Prime Numbers

In particular, when a, q are coprime, we obtain a beautiful estimate of the

where the overall error depends in complicated ways on a, q, n, and, of course, whatever is our theorem of choice on the distribution of primes in a residue class. We uncover thus a fundamental spectral property of primes: When q is small, the magnitude of the exponential sum is e ectively reduced, by an explicit factor µ/ϕ, below the trivial estimate π(n). Such reduction is due, of course, to cancellation among the oscillating summands; relation (1.39) quantifies this behavior.

Vinogradov was able to exploit the small-q estimate above in the following way. One chooses a cuto Q = lnB n for appropriately large B, thinking of q as “small” when 1 ≤ q ≤ Q. (It turns out to be enough to consider only the range Q < q < n/Q for “large” q.) Now, the integrand in (1.36) exhibits “resonances” when the integration variable t lies near to a rational a/q for the small q [1, Q]. These regions of t are traditionally called the “major arcs.” The rest of the integral—over the “minor arcs” having t ≈ a/q with q (Q, n/Q)—can be thought of as “noise” that needs to be controlled (bounded). After some delicate manipulations, one achieves an integral estimate in the form

where we see a resonance sum from the major arcs, while now contains all previous arithmetic-progression errors plus the minor-arc noise. Already in the above summation over q [1, Q] one can, with some additional algebraic e ort, see how the final ternary-Goldbach estimate (1.12) results, as long as the error and the finitude of the cuto Q and are not too troublesome (see Exercise 1.68).

It was the crowning achievement of Vinogradov to find an upper bound on the minor-arc component of the overall error . The relevant theorem is this: If gcd(a, q) = 1, q ≤ n, and a real t is near a/q in the sense |t − a/q| ≤ 1/q2, then

with absolute constant C. This result is profound, the proof di cult— involving intricate machinations with arithmetic functions—though having undergone some modern revision, notably by R. Vaughan (see references below). The bound is powerful because, for q (Q, n/Q) and a real t of the theorem, the magnitude of En(t) is reduced by a logarithmic-power factor below the total number π(n) of summands. In this way the minor-arc noise has been bounded su ciently to allow rigor in the ternary-Goldbach estimate. (Powerful as this approach may be, the binary Goldbach conjecture has so far

been beyond reach, the analogous error term , which includes yet noisier components, being so very di cult to bound.)

In summary: The estimate (1.39) is used for major-arc “resonances,” yielding the main-term sum of (1.40), while the estimate (1.41) is used to bound the minor-arc “noise” and control the overall error . The relation (1.40) leads finally to the ternary-Goldbach estimate (1.12). Though this language has been qualitative, the reader may find the rigorous and compelling details—on this and related additive problems—in the references [Hardy 1966], [Davenport 1980], [Vaughan 1977, 1997], [Ellison and Ellison 1985, Theorem 9.4], [Nathanson 1996, Theorem 8.5], [Vinogradov 1985], [Estermann 1952].

Exponential-sum estimates can be, as we have just seen, incredibly powerful. The techniques enjoy application beyond just the Goldbach problem, even beyond the sphere of additive problems. Later, we shall witness the groundwork of Gauss on quadratic sums; e.g., Definition 2.3.6 involves variants of the form (1.33) with quadratic f . In Section 9.5.3 we take up the issue of discrete convolutions (as opposed to continuous integrals) and indicate through text and exercises how signal processing, especially discrete spectral analysis, connects with analytic number theory. What is more, exponential sums give rise to attractive and instructive computational experiments and research problems. For reader convenience, we list here some relevant Exercises: 1.35, 1.66, 1.68, 1.70, 2.27, 2.28, 9.41, 9.80.

1.4.5Smooth numbers

Smooth numbers are extremely important for our computational interests, notably in factoring tasks. And there are some fascinating theoretical applications of smooth numbers, just one example being applications to a celebrated problem upon which we just touched, namely the Waring problem [Vaughan 1989]. We begin with a fundamental definition:

Definition 1.4.8. A positive integer is said to be y-smooth if it does not have any prime factor exceeding y.

What is behind the usefulness of smooth numbers? Basically, it is that for y not too large, the y-smooth numbers have a simple multiplicative structure,

yet they are surprisingly numerous. For example, though only a vanishingly

√

small fraction of the primes in [1, x] are in the interval [1, x], nevertheless

√

more than 30% of the numbers in [1, x] are x-smooth (for x su ciently large). Another example illustrating this surprisingly high frequency of smooth numbers: The number of (ln2 x)-smooth numbers up to x exceeds √x for all su ciently large numbers x.

These examples suggest that it is interesting to study the counting function

This kind of combinatorial reasoning can be used, as Legendre once did, to show that π(x) = o(x). To that end, show that

where the error term E is O(2π(y)). Now use this last relation and the fact that the sum of the reciprocals of the primes diverges to argue that π(x)/x → 0 as x → ∞. (Compare with Exercise 1.11.)

1.14. Starting with the fundamental Theorem 1.1.1, show that for any fixed> 0, the number d(n) of divisors of n (including always 1 and n) satisfies

d(n) = O(n ).

How does the implied O-constant depend on the choice of ? You might get started in this problem by first showing that for fixed , there are only finitely many prime powers q with d(q) > q .

1.15. Consider the sum of the reciprocals of all Mersenne numbers Mn = 2n − 1 (for positive integers n), namely,

∞

E = 1 .

n=1 Mn

Prove the following alternative form involving the divisor function d (defined in Exercise 1.14):

Actually, one can give this sum a faster-than-linear convergence. To that end

Incidentally, the number E has been well studied in some respects. For example, it is known [Erd˝os 1948], [Borwein 1991] that E is irrational, yet it has never been given a closed form. Possible approaches to establishing deeper properties of the number E are laid out in [Bailey and Crandall 2002].

If we restrict such a sum to be over Mersenne primes, then on the basis of Table 1.2, and assuming that said table is exhaustive up through its final entry (note that this is not currently known), to how many good decimal digits do we know

1

?

Mq P Mq

1.16. Euler’s polynomial x2 +x+41 has prime values for each integer x with −40 ≤ x ≤ 39. Show that if f (x) is a nonconstant polynomial with integer coe cients, then there are infinitely many integers x with f (x) composite.

1.17. It can happen that a polynomial, while not always producing primes, is very likely to do so over certain domains. Show by computation that a polynomial found in [Dress and Olivier 1999],

f (x) = x2 + x − 1354363,

has the astounding property that for a random integer x [1, 104], the number |f (x)| is prime with probability exceeding 1/2. An amusing implication is this: If you can remember the seven-digit “phone number” 1354363, then you have a mental mnemonic for generating thousands of primes.

1.18.Consider the sequence of primes 2, 3, 5, 11, 23, 47. Each but the first is one away from the double of the prior prime. Show that there cannot be an infinite sequence of primes with this property, regardless of the starting prime.

1.19.As mentioned in the text, the relation

is valid (the sum converges absolutely) for Re(s) > 1. Prove this. But the limit as s → 1, for which we know the remarkable PNT equivalence

∞

µ(n) = 0, n

n=1

is not so easy. Two good exercises are these: First, via numerical experiments, furnish an estimate for the order of magnitude of

µ(n)

n

n≤x

as a function of x; and second, provide an at least heuristic argument as to why the sum should vanish as x → ∞. For the first option, it is an interesting computational challenge to work out an e cient implementation of the µ function itself. As for the second option, you might consider the first few

terms in the form

1 − p1 + pq1 − · · ·

p≤x pq≤x

to see why the sum tends to zero for large x. It is of interest that even without recourse to the PNT, one can prove, as J. Gram did in 1884 [Ribenboim 1996], that the sum is bounded as x → ∞.

1.20.Show that for all x > 1, we have

p1 > ln ln x − 1,

p≤x

While the prime number theorem is equivalent to the assertion that the product of the primes in [1, x] is e(1+o(1))x, it is still of interest to have completely explicit inequalities as in this exercise and Exercise 1.25.

For a positive integer N , let

(6N )!N !

C(N ) = (3N )!(2N )!(2N )! .

(1)Show that C(N ) is an integer.

(2)Show that if p is a prime with p > (6N )1/k, then pk does not divide C(N ).

(3)Using Exercise 1.25 and the idea in Exercise 1.27, show that

p > C(N )/4(6N )1/2+(6N )1/3 lg(1.5N ).

p≤6N

(4) Use Stirling’s formula (or mathematical induction) to show that C(N ) >

√

108N /(4 N ) for all N .

(5)Show that p≤x p > 2x for x ≥ 212.

(6)Close the gap from 212 to 31 with a direct calculation.

1.29. Use Exercise 1.28 to show that π(x) > x/ lg x, for all x ≥ 5. Since we have the binary logarithm here rather than the natural logarithm, this inequality for π(x) might be humorously referred to as the “computer scientist’s prime number theorem.” Use Exercise 1.25 to show that π(x) < 2x/ ln x for all x > 0. In this regard, it may be helpful to first establish the

1.30. Here is an exercise involving a healthy mix of computation and theory. With σ(n) denoting the sum of the divisors of n, and recalling from the discussion prior to Theorem 1.3.3 that n is deemed perfect if and only if

σ(n) = 2n, do the following, wherein we adopt a unique prime factorization n = pt11 · · · ptkk :

(1)Write a condition on the pi, ti alone that is equivalent to the condition σ(n) = 2n of perfection.

(2)Use the relation from (1) to establish (by hand, with perhaps some minor machine computations) some lower bound on odd perfect numbers; e.g., show that any odd perfect number must exceed 106 (or an even larger bound).

(3)An “abundant number” is one with σ(n) > 2n, as in the instance σ(12) = 28. Find (by hand or by small computer search) an odd abundant number. Does an odd abundant number have to be divisible by 3?