Prime Numbers

158 Chapter 3 RECOGNIZING PRIMES AND COMPOSITES

block, and we also compute φ(r2k, pb) for each b, so as to use these for the next block. The computed values of φ(x/(mpb+1), pb) are not stored, but are multiplied by µ(m) and added into a running sum that represents the second term on the right of (3.23). The time and space required to do these tasks

[Lagarias et al. 1985] and [Del´eglise and Rivat 1996].

3.7.2Analytic method

Here we describe an analytic method, highly e cient in principle, for counting primes. The idea is that in [Lagarias and Odlyzko 1987], with recent extensions that we shall investigate. The idea is to exploit the fact that the Riemann zeta function embodies in some sense the properties of primes. A certain formal manipulation of the Euler product relation (1.18) goes like so. Start by taking the logarithm

where all manipulations are valid (and the double sum can be interchanged if need be) for Re(s) > 1, with the caveat that ln ζ is to be interpreted as a continuously changing argument. (By modern convention, one starts with the positive real ln ζ(2) and tracks the logarithm as the angle argument of ζ, along a contour that moves vertically to 2 + i Im(s) then over to s.)

In order to use relation (3.24) to count primes, we define a function reminiscent of—but not quite the same as—the prime-counting function π(x). In particular, we consider a sum over prime powers not exceeding x, namely

m>0

where θ(z) is the Heaviside function, equal to 1, 1/2, 0, respectively, as its argument z is positive, zero, negative. The introduction of θ means that the sum involves only prime powers pm not exceeding x, but that whenever the real x actually equals a power pm, the summand is 1/(2m). The next step is to invoke the Perron formula, which says that for nonnegative real x, positive integer n, and a choice of contour C = {s : Re(s) = σ}, with fixed σ > 0 and

It follows immediately from these observations that for a given contour (but now with σ > 1 so as to avoid any ln ζ singularity) we have:

This last formula provides analytic means for evaluation of π(x), because if x is not a prime power, say, we have from relation (3.25) the identity:

This analytic approach thus comes down to numerical integration, yet such integration is the problematic stage. First of all, one has to evaluate ζ with su cient accuracy. Second, one needs a rigorous bound on the extent to which the integral is to be taken along the contour. Let us address the latter problem first. Say we have in hand a sharp computational scheme for ζ itself, and we take x = 100, σ = 3/2. Numerical integration reveals that for sample integration limits T {10, 30, 50, 70, 90}, respective values are

which values exhibit poor convergence of the contour integral: The true value of π (100) can be computed directly, by hand, to be 428/15 ≈ 28.533 . . . .

Furthermore, on inspection the value as a function of integration limit T is rather chaotic in the way it hovers around the true value, and rigorous error bounds are, as might be expected, nontrivial to achieve (see Exercise 3.37).

The suggestions in [Lagarias and Odlyzko 1987] address, and in principle repair, the above drawbacks of the analytic approach. As for evaluation of ζ itself, the Riemann–Siegel formula is often recommended for maximum speed;

in fact, whenever s has a formidably large imaginary part t, said formula has been the exclusive historical workhorse (although there has been some modern work on interesting variants to Riemann–Siegel, as we touch upon at the end of Exercise 1.61). What is more, there is a scheme found in [Odlyzko and Sch¨onhage 1988] for a kind of “parallel” evaluation of ζ(s) values, along, say, a progression of imaginary ordinates of the argument s. This sort of simultaneous evaluation is just what is needed for numerical integration. For a modern compendium including variants on the Riemann– Siegel formula and other computational approaches, see [Borwein et al. 2000] and references therein. In [Crandall 1998] can be found various fast algorithms for simultaneous evaluation at various argument sets. The essential idea for acceleration of ζ computations is to use FFT, polynomial evaluation, or Newton-method techniques to achieve simultaneous evaluations of ζ(s) for a given set of s values. In the present book we have provided enough instruction—via Exercise 1.61—for one at least to get started on single evaluations of ζ(s + it) that require only O t1/2+ bit operations.

As for the problem of poor convergence of contour integrals, the clever ploy is to invoke a smooth (one might say “adiabatic”) turn-o function that renders a (modified) contour integral more convergent. The phenomenon is akin to that of reduced spectral bandwidth for smoother functions in Fourier analysis. The Lagarias–Odlyzko identity of interest is (henceforth we shall assume that x is not a prime power)

To understand the import of this scheme, take the turn-o function c(u, x) to be θ(x − u). Then F (s, x) = xs/s, the final sum in (3.28) is zero, and we recover the original analytic representation (3.27) for π . Now, however, let us contemplate the class of continuous turn-o functions c(u, x) that stay at 1 over the interval u [0, x − y), decay smoothly (to zero) over u (x − y, x],

√

Indeed, the last summation is rather easy, since it has just O ( x) terms. The

next-to-last summation, which just records the di erence between π(x) and

√

π (x), also has just O ( x) terms.

Let us posit a specific smooth decay, i.e., for u (x − y, x] we define

Observe that c(x − y, x) = 1 and c(x, x) = 0, as required for continuous c functions in the stated class. Mellin transformation of c gives

y3

6

(3.30)

−2xs+3 + (s + 3)xs+2y + (x − y)s(2x3 + (s − 3)x2y − 2sxy2 + (s + 1)y3) .

This expression, though rather unwieldy, allows us to count primes more e ciently. For one thing, the denominator of the second fraction is O(t4), which is encouraging. As an example, performing numerical integration as in relation (3.29) with the choices x = 100, y = 10, we find for the same trial set of integration limits T {10, 30, 50, 70, 90} the results

π(100) ≈ 25.3, 26.1, 25.27, 24.9398, 24.9942,

which are quite satisfactory, since π(100) = 25. (Note, however, that there is still some chaotic behavior until T be su ciently large.) It should be pointed out that Lagarias and Odlyzko suggest a much more general, parameterized form for the Mellin pair c, F , and indicate how to optimize the parameters. Their complexity result is that one can either compute π(x) with bit operation count O x1/2+ and storage space of O x1/4+ bits, or on the notion of limited memory one may replace the powers with 3/5 + , , respectively.

As of this writing, there has been no practical result of the analytic method on a par with the greatest successes of the aforementioned combinatorial methods. However, this impasse apparently comes down to just a matter of calendar time. In fact, [Galway 1998] has reported that values of π(10n) for n = 13, and perhaps 14, are attainable for a certain turn-o function c and (only) standard, double-precision floating-point arithmetic for the numerical integration. Perhaps 100-bit or higher precision will be necessary to press the analytic method on toward modern limits, say x ≈ 1021 or more; the required precision depends on detailed error estimates for the contour integral. The Galway functions are a clever choice of Mellin pair, and work out to be more e cient than the turn-o functions that lead to F of the type (3.30). Take

where erfc is the standard error function:

and a is chosen later for e ciency. This c function turns o smoothly at u x, but at a rate tunable by choice of a. The Mellin companion works out nicely to be

For s = σ+it the wonderful (for computational purposes) decay in F is e−t2a2 . Now numerical experiments are even more satisfactory. Sure enough, we can use relation (3.29) to yield, for x = 1000, decay function a(x) = (2x)−1/2, σ = 3/2, and integration limits T {20, 40, 60, 80, 100, 120}, the successive values

π(1000) ≈ 170.6, 169.5, 170.1, 167.75, 167.97, 167.998,

in excellent agreement with the exact value π(1000) = 168; and furthermore, during such a run the chaotic manner of convergence is, qualitatively speaking, not so manifest.

Incidentally, though we have used properties of ζ(s) to the right of the critical strip, there are ways to count primes using properties within the strip; see Exercise 3.50.

3.8Exercises

3.1.In the spirit of the opening observations to the present chapter, denote by SB (n) the sum of the base-B digits of n. Interesting phenomena accrue for specific B, such as B = 7. Find the smallest prime p such that S7(p) is itself composite. (The magnitude of this prime might surprise you!) Then, find all possible composite values of S7(p) for the primes p < 16000000 (there are very few such values!). Here are two natural questions, the answers to which are unknown to the authors: Given a base B, are there infinitely many primes p with SB (p) prime? (composite?) Obviously, the answer is “yes” for at least one of these questions!

3.2.Sometimes other fields of thought can feed back into the theory of prime numbers. Let us look at a beautiful gem in [Golomb 1956] that uses clever combinatorics—and even some “visual” highlights—to prove Fermat’s little Theorem 3.4.1.

For a given prime p you are to build necklaces having p beads. In any one necklace the beads can be chosen from n possible di erent colors, but you have the constraint that no necklace can be all one color.

(1)Prove: For necklaces laid out first as linear strings (i.e., not yet circularized) there are np − n possible such strings.

(2)Prove: When the necklace strings are all circularized, the number of distinguishable necklaces is (np − n)/p.

(3)Prove Fermat’s little theorem, that np ≡ n (mod p).

(4)Where have you used that p is prime?

3.3.Prove that if n > 1 and gcd(an − a, n) = 1 for some integer a, then not only is n composite, it is not a prime power.

3.4.For each number B ≥ 2, let dB be the asymptotic density of the integers that have a divisor exceeding B with said divisor composed solely of primes not exceeding B. That is, if N (x, B) denotes the number of positive integers up to x that have such a divisor, then we are defining dB = limx→∞ N (x, B)/x.

(1) Show that

≤

where the product is over primes.

(2)Find the smallest value of B with dB > d7.

(3)Using the Mertens Theorem 1.4.2 show that limB→∞ dB = 1 − e−γ ≈ 0.43854, where γ is the Euler constant.

3.5.Let c be a real number and consider the set of those integers n

whose largest prime factor does not exceed nc. Let c be such that the

√

asymptotic density of this set is 1/2. Show that c = 1/(2 e). A pleasantly interdisciplinary reference is [Knuth and Trabb Pardo 1976].

Now, consider the set of those integers n whose second-largest prime factor (if there is one) does not exceed nc. Let c be such that the asymptotic density of this set is 1/2. Show that c is the solution to the equation

and solve this numerically for c. An interesting modern approach for the numerics is to show, first, that this integral is given exactly by

in which the standard polylogarithm Li2(c) = c/12 + c2/22 + c3/32 + · · ·

appears. Second, using any of the modern packages that know how to evaluate Li2 to high precision, implement a Newton-method solver, in this way circumventing the need for numerical integration per se. You ought to be able to obtain, for example,

c ≈ 0.2304366013159997457147108570060465575080754 . . . ,

presumed correct to the implied precision.

Another intriguing direction: Work out a fast algorithm—having a value of c as input—for counting the integers n [1, x] whose second-largest prime

factor exceeds nc (when there are less than two prime factors let us simply not count that n). For the high-precision c value given above, there are 548 such n [1, 1000], whereas the theory predicts 500. Give the count for some much higher value of x.

3.6.Rewrite the basic Eratosthenes sieve Algorithm 3.2.1 with improvements. For example, reduce memory requirements (and increase speed) by observing that any prime p > 3 satisfies p ± 1 (mod 6); or use a modulus greater than 6 in this fashion.

3.7.Use the Korselt criterion, Theorem 3.4.6, to find by hand or machine some explicit Carmichael numbers.

3.8.Prove that every composite Fermat number Fn = 22n + 1 is a Fermat pseudoprime base 2. Can a composite Fermat number be a Fermat pseudoprime base 3? (The authors know of no example, nor do they know a proof that this cannot occur.)

3.9.This exercise is an exploration of rough mental estimates pertaining to the statistics attendant on certain pseudoprime calculations. The great computationalist/theorist team of D. Lehmer and spouse E. Lehmer together pioneered in the mid-20th century the notion of primality tests (and a great many other things) via hand-workable calculating machinery. For example,

they proved the primality of such numbers as the repunit (1023 − 1)/9 with a mechanical calculator at home, they once explained, working a little every day over many months. They would trade o doing the dishes vs. working on the primality crunching. Later, of course, the Lehmers were able to handle much larger numbers via electronic computing machinery.

Now, the exercise is, comment on the statistics inherent in D. Lehmer’s (1969) answer to a student’s question, “Professor Lehmer, have you in all your lifetime researches into primes ever been tripped up by a pseudoprime you had thought was prime (a composite that passed the base-2 Fermat test)?” to which Lehmer’s response was as terse as can be: “Just once.” So the question is, does “just once” make statistical sense? How dense are the base-2 pseudoprimes in the region of 10n? Presumably, too, one would not be fooled, say, by those base-2 pseudoprimes that are divisible by 3, so revise the question to those base-2 pseudoprimes not divisible by any “small” prime factors. A reference on this kind of question is [Damg˚ard et al. 1993].

3.10.Note that applying the formula in the proof of Theorem 3.4.4 with a = 2, the first legal choice for p is 5, and as noted, the formula in the proof gives n = 341, the first pseudoprime base 2. Applying it with a = 3, the first legal choice for p is 3, and the formula gives n = 91, the first pseudoprime base 3. Show that this pattern breaks down for larger values of a and, in fact, never holds again.

3.11.Show that if n is a Carmichael number, then n is odd and has at least three prime factors.

3.19.[Lenstra, Granville] Show that if an odd number n be divisible by the square of some prime, then W (n), the least witness for n, is less than ln2 n. (Hint: Use (1.45).) This exercise is re-visited in Exercise 4.28.

3.20.Describe a probabilistic algorithm that gives nontrivial factorizations of Carmichael numbers in expected polynomial time.

3.21.We say that an odd composite number n is an Euler pseudoprime base a if a is coprime to n and

a(n−1)/2 ≡ a (mod n), (3.32) n

where na is the Jacobi symbol (see Definition 2.3.3). Euler’s criterion (see Theorem 2.3.4) asserts that odd primes n satisfy (3.32). Show that if n is a strong pseudoprime base a, then n is an Euler pseudoprime base a, and that if n is an Euler pseudoprime base a, then n is a pseudoprime base a.

3.22.[Lehmer, Solovay–Strassen] Let n be an odd composite. Show that

the set of residues a (mod n) for which n is an Euler pseudoprime is a proper subgroup of Zn. Conclude that the number of such bases a is at most ϕ(n)/2.

3.23.Along the lines of Algorithm 3.5.6 develop a probabilistic compositeness test using Exercise 3.22. (This test is often referred to as the Solo- vay–Strassen primality test.) Using Exercise 3.21 show that this algorithm is majorized by Algorithm 3.5.6.

3.24.[Lenstra, Robinson] Show that if n is odd and if there exists an integer b with b(n−1)/2 ≡ −1 (mod n), then any integer a with a(n−1)/2 ≡ ±1 (mod n)

also satisfies a(n−1)/2 ≡ na (mod n). Using this and Exercise 3.22, show that if n is an odd composite and a(n−1)/2 ≡ ±1 (mod n) for all a coprime to n, then in fact a(n−1)/2 ≡ 1 (mod n) for all a coprime to n. Such a number must be a Carmichael number; see Exercise 3.12. (It follows from the proof of the infinitude of the set of Carmichael numbers that there are infinitely many odd composite numbers n such that a(n−1)/2 ≡ ±1 (mod n) for all a coprime to n. The first example is Ramanujan’s “taxicab” number, 1729.)

3.25.Show that there are seven Fibonacci pseudoprimes smaller than 323.

3.26.Show that every composite number coprime to 6 is a Lucas pseudoprime with respect to x2 − x + 1.

3.27.Show that if (3.12) holds, then so does

In particular, conclude that a Frobenius pseudoprime with respect to f (x) = x2 − ax + b is also a Lucas pseudoprime with respect to f (x).

3.28. Show that the definition of Frobenius pseudoprime in Section 3.6.5 for a polynomial f (x) = x2 − ax + b reduces to the definition in Section 3.6.2.

3.29. Show that if a, n are positive integers with n odd and coprime to a, then n is a Fermat pseudoprime base a if and only if n is a Frobenius pseudoprime with respect to the polynomial f (x) = x − a.

3.30.Let a, b be integers with ∆ = a2−4b not a square, let f (x) = x2−ax+b,

let n be an odd prime not dividing b∆, and let R = Zn[x]/(f (x)). Show that if (x(a − x)−1)2m = 1 in R, then (x(a − x)−1)m = ±1 in R.

3.31.Show that a Frobenius pseudoprime with respect to x2 − ax + b is also an Euler pseudoprime (see Exercise 3.21) with respect to b.

3.32.Prove that the various identities in Section 3.6.3 are correct.

3.33.Prove that the recurrence (3.22) is valid.

3.34.Show that if a = π x1/3 , then the number of terms in the double

sum in (3.23) is O x2/3/ ln2 x .

3.35.Show that with M computers where M < x1/3, each with the capacity for O x1/3+ space, the prime-counting algorithm of Section 3.7 may be speeded up by a factor M .

3.36.Show that instead of using analytic relation (3.27) to get the modified count π (x), one could, if desired, use the “prime-zeta” function

P(s) =

p P ps

in place of ln ζ within the integral, whence the result on the left-hand side of (3.27) is, for noninteger x, the π function itself. Then show that this observation is not entirely vacuous, and might even be practical, by deriving

the relation

∞

P(s) = µ(n) ln ζ(ns), n

n=1

3.37.By establishing theoretical bounds on the magnitude of the real part

of the integral

where T, α, β are positive reals, determine a bound on that portion of the integral in relation (3.27) that comes from Im(s) > T . Describe, then, how large T must be for π (x) to be calculated to within some ± of the true value. See Exercises 3.38, 3.39 involving the analogous estimates for much more e cient prime-counting methods.