Gaps between Consecutive Primes

We show first that there occur arbitrarily large gaps between consecutive primes:

Theorem 5.5.1. For any positive integer𝐾there exist𝐾consecutive composite numbers.

♣ Proof. Take any𝑁 > 𝐾, and consider the integers𝑎𝑖 = 𝑁! +𝑖,𝑖 = 2,3, . . . , 𝐾 + 1.

Clearly,𝑖 ∣ 𝑎𝑖and𝑎𝑖 > 𝑖, hence every𝑎𝑖is composite. □ Remark: We can replace𝑁!in the proof by the product of primes not exceeding𝑁.

Generalizing Theorem 5.5.1, we prove now that even both of two consecutive gaps can be arbitrarily large, i.e. there exist primes surrounded by many composite numbers from both sides (these are calledsolitaryprimes).

5.5. Gaps between Consecutive Primes 135

Theorem 5.5.2. For any positive integer𝐾there exists a prime𝑝such that all numbers

𝑝 ± 1,𝑝 ± 2, . . . ,𝑝 ± 𝐾are composite. ♣

Proof. We choose a prime𝑞 ≥ 𝐾 + 2, and consider

𝑑 = 2 ⋅ 3 . . . (𝑞 − 2)(𝑞 − 1)(𝑞 + 1)(𝑞 + 2) . . . (2𝑞 − 2) =(2𝑞 − 2)!

𝑞 .

Here(𝑞, 𝑑) = 1, thus there exist (infinitely many)𝑘 > 0for which𝑝 = 𝑞 + 𝑑𝑘is a prime. We show that such a𝑝meets the requirements. For any1 ≤ 𝑗 ≤ 𝑞 − 2we have

𝑝 ± 𝑗 = 𝑞 + 𝑘𝑑 ± 𝑗 = (𝑞 ± 𝑗) +𝑘(2𝑞 − 2)!

𝑞 = (𝑞 ± 𝑗)(1 + 𝑐𝑗),

where𝑐𝑗is a positive integer. Thus every𝑝 ± 𝑗is composite. □ Now we prove Chebyshev’s Theorem stating that there must occur a prime between any number and its double.

Theorem 5.5.3(Chebyshev’s Theorem). For any integer𝑛 ≥ 1there exists a prime𝑝

satisfying𝑛 < 𝑝 ≤ 2𝑛. ♣

This obviously implies that the theorem remains valid for any real numbers𝑛 ≥ 1 (instead of integers).

Another name for this result isBertrand’s postulate, because the conjecture was first formulated in 1845 by Bertrand in a slightly stronger form: To every𝑛 > 3there is a prime𝑝satisfying𝑛 < 𝑝 ≤ 2𝑛 − 2. (This version is true as well, and even much stronger results hold, see assertions (A) in Theorems 5.5.4 and 5.5.5.) Theorem 5.5.3 was proved by Chebyshev in 1852. The proof below was found by Erdős when he was 19years old.

Proof. The basic idea is to observe that the product of primes between𝑛and2𝑛is closely related to the binomial coefficient(2𝑛𝑛). We assume𝑛 ≥ 5from now on.

I. We write the standard form of(2𝑛𝑛)and break it into the product of three factors according to the size of the primes the following way:

(5.5.1) (2𝑛 𝑛) = ∏

𝑝≤2𝑛

𝑝𝜈𝑝= ∏

𝑝≤√2𝑛

𝑝𝜈𝑝⋅ ∏

√2𝑛<𝑝≤𝑛

𝑝𝜈𝑝⋅ ∏

𝑛+1≤𝑝≤2𝑛

𝑝𝜈𝑝.

We denote the three subproducts on the right-hand side of (5.5.1) by𝐴,𝐵, and 𝐶. It is sufficient to show 𝐶 > 1, since then there must exist a prime𝑝satisfying 𝑛 + 1 ≤ 𝑝 ≤ 2𝑛. (It can be easily shown that every exponent𝜈𝑝in𝐶is1, so𝐶equals the product of primes between𝑛and2𝑛, see Exercise 5.5.7a.)

To verify𝐶 > 1, we establish upper bounds for𝐴and𝐵, and a lower bound for (2𝑛𝑛).

II. Lower bound for(2𝑛𝑛): Since(2𝑛𝑘) ≤ (2𝑛𝑛)for every0 ≤ 𝑘 ≤ 2𝑛(see Exercise 5.5.5) (2𝑛 + 1)(2𝑛

𝑛) >

2𝑛

∑

𝑘=0

(2𝑛 𝑘) = 22𝑛,

(5.5.2) (2𝑛

𝑛) > 4𝑛 2𝑛 + 1.

III. Upper bound for𝐴: By Lemma 5.4.4, we have𝑝𝜈𝑝 ≤ 2𝑛, hence

(5.5.3) 𝐴 = ∏

𝑝≤√2𝑛

𝑝𝜈𝑝≤ (2𝑛)𝜋(√2𝑛)< (2𝑛)√2𝑛.

IV. Upper bound for𝐵: Again,𝑝𝜈𝑝≤ 2𝑛, by Lemma 5.4.4, and since𝑝 > √2𝑛, this implies𝜈𝑝≤ 1.

We show that𝜈𝑝= 0for (𝑝 > 2and)2𝑛/3 < 𝑝 ≤ 𝑛. Indeed, such a𝑝occurs exactly to the first power both in the numerator and denominator of

(2𝑛

𝑛) = 2𝑛(2𝑛 − 1) . . . (𝑛 + 1)

𝑛! ,

it appears only in the factor2𝑝in the numerator, and in the factor𝑝in the denominator.

Hence

(5.5.4) 𝐵 = ∏

√2𝑛<𝑝≤𝑛

𝑝𝜈𝑝 = ∏

√2𝑛<𝑝≤2𝑛/3

𝑝𝜈𝑝 ≤ ∏

√2𝑛<𝑝≤2𝑛/3

𝑝.

This and Lemma 5.4.5 imply

(5.5.5) 𝐵 < ∏

𝑝≤2𝑛/3

𝑝 < 42𝑛/3.

V. Substituting (5.5.2), (5.5.3), and (5.5.5) into (5.5.1), and expressing𝐶, we get

(5.5.6) 𝐶 > 4𝑛

(2𝑛 + 1)(2𝑛)√2𝑛⋅ 42𝑛/3 > 4𝑛/3 (2𝑛 + 1)1+√2𝑛.

To prove𝐶 > 1it is sufficient to verify that the logarithm of the expression𝑠𝑛on the right-hand side of (5.5.6) is positive. Since

(5.5.7) log 𝑠𝑛= 𝑛 log 4

3 − (1 + √2𝑛) log(2𝑛 + 1) → ∞ as𝑛 → ∞,

log 𝑠𝑛> 0if𝑛is large enough. A calculation shows that𝑛 > 511guarantees positivity, hence𝐶 > 1for𝑛 > 511.

VI. Finally, we verify the statement directly for𝑛 ≤ 511. This can be done by generating a sequence of primes starting with2where every element is less than the double of the previous element:2,3,5,7,13,23,43,83,163,317,631is such a sequence.

(It is Chebyshev’s Theorem which guarantees the existence of aninfinitesequence with

this property.) □

Related to Chebyshev’s Theorem the following more general problem arises concerning the “gap function”:

For which functionsℎ(𝑛)is it true that the open interval(𝑛, 𝑛 + ℎ(𝑛))always contains a prime if𝑛is large enough?

5.5. Gaps between Consecutive Primes 137

Chebyshev’s Theorem asserts thatℎ(𝑛) = 𝑛works, but according to Theorem 5.5.1, a constantℎ(𝑛)is not suitable, as the interval(𝑛, 𝑛 + 𝐾)is primefree for infinitely many 𝑛however we fix𝐾.

The order of magnitude of the bestℎ(𝑛)is a famous unsolved problem. We state the related strongest results without proof:

Theorem 5.5.4. (A) Let𝜃 = 0.525. Then the interval(𝑛, 𝑛 + 𝑛𝜃)contains a prime for every𝑛is large enough.

(B) There exists a constant𝑐 > 0such that the interval

(𝑛, 𝑛 +𝑐 ⋅ log 𝑛 ⋅ log log 𝑛 ⋅ log log log log 𝑛 log log log 𝑛 )

is primefree for infinitely many positive integers𝑛. ♣ Both assertions in Theorem 5.5.4 are very deep results (they are much sharper than the ones deducible from the Prime Number Theorem, see Theorem 5.5.5). There is, however, an enormous gulf between them:ℎ(𝑛)can be chosen as𝑛𝜃, and cannot be chosen as a function not much bigger than the logarithm. Some probabilistic consid- erations suggest that the boundary should be around(log 𝑛)2.

It is interesting to note that (A) does not imply even the innocent looking conjecture mentioned in Section 5.1 claiming that every interval between two consecutive squares contains a prime. To prove this conjecture one has to reduce the exponent𝜃 to1/2which could not be verified even assuming the famous unproved Riemann Hy- pothesis.

Another remarkable fact about the difficulties in this field is that the previous best result concerning primefree intervals was achieved in 1936(!), which differed from(𝐵) just in the denominator being squared, and there was no progress at all for nearly 80(!) years, in spite of all efforts and a prize of 10000(!) US dollars offered by Erdős.

The five authors of this slight improvement thus got the biggest prize ever given (now with the contribution of Ron Graham) for the solution of an Erdős problem.

In what follows, we show how the results of Theorems 5.5.3 and 5.5.1 can be sharp- ened using the Prime Number Theorem.

Theorem 5.5.5. (A) For any𝜀 > 0there exists an𝑛0(depending on𝜀) such that the interval(𝑛, (1 + 𝜀)𝑛)contains a prime for every𝑛 > 𝑛0.

(B) For any0 < 𝜀 < 1there exist infinitely many positive integers so that the interval

(𝑛, 𝑛 + (1 − 𝜀) log 𝑛)is primefree. ♣

Proof. To prove (A), we have to verify

(5.5.8) 𝜋((1 + 𝜀)𝑛) − 𝜋(𝑛) > 0

for every𝑛large enough. Using the Prime Number Theorem in two different directions, we get

(5.5.9) 𝜋(𝑛) < (1 + 𝜀

4) ⋅ 𝑛 log 𝑛

on the one hand, and

(5.5.10) 𝜋((1 + 𝜀)𝑛) > (1 − 𝜀

4) ⋅ (1 + 𝜀)𝑛 log((1 + 𝜀)𝑛) on the other hand, if𝑛is sufficiently large. Further,

(5.5.11) log((1 + 𝜀)𝑛) = log(1 + 𝜀) + log 𝑛 < (1 +𝜀 4) log 𝑛.

From (5.5.9), (5.5.10), and (5.5.11) we get

(5.5.12) 𝜋((1 + 𝜀)𝑛) − 𝜋(𝑛) > ((1 −𝜀4)(1 + 𝜀)

1 + 4𝜀 − (1 + 𝜀 4)) 𝑛

log 𝑛. The coefficient of𝑛/ log 𝑛on the right-hand side of (5.5.12) is

(1 − 𝜀4)(1 + 𝜀) − (1 + 4𝜀)2

1 +4𝜀 =

𝜀

4(1 −5𝜀4) 1 +4𝜀 > 0 (since we may assume𝜀 < 4/5), hence (5.5.8) follows from (5.5.12).

We apply a proof by contradiction for (B): we assume that the interval (𝑛, 𝑛 + (1 − 𝜀) log 𝑛)contains a prime for every𝑛 > 𝑛0 for some given𝜀 > 0and 𝑛0.

We fix a large integer𝑁 and consider all primes between𝑛0and𝑁: 𝑛0 < 𝑝𝑟 <

𝑝𝑟+1< ⋯ < 𝑝𝑘≤ 𝑁. Using our assumption, we obtain the inequalities

(5.5.13)

𝑝𝑟+1< 𝑝𝑟+ (1 − 𝜀) log 𝑝𝑟 𝑝𝑟+2< 𝑝𝑟+1+ (1 − 𝜀) log 𝑝𝑟+1

⋮

𝑝𝑘+1< 𝑝𝑘+ (1 − 𝜀) log 𝑝𝑘.

Summing the inequalities in (5.5.13), the terms𝑝𝑟+1, . . . ,𝑝𝑘get cancelled, and we obtain

(5.5.14) 𝑝𝑘+1< 𝑝𝑟+ (1 − 𝜀)

𝑘

∑

𝑗=𝑟

log 𝑝𝑗.

By the definition of𝑝𝑘, we have𝑝𝑘+1> 𝑁, thus to get a contradiction it is sufficient to show that the right-hand side of (5.5.14) is less than𝑁.

To achieve this, we estimate the right-hand side of (5.5.14) from above the following way:

(5.5.15) 𝑝𝑟+ (1 − 𝜀)

𝑘

∑

𝑗=𝑟

log 𝑝𝑗< 𝑝𝑟+ (1 − 𝜀)𝜋(𝑁) log 𝑁.

If𝑁is large enough, then

(5.5.16) 𝜋(𝑁) < (1 +𝜀

4) 𝑁 log 𝑁 by the Prime Number Theorem and

(5.5.17) 𝑝𝑟< 𝜀𝑁

Exercises 5.5 139

if𝑁is large enough. Substituting (5.5.16) and (5.5.17) into (5.5.15), we obtain that the right-hand side of (5.5.14) is less than

((1 − 𝜀)(1 +𝜀 4) +𝜀

4) 𝑁 < (1 − 𝜀

2)𝑁 < 𝑁,

yielding the desired contradiction. □

Exercises 5.5

1. Prove that𝑛!is not a perfect power if𝑛 > 1.

2. Verify that at least one of any two consecutive integers is representable as the sum of distinct primes (we allow sums consisting of a single term).

3. Demonstrate that infinitely many primes have (a) 1as first digit

(b) 4as the first thousand digits in decimal system.

4. Prove that neither of the following sums is an integer for1 ≤ 𝑘 < 𝑛:

(a)

𝑛

∑

𝑗=1

1 𝑗 (b)

𝑛

∑

𝑗=𝑘

1 𝑗.

5. Show that(2𝑛𝑛)is the largest among the binomial coefficients(2𝑛𝑘),0 ≤ 𝑘 ≤ 2𝑛.

6. Give another proof for Theorem 5.5.2 on the following lines: Choose2𝐾primes greater than𝐾,𝑝1, . . . ,𝑝𝐾,𝑞1, . . . ,𝑞𝐾, and consider the system of simultaneous congruences

𝑥 ≡ 𝑗 (mod 𝑝𝑗) , 𝑥 ≡ −𝑗 (mod 𝑞𝑗) , 𝑗 = 1, 2, . . . , 𝐾.

Show that the solutions contain (infinitely many) primes𝑝and they meet the requirements of the theorem.

7. (a) Prove that(2𝑛𝑛)is divisible by exactly the first power of every prime𝑛 + 1 ≤ 𝑝 ≤ 2𝑛.

(b) Show that if𝑝 > 3is a prime and2𝑛/5 < 𝑝 ≤ 𝑛/2, then(2𝑛𝑛)is not divisible by 𝑝. How can we generalize this observation?

8. Show that the proof of Chebyshev’s Theorem yields the following stronger result (for𝑛 ≥ 2): There are more than𝑐𝑛/ log 𝑛primes between𝑛and2𝑛where𝑐is a suitable positive constant.

S 9. (a) Using (A) in Theorem 5.5.4, verify that there is a prime between any two sufficiently large consecutive cubes.

* (b) Prove the existence of a real number𝛼 > 1such that⌊𝛼3𝑛⌋is a prime for every positive integer𝑛.

10. Establish results similar to (B) in Theorem 5.5.5 using the following facts or meth- ods instead of the Prime Number Theorem:

(a) Theorem 5.4.3

(b) the proof of Theorem 5.5.1

* 11. (Cf. with Remark2on twin primes in Section 5.1.) Prove that for any𝜀 > 0there exist infinitely many positive integers𝑛satisfying𝑝𝑛+1− 𝑝𝑛 < (1 + 𝜀) log 𝑛. (As usual,𝑝𝑛denotes the𝑛th prime.)

The Fundamental Theorem of Arithmetic

Residue Systems and Residue Classes