Math 782: Analytic Number Theory (Instructor’s Notes)*

Analytic Versus Elementary:

• Terminology (Analytic Number Theory makes use of Complex Analysis and Elemen- tary Number Theory does not; but it isn’t so simple to distinguish.)

• Writing an integer as a sum of two squares. This is the first of a few examples of how Complex Analysis can be used to answer a question seemingly unrelated to it: if a and b are sums of 2 squares, why must ab be also?

• Coverings. A covering of the integers is a system of congruences x ≡ aj (mod mj) such that every integer satisfies at least one of these congruences. A perfect covering is one in which each integer satisfies exactly one of the congruences. Suppose we have a finite perfect covering of the integers with each modulus mj > 1. Show that the largest modulus occurs twice. Use that

1 1 − z =

r∑ j=1

zaj

1 − zmj for |z| < 1,

where 1 < m1 ≤ m2 ≤ · · · ≤ mr and 0 ≤ aj < mj . Suppose that mr−1 6= mr. Then the right-hand side, and not the left-hand side, gets large (in absolute value) as z approaches exp(2πi/mr). The contradiction implies mr−1 = mr.

• Preliminaries on exponentials. The following are useful formulas: n−1∑ j=0

e(i2πk/n)j = {

0 if k 6= 0 n if k = 0,

∫ 1 0

ei2πkθ dθ = 1 2π

∫ 2π 0

eikθ dθ = {

0 if k 6= 0 1 if k = 0.

• Putnam Problem A-6, 1985. If p(x) = a0 + a1x+ · · · + amxm is a polynomial with real coefficients aj , then set Γ(p(x)) = a20 + a

2 1 + · · · + a2m. Let f(x) = 3x2 + 7x + 2.

Find, with proof, a polynomial g(x) with real coefficients such that (i) g(0) = 1, and (ii) Γ(f(x)n) = Γ(g(x)n) for all positive integers n.

There are perhaps better ways to do this problem, but we use what we just learned to establish

Γ(p(x)) = ∫ 1

0

p ( e2πiθ

) p ( e−2πiθ

) dθ.

Note that f(x) = (3x+ 1)(x+ 2). That g(x) = 6x2 + 5x+ 1 = (3x+ 1)(2x+ 1) provides a solution follows from

Γ (f(x)n) = ∫ 1

0

f ( e2πiθ

)n f ( e−2πiθ

)n dθ

*These notes are from a course taught by Michael Filaseta in the Fall of 1996.

1

2

= ∫ 1

0

( 3e2πiθ + 1

)n ( e2πiθ + 2

)n ( 3e−2πiθ + 1

)n ( e−2πiθ + 2

)n dθ

= ∫ 1

0

( 3e2πiθ + 1

)n ( 2e2πiθ + 1

)n ( 3e−2πiθ + 1

)n ( 2e−2πiθ + 1

)n dθ

= ∫ 1

0

g ( e2πiθ

)n g ( e−2πiθ

)n dθ

= Γ (g(x)n) .

• A Geometric Problem. Let R be a rectangle, and suppose R can be expressed as a union of rectangles Rj with edges parallel to R and common points only along these edges. Suppose each Rj has at least one edge of integer length. Then R itself must have an edge of integer length.

Let R be positioned so that it is in the first quadrant of the xy-plane, with an edge on each of the x and y axes. Let Rj have bottom left-hand corner (uj , vj), and let αj and βj be the horizontal and vertical dimensions of Rj respectively. Observe that∣∣∣∣

∫ w+γ w

e2πit dt

∣∣∣∣ = 12π ∣∣e2πiγ − 1∣∣ = 0 ⇐⇒ γ ∈ Z.

One uses that one of αj and βj is an integer for each j and that∣∣∣∣ ∫ ∫ R

e2πi(x+y) dx dy

∣∣∣∣ = ∣∣∣∣

r∑ j=1

∫ ∫ Rj

e2πi(x+y) dx dy

∣∣∣∣ to obtain the desired result.

Homework:

(1) (a) If a and b are integers which can each be expressed in the form x2 + 5y2 for some integers x and y, explain why it is possible to express ab in this form as well.

(b) Determine whether the following is true: if a and b are integers and ab can be expressed in the form x2 + 5y2 with x and y integers, then either a or b can be as well. Either prove the result or give a counterexample.

(2) (Putnam Problem A-5, 1985) For m a positive integer, define

Im = ∫ 2π

0

cosx cos(2x) cos(3x) · · · cos(mx) dx.

(a) Use one of the preliminary results above (yes, you must to get credit for the problem) to determine with proof for which integers m with 1 ≤ m ≤ 10 we have Im 6= 0. (Hint: Use that cosx = (eix + e−ix)/2.)

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(b) Generalize part (a). In other words, determine which positive integers m are such that Im 6= 0. The answer should be in the form: Im 6= 0 if and only if m ≡ u (mod v) where v is an integer and u is a list of integers. Justify your answer.

Elementary Aspects:

• The Fundamental Theorem of Arithmetic (the atoms of which the matter called integers is made are “primes”)

• There are infinitely many primes (Give three proofs: (i) Euclid’s, (ii) 22n + 1 is divisible by a prime > 2n+1, and (iii)

∏ p(1 − 1/p2)−1 = π2/6, an irrational number.)

• Explain the notion of rearranging the terms of a series as well as the following lemmas about absolute convergence.

Lemma 1. If the terms of an absolutely converging series ∑∞

n=1 an with an ∈ C are rearranged, then the value of the series remains unchanged. On the other hand, if the series

∑∞ n=1 an with an ∈ R converges but not absolutely, then any real value can be

obtained from an appropriate rearrangement of the series.

Use the example of the alternating harmonic series ∑∞

n=1(−1)n+1/n to illustrate the second half of the lemma (and to give the first half more meaning). For the first half, let S =

∑∞ n=1 an and let

∑∞ n=1 a

′ n be a rearrangement of it. Let ε > 0. Fix n0 such that∑

n>n0 |an| < ε. Let N be sufficiently large so that the terms a1, . . . , an0 appear among

a′1, . . . , a ′ N . Then ∣∣∣∣

N∑ n=1

a′n − S ∣∣∣∣ ≤ ∑

n>n0

|an| < ε.

The first part of the lemma follows.

Lemma 2. The product of two absolutely converging series converges absolutely.

The product of ∑∞

n=1 an and ∑∞

n=1 bn is ∑∞

n=2 cn where cn = ∑n−1

k=1 akbn−k. The result follows since

∑N n=2 |cn| is an increasing function of N bounded above by

( N∑ n=1

|an| )( N∑

n=1

|bn| )

≤ ( ∞∑

n=1

|an| )( ∞∑

n=1

|bn| ) .

In fact, ∑∞

k=1

∑∞ `=1 |akb`| converges. Lemma 2 is of interest but is not really needed in

what follows.

Homework:

(1) Let f : (Z+)2 7→ C. We say that the iterated series ∞∑

n=1

∞∑ m=1

f(m,n) = ∞∑

n=1

( ∞∑ m=1

f(m,n) )

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converges if for each positive integer n, the series ∑∞

m=1 f(m,n) converges, and if

lim N→∞

N∑ n=1

( ∞∑ m=1

f(m,n) )

exists. The value of this limit is called the value of the iterated series. We say that the double series ∑

1≤n

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for any M > 0. This is a contradiction (the left-hand side is finite).

• Logarithms and ∏∞n=1(1 − an). Suppose that 0 ≤ an ≤ 1/2. This condition implies

(∗) ∞∑

k=2

akn k

≤ ∞∑

k=2

akn ≤ 2a2n ≤ an.

We use

log(1 − x) = − ∞∑

k=1

xk

k for |x| < 1 (actually for − 1 ≤ x < 1).

Observe that

log N∏

n=1

(1 − an) = N∑

n=1

log(1 − an) ≥ −2 N∑

n=1

an.

The left-hand side is decreasing so that if ∑∞

n=1 an converges, then so does ∏∞

n=1(1− an). The inequality log

∏N n=1(1 − an) ≤ −

∑N n=1 an implies that if

∏∞ n=1(1 − an) converges,

then so does ∑∞

n=1 an.

• The estimate ∏p≤x (

1 − 1 p

) ≤ 1

log x .

Use ∏ p≤x

( 1 − 1

p

)−1 ≥

[x]∑ n=1

1 n ≥ ∫ x

1

1 t dt = log x.

• Second proof that ∑p 1/p diverges. Let P (x) =

∏ p≤x(1 − (1/p))−1 and S(x) =

∑ p≤x 1/p. Then by (∗)

logP (x) = S(x) + E where 0 ≤ E ≤ 2 ∑ p≤x

1 p2

≤ 2.

The result follows since P (x) ≥ log x. (Note: we have found a “good” lower bound for∑ p≤x 1/p.)

• A noteworthy example. The product

∏ p(1 − (1/p)) and

∑∞ n=1 µ(n)/n are related (if you expand the product

and rearrange the terms, you get the sum). The value of the product is 0 and the value of the sum is 0; however, it is considerably harder to show the latter. (It is equivalent to the Prime Number Theorem).

• The notation π(x) and a proof that limx→∞ π(x)/x = 0. Show that in fact π(x) < Cx/ log log x for some constant C by using the sieve of Er-

atosthenes. (Discuss sieves in general.)

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Homework:

(1) (a) Modify the previous argument to show that almost all positive integers n have a prime factor ≤ log n. More precisely (and moreover) show that

|{n ≤ x : n does not have a prime factor ≤ logn}|

is bounded above by Cx/ log log x for some constant C. (Hint: Most n ≤ x which do not have a prime factor ≤ log n also do not have a prime factor ≤ (log x)/2.) (b) Suppose w(n) is any function defined for n ∈ Z+ which tends to infinity with n. Prove t