$150000 prize is offered to the first person who discovers a 100 million digit prime number and $250000 for a billion digit prime number.
There exist very few Mersenne prime numbers. In fact, only 47 have so far been found. In their book [27], Crandall and Pomerance provide a heuris- tic argument (see Problem 5.13) which yields that the number of Mersenne primes 2p−1 withp≤xshould be approximately eγ
log 2logx+O(1). Thus, even though it is believed that there exist infinitely many Mersenne prime numbers, most believe they are quite rare.
Interestingly, not much is known about the size of the largest prime factor of Mersenne numbers. Nevertheless, recently, Ford, Luca, and Shpar- linski [59] proved that, if P(k) stands for the largest prime factor ofk, then the series
∞ n=1
(logn)α
P(2n−1) is convergent for each α <1/2. The fact that the above series converges for all positive numbersαfollows from a recent result of Stewart [138].
2.10. Conjectures on the distribution of prime numbers Let b1 <ã ã ã < bk be positive integers. For each prime number p, let ν(p) be the number of distinct residue classes modulop occupied by the integers b1, . . . , bk. In other words, let
ν(p) = #{bi (mod p) : i= 1, . . . , k}.
The following conjecture, due to Dickson [39], is known as the Prime k- tuples conjecture.
Conjecture 2.11. If ν(p) < p for all prime numbers p, then there exist infinitely many positive integers n such that
(2.15) n+b1, n+b2, . . . , n+bk are all prime numbers.
Settingk= 2, b1 = 0, b2= 2, we obtain the twin prime conjecture. Note that the condition ν(p) < p is necessary if we require that each number in the list (2.15) to be prime for infinitely many integers n. Moreover, it is easy to see that it is sufficient to verify this condition only forp < k. Some 30 years later, Hardy and Littlewood [76] proposed a quantitative version of this conjecture.
Hardy and Littlewood have also conjectured the following inequality.
Conjecture 2.12.
(2.16) π(x+y)≤π(x) +π(y) for all x >1, y >1.
At first one might think that this inequality follows naturally from the Prime Number Theorem. Indeed, if one naively replaces the π(x) function by the functionx/logx, inequality (2.16) would then read as follows:
x+y
log(x+y) < x
logx + y
logy for all x >1, y >1,
which clearly holds. However, in 1972, Hensley and Richards [81] surprised the mathematical community by proving that Conjectures 2.11 and 2.12 are incompatible. Their proof goes as follows.
We say that a sequence of positive integersb1 <ã ã ã< bk isadmissibleif ν(p)< pfor all primes p. Let
ρ∗(y) = max{k∈N : sequence x < b1 <ã ã ã< bk≤x+y is admissible}. Thus,ρ∗(y) is the maximal length (≤y) of an admissible sequence of positive integers that one can insert inside an interval of lengthy. But Hensley and Richards proved that
ylim→∞(ρ∗(y)−π(y)) =∞.
Assume that y is such that ρ∗(y) > π(y) and that the sequence x < b1 <
ã ã ã< bk ≤x+y is admissible, where k=ρ∗(y). If Conjecture 2.11 is true, then for a certain nwe will have that n+b1 <ã ã ã< n+bk are all primes.
Since all these belong to (n+x, n+x+y], we obtain that ρ∗(y) =k≤π(x+n+y)−π(x+n), so that
π(x+n+y)≥π(x+n) +ρ∗(y)> π(x+n) +π(y), contradicting inequality (2.16).
Most mathematicians believe that Conjecture 2.11 is true, which would imply that Conjecture 2.12 is false. Some calculations seem to indicate that indeed Conjecture 2.12 could be false. Indeed, in 1979, Vehka showed that there exists a numberx0 such that the interval I = [x0, x0+ 11763] contains more prime numbers than the interval [1,11763], sinceI contains 1412 prime numbers, whileπ(11763) = 1409. How large is that numberx0? Most likely, very large !
In 1996, in his master’s thesis, N. Jarvis proved that the number “11763”
can be replaced by “4930”. In 1998, Gordon and Rodemich [67] showed that the smallesty for which there existsx0 such that the interval [x0, x0+ y] contains more prime numbers than the interval [1, y] must be ≥ 1731.
Recently, S. Wagon was able to show that the number “4930” (of Jarvis) can be replaced by “4893”. Hence the smallest number y such that the
2.10. Conjectures on the distribution of prime numbers 35
interval [x0, x0 +y] contains more prime numbers than the interval [1, y]
must satisfy
(2.17) 1731≤y≤4893.
Now it is conjectured that the number of k-tuples of prime numbers≤x is
of order x
0
dt
logkt ∼ x
logkx asx→ ∞. Taking k=π(1731) = 269, we have that
x
logkx >1 if x >10891.
In the end, it follows that in order for (2.17) to hold, we must have that 10891< x <102454,
an indication that the value of the number x0 is out of range of today’s computational capabilities.
Conjecture 2.11 was generalized by Schinzel in a paper written jointly with Sierpi´nski [132] : it is now called the Schinzel Hypothesis or at times Schinzel’s Hypothesis H.
Conjecture 2.13. Let f1(X), . . . , fk(X)∈Z[X] be nonconstant polynomi- als, each one having a positive leading coefficient. Assume that:
(i) fi(X) is irreducible for all i= 1, . . . , k.
(ii) there exists no prime number p such that p | f1(n)f2(n)ã ã ãfk(n) for alln≥0.
Then there exist infinitely many positive integers n such that f1(n), f2(n), . . . , fk(n)
are all prime numbers.
Bateman and Horn [10] proposed a quantitative version of this conjec- ture. More precisely, for each prime number p, let
ω(p) = #{0≤n≤p−1 : f1(n)f2(n)ã ã ãfk(n)≡0 (modp)}, and set
πf1,...,fk(x) = #{n≤x : f1(n), . . . , fk(n) are all primes}. Then πf1,...,fk(x) should be asymptotically equal to
C(f1, . . . , fk) 1 d1ã ã ãdk
x (logx)k,
where di = deg(fi), and where the constant C(f1, . . . , fk) is given by (2.18) C(f1, . . . , fk) =
p≥2
1−ω(p)/p (1−1/p)k.
It is not even clear that in formula (2.18) the expression C(f1, . . . , fk) rep- resents a product which converges to a positive limit. In any event, one can easily show that this is indeed the case when fi(X) = aiX +bi (ai > 0, gcd(ai, bi) = 1) is linear for each i= 1, . . . , k, thereby providing an effective form of the Hardy and Littlewood Conjecture 2.11.
It follows from the argument detailed in this section that there are ap- proximately 2Cx/(logx)2 twin prime pairs not exceedingx, where
C=
p>2
1− 1
(p−1)2
≈0.6601618158.
In 1920, Viggo Brun [19] established an upper bound for the number of twin prime pairsp, p+ 2 withp≤x. In the same paper, he also proved that there exist infinitely many prime numbers p withp+ 2 a P9. The record belongs to Chen [23] who in 1973 proved that there exists a positive constantcsuch that there are more than cx/(logx)2 primes p < xsuch that p+ 2 is aP2, provided xis sufficiently large.
Problems on Chapter 2
Problem 2.1. Show that if n > 1 is not prime then n has a prime factor p≤√
n.
Problem 2.2. Show that if n >1 thenn4+ 4n is composite.
Problem 2.3. Show that the set S = {logp : p prime} consists of real numbers which are linearly independent over the rational numbers Q. Problem 2.4. Prove that if f(X)∈Z[X]is nonconstant, then the set
S ={p : p is prime andp |f(n) for some positive integer n} is infinite.
Problem 2.5. Show that if P(X1, . . . , Xn) ∈ C[X1, . . . , Xn] takes only prime values at all nonnegative integer values Xi, then P is constant.
Problem 2.6. Show that ifP(X)∈Z[X]is a nonconstant polynomial, then there exists n such thatP(n!) is composite.
Problem 2.7. Show that if 2n+ 1is prime, then n is a power of 2.
Problems on Chapter 2 37
Problem 2.8. Let a >1 be an integer. Show that an−1 dividesam−1 if and only if ndivides m.
Problem 2.9. A positive integer n is a pseudoprime to base 2 if it is composite and if the congruence 2n−1 ≡ 1 (mod n) holds. Show that if k+ 1 ≤ n1 < n2 <ã ã ã < ns < 2k, then Fn1ã ã ãFns is either a prime or a base 2 pseudoprime. Deduce that there are infinitely many base 2 pseudo- primes.
Problem 2.10. Show that (n−1)! ≡ −1 (mod n) if n is prime and n | (n−1)!if n >4 is composite. Use this to prove that
pn= 1 +
2n
m=1
⎢⎢
⎢⎢
⎣
⎢⎢
⎢⎣ n 1 +m
j=2
&
(j−1)!+1 j −&
(j−1)!
j
''
⎥⎥
⎥⎦
1/n⎥⎥⎥⎥⎦.
Problem 2.11. Prove that
lcm[1,2, . . . , n]≥2n holds for all integers n≥9.
Problem 2.12. Show that Li(x) =
x
2
dt
logt = x logx +O
x log2x
.
Problem 2.13. Let pn be the n-th prime. Show that, for each positive integer n,
2
9nlogn < pn.
Problem 2.14. Show that, for each positive integer n, pn<12
nlogn+ log 12
e
.
Problem 2.15. Show that there do not exist polynomials P(x) and Q(x) such that π(n) =P(n)/Q(n) for infinitely many positive integers n.
Problem 2.16. Show that for every n >1, there exist n consecutive com- posite numbers.
Problem 2.17. Let Sn = n
i=1pi. Prove that the interval [Sn, Sn+1] con- tains a perfect square.
Problem 2.18. Let pn be the n-th prime. Euclid’s proof shows that there is always a prime in the interval(pn, p1ã ã ãpn+ 1]. Show, using Chebyshev’s estimates, that for any constant K, we have
π(p1p2ã ã ãpn+ 1)nK.
Problem 2.19. Using the Prime k-tuples conjecture, show that for each positive integer K, there exists a positive integer A such that n2−n+A is prime for all n= 0, 1, . . . , K. (Hint: Use bk=k2−k.)
Problem 2.20. Assuming the Prime Number Theorem, show that pn = nlogn+o(nlogn) as n→ ∞.
Problem 2.21. Assume the Prime Number Theorem. Show that the set of ratios of primes {p/q : p, q primes} is dense in[0,∞). Recall that a subset S of the positive real numbers is called dense in [0,∞) if every positive real number is the limit of a sequence {sn}n≥1, whose members sn are in S for all n≥1. (Hint: Given positive integers a and b, examine the limit of the sequence pan/pbn.)
Problem 2.22. Using Chebyshev’s theorem, show that there exists a positive constantCsuch that for each positive integerx, there exists a positive integer K (which may depend on x) such that
#{p≤x prime : p=n2+K for some positive integer n}> C
√x logx. Problem 2.23. Show that there exists a positive constant C such that
x
2
logt
t2 dt=C+O 1
x1/2
.
Problem 2.24. Recall the statement (2.4) made by Legendre in 1798. Prove that it follows from the Prime Number Theorem in the form π(x) ∼ li(x) that Legendre’s statement is accurate if one chooses A= 1 and B =−1.
Problem 2.25. Prove that there exists an interval of the form[n2,(n+ 1)2] which contains at least 1 000 prime numbers. (Hint: Assume the contrary and establish that it would then imply that
+∞=
p
1 p =
∞ n=1
n2<p≤(n+1)2
1
p <1 000 ∞ n=1
1
n2 <+∞, thereby creating a contradiction.)
Problem 2.26. Charles Hermite (1822–1901, France) provided the follow- ing very simple proof of the infinitude of primes: For each integer n ≥ 1, let qn be the smallest prime factor of n! + 1; since it is clear that qn > n, we have thus generated an infinite sequence of primes. Now, consider the sequence {qn}n≥1, whose first 40 terms are 2, 3, 7, 5, 11, 7, 71, 61, 19, 11, 39916801, 13, 83, 23, 59, 17, 661, 19, 71, 20639383, 43, 23, 47, 811, 401, 1697, 10888869450418352160768000001, 29, 14557, 31, 257, 2281, 67, 67411, 137, 37, 13763753091226345046315979581580902400000001, 14029308060317546154181, 79 and 41. Show that all prime numbers will eventually appear in this sequence.
Chapter 3
The Riemann Zeta Function