Download -8- 6CCM320A (CM320X) SECTION D – Distribution of Prime

-8-
6CCM320A (CM320X)
SECTION D – Distribution of Prime Numbers
7.
(a) How many zeros are there at the end of the decimal expansion of
90
45
?
(Hint: You may want to check the highest powers of 2 and 5 that divide
the numerator and the denominator of the corresponding expression.)
(b)
i. Prove that there exists precisely one prime number p for which p + 8
and p + 16 are also both prime and find it.
ii. Prove that there are infinitely many primes p such that p + 8 is composite. You may use any result presented in class.
(Hint: You may either want to argue by contradiction and part (i),
or use Dirichlet’s Theorem, in which case you may find inspecting the
remainders of division by 3 beneficial.)
(c) Let {pn } be the list of all primes in increasing order, so that for all n ≥ 1,
pn < pn+1 , and let gn = pn+1 − pn be the corresponding gaps.
i. State Bertrand’s Postulate and show that it implies that for every n,
gn ≤ pn .
ii. Show that for every δ > 0, gn ≤ δpn for n sufficiently large. You may
use any result from the course, provided you quote it correctly.
iii. Let x > 3 be a parameter and M (x) be the average gap for all primes
p ≤ x, i.e. the average of all gn with pn+1 ≤ x. Explain why M (x) =
P
1
(pn+1 − pn ) and use the latter expression to show that as
π(x)−1
pn+1 ≤x
x → ∞,
M (x) ∼ log x.
(Hint: Let p(x) be the largest prime with p(x) ≤ x. You may want to
prove that p(x) ∼ x, for which end you may use any result from the
course.)
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-98.
6CCM320A (CM320X)
(a) State Chebyshev’s Theorem and the Prime Number Theorem carefully,
explaining all the notation you use. State which of the results above is
stronger, and briefly explain the difference between them.
(b) Use the Prime Number Theorem in order to determine whether for all
sufficiently large x, xπ(x) > 2x or xπ(x) < 2x , and rigorously prove your
claim.
(Hint: You may want to take logarithms of both sides.)
(c) Let ϕ(n) be the Euler totient function defined as
ϕ(n) = #{m ∈ N, m ≤ n : (m, n) = 1},
i.e. the number of positive integers m ≤ n co-prime to n.
i. Compute ϕ(pk ) for p prime and k ≥ 1.
ii. Assuming that ϕ is multiplicative deduce that given a prime factorization of a number n = pk11 · . . . · pke e ,
ϕ(n) = (pk11 − pk11 −1 ) · . . . · (pke e − pke e −1 ) = n ·
Y
1
(1 − ).
p
p|n
Is ϕ completely multiplicative?
(d)
i. Determine the (necessary and sufficient) criterion for a sequence
Sa = {a + 2013 · n}n≥1
to contain infinitely many primes, depending on an integer a ≥ 0. You
may use any results of the course, provided you quote them correctly.
ii. Construct 3 concrete examples of sequences as above that contain infinitely many primes, 2 examples with precisely one prime and 3 examples with no primes at all. Are there sequences with precisely 2
primes?
iii. Using parts d(i) and (c) compute how many of the sequences Sa , with
0 ≤ a ≤ 2012, contain infinitely many primes.
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6CCM320A (CM320X)
SECTION C – Distribution of Prime Numbers
1.
(a). First we find the highest power of 2 that divides 90! and 45!. That would
be
[90/2]+[90/4]+[90/8]+[90/16]+[90/32]+[90/64] = 45+22+11+5+2+1 = 86
and
[45/2] + [45/4] + [45/8] + [45/16] + [45/32] = 22 + 11 + 5 + 2 + 1 = 41,
and thus the highest power of 2 that divides 90
is 86 − 2 · 41 = 4.
45
Concerning the power of 5, the situation is the following:
[90/5] + [90/25] = 18 + 3 = 21
and
[45/5] + [45/25] = 9 + 1 = 10,
and thus the highest power of 5 that divides 90
is 21 − 2 · 10 = 1. All in
45
all there is 1 = min 4, 1 zero at the end of decimal expansion of 90
.
45
(b).
(i) First, the number we are after is 3, whence the triple is (3, 11, 19). If
p 6= 3, then it is necessarily not divisible by 3, hence either p = 3k +1
or p = 3k + 2. In the former case p + 8 = 3(k + 3) cannot be prime
and in the latter case p + 16 = 3(k + 6) cannot be prime.
(ii) Arguing by contradiction, if there are only finitely many primes p
with p + 8 composite, that would mean that for all p sufficiently
large p + 8 is prime. Let p be any sufficiently large (exists by the
infinitude of primes), then p + 8 is again a prime of the same form,
whence p + 16 = (p + 8) + 8 is a prime too. But then the numbers
(p, p + 8, p + 16) form a prime triple, which is contradictory to part
(i).
Otherwise, consider the sequence {3 · n + 1}. By Dirichlet, there exist
infinitely many primes p in this sequence; p + 8 = (3n + 1) + 8 =
3(n + 3) is necessarily composite.
(c).
(i) Bertrand’s Postulate states that for every m ≥ 1, there exists at least
one prime p, in the range m < p ≤ 2m. Taking m = pn guarantees
the existence of a prime pn < p ≤ 2pn (which excludes pn itself), and
then gn = pn+1 − pn ≤ p − pn ≤ 2pn − pn = pn .
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6CCM320A (CM320X)
(ii) It is known that for x sufficiently large there exists a prime p in the
range x < p ≤ (1 + δ)x. Taking x = pn , we have gn = pn+1 − pn ≤
p − pn ≤ (1 + δ)pn − pn = δpn .
(iii) Since there are precisely π(x) − 1 tuples (pn , pn+1 ) with pn+1 ≤ x, we
have
X
X
1
1
M (x) =
gn = M (x) =
(pn+1 − pn )
π(x) − 1 p ≤x
π(x) − 1 p ≤x
n+1
n+1
Let x be given, and p(x), as defined in the hint, be equal to p(x) = pk .
Then
X
(pn+1 −pn ) = (p2 −p1 )+(p3 −p2 )+. . .+(pn −pn−1 ) = pk −p1 = p(x)−2
pn+1 ≤x
by telescoping series (i.e. all the summands cancel out but the last
and the first). Therefore
M (x) =
p(x) − 2
x
∼
= log x,
π(x) − 1
x/ log x
provided that we show that p(x) ∼ x.
To this end let us assume by contradiction that p(x) 6∼ x, i.e. lim inf
x→∞
δ0 < 1 (note that since p(x) ≤ x, lim inf
x→∞
p(x)
x
p(x)
x
p(x)
x
≤ 1, and thus if the lat-
ter is an equality then p(x) ∼ x). Then
< δ1 < 1 for a sequence
of numbers x growing to infinity (and thus arbitrarily large), for any
choice of a fixed δ1 > δ0 , i.e. p(x) < δ1 x, or, equivalently, δ11 p(x) < x,
which means that there is no prime in the range (p(x), δ11 p(x)), contradicting the existence of prime in that range (that was already
mentioned in part (ii)).
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2.
6CCM320A (CM320X)
(a). Chebyshev’s Theorem: π(x) logx x , meaning a ≤ x/π(x)
≤ A for suflog x
ficiently large x, for some 0 < a < A < ∞. Prime Number Theorem:
π(x) ∼ logx x , meaning lim x/π(x)
= 1, which is stronger than Chebyshev’s.
log x
x→∞
to flucThe difference is that Chebyshev’s Theorem allows the ratio x/π(x)
log x
tuate as long as it is bounded away from 0 and ∞, whereas the Prime
Number Theorem imposes for this ratio to concentrate around 1.
(b). By taking logarithms, the statement xπ(x) > 2x is equivalent to π(x) log x >
x log 2. Now, by the PNT,
π(x) log x
→ 1 > log 2,
x
and thus, for x sufficiently large, the left hand side is larger than the right
hand side, which then implies that xπ(x) > 2x indeed.
(c).
(i) A number m ≤ pk is co-prime to pk if and only if it is not divisible
k
by p; thus the total number of such is pk − pp = pk − pk−1 .
(ii) By the multiplicativity,
ϕ(n) = ϕ(pk11 ) · . . . · ϕ(pke e ) = (pk11 − pk11 −1 ) · . . . · (pke e − pke e −1 )
so that
ϕ(n) = pk11 (1 − 1/p1 ) · . . . · pke e (1 − 1/pe ) = n ·
Y
1
(1 − ).
p
p|n
The function ϕ is not completely multiplicative, since, for example,
ϕ(4) = 2 6= 1 = ϕ(2) · ϕ(2).
(d).
(i) By Dirichlet’s Theorem (stating that {r + n · q}n≥1 contains infinitely
many primes, if and only if (r, n) = 1), Sa contains infinitely many
primes, if and only if (a, 2013) = 1. Since (post-factum) 2013 =
3 · 11 · 61, this happens if and only if 3 -, 11 -, 61 - a, but this is not
expected from a student.
(ii) Infinitely many primes: a = p any prime not dividing 2013, e.g.
a = 5, 7, 13.
Precisely one prime: a = 3, 11 - by trying small primes to divide
2013, or factoring 2013 in the first place.
No primes: a = 6, 9, 12.
There are either 1 or infinitely many primes, so no sequences with
precisely 2 primes.
(iii) The number of such sequences is, by Dirichlet (part (i)) ϕ(2013) =
(3 − 1) · (11 − 1) · (61 − 1) = 1200.
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