Prime Numbers - Homework 3 Solutions
Yankı Lekili, Winter 2017
1) (i) Prove that for any natural number n
22n
≤
2n
(ii) Prove that
(iii) Prove that
2n
n
2n
n
≤ 22n .
is not divisible by any prime p > 2n.
2n
n
is divisible by all primes n < p ≤ 2n.
P
2n
. It is immediate from this
(i) Consider the binomial expansion 22n = (1 + 1)2n = 2n
k=0
k
2n
2n
that 2 > n since the latter appears in the sum.
2n < 2n 2n since the binomial
Also, 2n
n is the largest term in the binomial expansion, and so 2
n
expansion has (2n + 1) terms, two of which are equal to 1.
(ii) Prime divisors of
divide 2n
n .
(2n)!
,
(n!)2
(iii) If n < p ≤ 2n, then
h
in particular divide (2n)!, hence a prime p satisfying p > 2n does not
2n
p
i
≥ 1 and
h i
n
p
= 0. Hence, p |
2n
n
.
2) We have seen a proof of Chebyshev’s theorem in class which states that there exist constants
c1 , c2 > 0 such that
x
x
c1
< π(x) < c2
log x
log x
To write this more succinctly, we introduce the notation: f g, which means that there there
exist positive constants c1 , c2 such that c1 f < g < c2 f. In this notation, Chebyshev’s theorem
is stated as
x
π(x)
log x
Relatedly, the prime number theorem states that
x
lim π(x)/
= 1.
x→∞
log x
1
To write this, we introduce the notation f ∼ g to signify that limx→∞
to 1.
f (x)
g(x)
exists and is equal
Show that the relations and ∼ are equivalence relations:
(a) Reflexivity: f f and f ∼ f .
(b) Symmetry : If f g then g f , and its analogue for ∼.
(c) Transitivity: If f g and g h, then f h, and its analogue for ∼.
(a) Reflexivity: follows from
f (x)
f (x)
= 1.
(b) Symmetry: For , it follows from the fact that if a <
For ∼, it follows from the fact that limx→∞
f (x)
g(x)
g(x)
f (x)
< A then
= 1 implies limx→∞
g(x)
f (x)
1
A
<
f (x)
g(x)
< a1 .
= 1.
(c) Transitivity: We have
h(x) g(x)
h(x)
=
·
.
f (x)
g(x) f (x)
Thus, if a1 <
limx→∞
g(x)
f (x)
h(x)
g(x)
< A1 and a2 <
= 1, then limx→∞
h(x)
f (x)
g(x)
f (x)
< A2 , then a1 a2 <
h(x)
f (x)
< A1 A2 , and if limx→∞
h(x)
g(x)
=
= 1.
3) Use Chebyshev’s theorem to show that π(x) <
x
1000
for x sufficiently large.
By Chebyshev (note that since is symmetric, also logx x π(x)), there exists a constant A > 0,
1
1
so that π(x) ≤ A logx x for x sufficiently large. Thus if logA x < 1000
, then π(x) < 1000
x. This will
1000A
happen for x > e
.
4) Use Chebyshev’s theorem to show that there do not exist polynomials f, g ∈ Z[X] such that
(n)
for all integers n ≥ 1.
π(n) = fg(n)
(n)
If π(n) = fg(n)
, then, since π(x) ≤ x everywhere, it forces degf ≤ degg + 1. Now, if degf ≤ degg,
then π(x) would be bounded, contracdicting the infinitude of primes. If otherwise, degf =
degg + 1, then (note that since π(x) is positive, by changing sign of f, g if necessary we may
assume that both are positive for x sufficiently large), by using the leading powers of f and g
(x)
x 6 logx x , contradicting Chebyshev’s Theorem.
only, fg(x)
5) Show that asymptotically, “there exists more primes than squares”. More precisely, use
Chebyshev’s theorem to show that
π(x) = #{p ∈ N : p is prime and p ≤ x} > 1000 · #{n ∈ N : n is a square and n ≤ x}
for all x sufficiently large, where by “square” we mean “square of an integer”. Can you replace
1000 in the latter inequality by an arbitrary constant C > 1000?
We have
#{n ∈ N : n is a square and n ≤ x} = #{m ∈ N : m2 ≤ x} = #{m ∈ N : m ≤
2
√
√
x} = [ x].
√
Thus, we are to compare π(x) and 1000 · [ x]. By Chebyshev’s Theorem, there exists a constant
a > 0 so that
√
√
x
a
x
π(x) > a
=
· 1000 x,
log x
1000 log x
√
which
is larger than 1000.[ x] for x sufficiently large, no matter how small a is, since as x → ∞,
√
x
log x → ∞.
The constant 1000 can be replaced by arbitrary large on at the expense of increasing “x sufficiently large”, by using the same analysis as above.
6) Use Bertrand’s postulate to show that for every integer n ≥ 3, there exists a prime number
such that p < n < 2p.
Given an integer n ≥ 3, consider p the largest prime strictly smaller than n. We claim that for
this choice of p, necessarily, p < n < 2p. Since the former inequality is satisfied by construction,
it remains to prove the latter. To see that, by Bertrand’s postulate, applied on p, there exists
a prime q with p < q < 2p. If q < n, that would contradict the maximality of p with the same
property. Thus necessarily n ≤ q, whence p < n ≤ q < 2p.
7) Prove that [x] + [x + 12 ] = [2x] and, more generally
n−1
X
k=0
k
x+
= [nx] .
n
Write x = [x] + {x} where [x] is the integer part of x and {x} is the fractional part of x. Suppose
i+1
i
n ≤ {x} < n for some 0 ≤ i ≤ n − 1. Then,
n[x] + i ≤ nx = n[x] + n{x} < n[x] + i + 1
Hence, [nx] = n[x] + i. On the other hand,
i+k
k
i+1+k
≤ x + < [x] +
n
n
n
Hence x + nk = [x] if 0 ≤ k < n − i and x + nk = [x] + 1 if n − i ≤ k ≤ n − 1. It follows
that
n−1
X
k
x+
= n[x] + i.
n
[x] +
k=0
3