 1 p 2 + 1) so (p+1) is a power of 2 say p+1 = 2 l Then 2n= p k+1 = (2 l1) +1 = (2l)k (2l)k 1+ +(2 ) 1 +1 = (2l)k (2l) 1+ +(2l) and this is an odd multiple of 2 l Since 2n is an odd multiple of 2 we must have 2n = 2l So we have 2n = p k+ 1 = (2n 1) + 1 This only happens when n= 1 or when k= 1 neither of which

n p has a prime factor q with q < r n p < 3 p n < p and this prime factor q is also a divisor of n; which contradicts the de nition of p: Therefore n p must be prime Question 6 [p 87 #12] Show that every integer greater than 11 is the sum of two composite integers Solution: If n > 11 and n is even then n 4 is even and n 4 > 7; so that n 4

that there is at least one prime factor pof the form 8n+ 5 If not a= Q q i i 1(mod 8) (as a product of numbers congruent to 1(mod 8) is still 1(mod 8)) But (2p 1p 2:::p k)2 4(mod 8) as p2 j 1(mod 8) for every j and a= (2p 1p 2:::p k)2 + 1 5(mod 8) so this is a contradiction At least one of the prime factors of a say p must be of the

(11) Show that if 2n ?1 is prime then n is prime For if n = pq say with pq > 1 then since y ? 1 divides yq ? 1 we have (y = xp) that xp ?1 divides xn ?1 = (xp)q ?1 Hence (x = 2) 2p ?1 divides 2n ?1 and 2p ? 1 6= 1 6= 2 n ? 1 (12) Show that if 2n +1 is prime then n is a power of 2 For suppose n = m? with ? > 1

show that second equals {z factor the ?rst p} 1 (1)1 (mod p) p 2 (1)2 (mod p) p+ 1 p 1 (1) (mod p) 2 2 p+ 1 p1 p 1:::(p 1) ( 1) 2 1 2::: 2 second {z factor} 2 (mod p) {z x} p1 2 is even since p 1 mod 4 and so second factor equals the ?rst factor so x= p1! 2 solves x 2 1 mod pif p 1 mod 4 Theorem 23 There are in?nitely

If n ‚ 2 and an ¡1 is prime we call an ¡1 a Mersenne prime For which integers a can an ¡1 be prime? We take n ‚ 2 as if n = 1 then a is just one more than a prime We know using the geometric series that an ¡1 = (a¡1)(an¡1 +an¡2 +¢¢¢ +a+1): (1) So a¡1 j an¡1and therefore an¡1will be composite unless a¡1 = 1 or