prime divisors in an arithmetic progression

Prime divisors in an arithmetic progression

Shanta Laishram

Associate Professor of MathematicsTheoretical Statistics and Mathematics UnitIndian Statistical Institute, New Delhi, India

E-mail: [email protected], [email protected]: http://www.isid.ac.in/∼shanta

Webinar organised by GC College, Silchar in association with AAM andgonitsora

July 7, 2020

Shanta Laishram Prime divisors in an arithmetic progression

Observation

Well-known: A product of k ≥ 1 consecutive positiveintegers is divisible by k !.One of the combinatorial proofs is given by the fact that thebinomial coefficient

(nk

)∈ Z.

nCk =

(nk

)=

n(n − 1) · · · (n − k + 1)

k !.

Each prime p ≤ k divides a product of k ≥ 1 consecutivepositive integers.We can ask if there is a prime > k dividing a product of kconsecutive positive integers.


A result of Sylvester-Erdos

A well-known theorem of Sylvester-Erdos answers thisquestion.A product of k consecutive integers each of which exceedsk is divisible by a prime greater than k .In other words,

P((n + 1)(n + 2) · · · (n + k)) > k when n ≥ k .

Here P(m) stands for the largest prime divisor of m withthe convention P(1) = 1.



This is a non trivial result.In fact if n = 0, then

(n + 1)(n + 2) · · · (n + k) = k !

and hence the condition n ≥ k cannot be avoided.A consequence of this result is the Bertrand’s Postulate:Given k > 1, there is a prime p with k < p < 2k .We can observe it easily by taking n = k . There is a prime> k dividing

(k + 1)(k + 2) · · · (2k − 1)2k

and this prime is of the form k + i with 1 ≤ i < k .



Let p1 = 2,p2 = 3,p3 = 5, · · · be the sequence of primes.So, pm is the m−th prime.Also denote by π(m) the number of primes ≤ m. Forexample, π(100) = 25 as there are 25 primes upto 100.Hence given k , the least prime exceeding k is pπ(k)+1.The assertion P((n + 1) · · · (n + k)) > k does not hold if

n < pπ(k)+1 − k .


Suffices for k prime

Lemma 1.

Suppose that

P ((n + 1) · · · (n + k)) > k for n ≥ k .(1)

holds for all primes k. Then it holds for all k .

Proof.Assume that (1) holds for all primes k . Let k1 ≤ k < k2 withk1, k2 consecutive primes. Let n ≥ k . Then n ≥ k1 and(n + 1) · · · (n + k1) has a prime factor p > k1 by our assumption.Further we observe that p ≥ k2 > k since k1 and k2 areconsecutive primes. Hence p divides(n + 1) · · · (n + k1) · · · (n + k) and (1) holds for k .


Proof of the result of Sylvester-Erdos

We now prove

P ((n + 1) · · · (n + k)) > k for k prime and n ≥ k .(2)

Putting x = n + k , we see from n ≥ k that x ≥ 2k and (2) isequivalent to

P((

xk

))= P

(x(x − 1) · · · (x − k + 1)

k !

)> k .

Therefore we will prove

P((

xk

))> k for k prime and x ≥ 2k .(3)

We will prove in an elementary way without using primenumber theory.


Proof of the result of Sylvester-Erdos

Suppose for some k prime and x ≥ 2k ,

P((

xk

))= P

(x(x − 1) · · · (x − k + 1)

k !

)≤ k .(4)

Then

P (x(x − 1) · · · (x − k + 1)) ≤ k .(5)


Power of prime dividing a binomial coefficient

Lemma 2.

Let pa∣∣(x

k

). Then pa ≤ x.

Proof.We observe that

ordp

(xk

)=∞∑

l=1

([xpl

]−[

x − kpl

]−[

kpl

]).

Each of the summand is at most 1 if pl ≤ x and 0 otherwise.Therefore ordp

(xk

)≤ s where ps ≤ x < ps+1. Thus

pa ≤ pordp(xk) ≤ ps ≤ x .


x bounded in terms of k

Lemma 3.

We have

P((

xk

))≤ k implies x < k

kk−π(k) .(6)

Further

x < k2 for k ≥ 11 and x < k32 for k ≥ 37.



Proof.We have (

xk

)=

xk

x − 1k − 1

· · · x − k + 11

>(x

k

)k.

From P((x

k

))≤ k and pa ≤ x for pa|

(xk

), we get

(xk

)≤ xπ(k).

Comparing the bounds for(x

k

), we have

xk

kk <

(xk

)< xπ(k) implying x < k

kk−π(k) .

For k ≥ 11, we exclude 1 and 9 to see that there are at most[k+1

2 ]− 2 odd primes upto k . Hence π(k) ≤ 1 + [ k+12 ]− 2 ≤ k

2for k ≥ 11 giving x < k2 for k ≥ 11.



Proof.Further the number of composite integers ≤ k and divisible by2 or 3 or 5 is

bk2c+ bk

3c+ bk

5c − bk

6c − b k

10c − b k

15c+ b k

30c − 3

≥k2

+k3

+k5

+k30− k

6− k

10− k

15− 7

=1115

k − 7.

Thus we have π(k) ≤ k − 1− (1115k − 7) ≤ k

3 for k ≥ 90. Bydirect computation, we see that π(k) ≤ k

3 for 37 ≤ k < 90.Hence k

k−π(k) ≤32 for k ≥ 37.



Lemma 4.

We have

P((

xk


kk−π(k) .(7)

Further



x < k32

Lemma 5.

Let x < k32 . Then

P((

xk

))≤ k implies

(xk

)< 4k+

√x .(8)


Proof of Lemma 5

Assume ∏p≤k

p∏

p≤√

k

p∏

p≤ 3√k

p · · · < 4k .(9)

Taking k =√

x ,∏p≤√

x

p∏

p≤ 4√x

p∏

p≤ 6√x

p · · · < 4√

x .

Since x < k32 , we have 2l−1

√x ≤ l√

k for l ≥ 2 and(xk

)=∏

pa||(xk)

pa ≤∏p≤k

p∏

p≤√

x

p∏

p≤ 3√x

p · · ·

≤

∏p≤k

p∏

p≤√

k

p∏

p≤ 3√k

p · · ·

∏p≤√

x

p∏

p≤ 4√x

p∏

p≤ 6√x

p · · ·

< 4k+

√x .


Proof continued: Upper bounds for prime divisors

We prove ∏p≤k

p∏

p≤√

k

p∏

p≤ 3√k

p · · · < 4k .(10)

For every prime p and a positive integer a withk < pa ≤ 2k , we have

ordp

((2kk

))= ordp

((2k)!

(k !)2

)≥

a∑i=1

{b2k

pi c − 2b kpi c}≥ 1

since

b2kpi c − 2b k

pi c ≥ 0 and b2kpa c − b

kpa c = 1.


Upper bounds for prime divisors

Let⌈ν⌉

denote the least integer greater than or equal to ν.Let 2m−1 ≤ k < 2m and we put

a1 =⌈k

2⌉, · · · , ah =

⌈ k2h

⌉, · · · , am =

⌈ k2m

⌉= 1.

Then

a1 > · · · > am and ah <k2h + 1 =

2k2h+1 + 1 ≤ 2ah+1 + 1

implying ah ≤ 2ah+1.Also 2a2 <

k2 + 2 ≤ a1 + 2 and therefore 2a2 ≤ a1 + 1.



Since 2a1 ≥ k , we see that

(1, k ] ⊆ ∪mh=1(ah,2ah].

Let p and r be such that pr ≤ k < pr+1 and let 1 ≤ i ≤ r .Then pi ≤ k and there exists hi such that

ahi < pi ≤ 2ahi and p∣∣(2ahi

ahi

).

Observe: ahi 6= ahj for 1 ≤ j < i ≤ r sincepahj < pj+1 ≤ pi ≤ 2ahi . Thus

pr ∣∣(2a1

a1

)(2a2

a2

)· · ·(

2am

am

).



Hence∏p≤k

p∏

p≤k12

p · · · =∏

pr≤k<pr+1

pr ≤(

2a1

a1

)(2a2

a2

)· · ·(

2am

am

).

Suffices to show:(2a1

a1

)(2a2

a2

)· · ·(

2am

am

)< 4k .(11)

Direct calculation: (11) holds for k ≤ 10.For example, when k = 5, we have a1 = 3,a2 = 2,a3 = 1so that (

2a1

a1

)(2a2

a2

)(2a3

a2

)= 20× 6× 2 < 45.



Let k > 10 and (11) holds for any integer less than k . Then(2a1

a1

)(2a2

a2

)· · ·(

2am

am

)<

(2a1

a1

)42a2−1.(12)

For k ≥ 8, we have (2kk

)< 4k−1.

Hence (2a1

a1

)(2a2

a2

)· · ·(

2am

am

)< 4a1−1+2a2−1

and (11) follows since 2a1 ≤ k + 1 and 2a2 ≤ a1 + 1.


Upper bounds for middle binomial coefficient

Lemma 6.

For k > 1, we have (2kk

)<

4k√

2k.(13)

Proof.

1 >(

1− 122

)(1− 1

42

)· · ·(

1− 1(2k − 2)2

)=

1 · 322

3 · 542 · · ·

(2k − 3)(2k − 1)

(2k − 2)2

>1

2k − 1

(3 · 5 · · · (2k − 1)

2kk !· 2k

)2

>4k2

2k

((2k)!

4k (k !)2

)2

.


Lower bounds for middle binomial coefficient

Lemma 7.

For k > 1, we have (2kk

)>

4k

2√

k(14)

Proof.For k > 1, we have

1 >(

1− 132

)(1− 1

52

)· · ·(

1− 1(2k − 1)2

)=

2 · 432

4 · 652 · · ·

(2k − 2)2k(2k − 1)2

>1

4k

(2kk !

3 · 5 · · · (2k − 1)

)2

=1

4k

(4k (k !)2

(2k)!

)2

.



We have

P((

xk


kk−π(k) .(15)

Further


Let x < k32 . Then

P((

xk

))≤ k implies

(xk

)< 4k+

√x < 4k+k

34 .(16)

For k > 1, we have

4k

2√

k<

(2kk

)<

4k√

2k.(17)


Lower bounds for(x

k

)for x < k

32

x ≥ 4k : We have(4k

k

)≤(x

k

)< 4k+

√x < 4k+k

34 and

4k+k34 >

(4kk

)=

(2kk

)4k(4k − 1) · · · (3k + 1)

2k(2k − 1) · · · (k + 1)

>4k

2√

k2k =

8k

2√

k

52k < x < 4k : Here

(⌈ 52 k⌉

k

)≤(x

k

)< 4k+

√x < 4k+2

√k and

4k+2√

k >

(2kk

)⌈52k⌉(⌈5

2k⌉− 1) · · · (

⌈52k⌉− k + 1)

2k(2k − 1) · · · (k + 1)>

4k

2√

k

(54

)k

.

2k ≤ x ≤ 52k : Here

(2kk

)≤(x

k

)< 4k+

√x ≤ 4k+

√5k/2 and

4k+√

5k/2 >

(2kk

)>

4k

2√

k.


Final Bounds

Comparing the bounds, we obtain for k ≥ 11 and x < k32

that x < 4k .Further k < 103 for 5

2k < x < 4k and k < 113 for2k ≤ x ≤ 5

2k .

Recall x < kk

k−π(k) and x < k2 for k ≥ 11 and x < k32 for

k ≥ 37.Combining all those: k ≤ 113;

x < k2 for 11 ≤ k ≤ 31; x < 4k for 37 ≤ k ≤ 113

andx < k

kk−π(k) for k < 11.


Idea of Sylvester-Erdos

Let S = {x , x − 1, · · · , x − k + 1}.For each prime p ≤ k , let 0 ≤ ip < k be such thatordp(x − ip) is maximal.Then for i 6= ip, we have

ordp(x − i) ≤ ordp(x − i − (x − ip)) = ordp(i − ip).

Put S0 = {x , x − 1, · · · , x − k + 1} \ {x − ip : p ≤ k}.Then |S0| ≥ k − π(k) and

ordp

∏x−i∈S0

(x − i)

≤ ordp (ip! (k − 1− ip)!) ≤ ordp((k−1)!)


Idea of Sylvester-Erdos

Thus ∏x−i∈S0

(x − i)

≤∏p<k

pordp((k−1)!) ≤ (k − 1)!.

Hence from |S0| ≥ k − π(k), we get

(x − k + 1)k−π(k) ≤ (k − 1)! or x ≤ k − 1 + ((k − 1)!)1

k−π(k) .

with the equality only when k = 2.We get x ≤ 2,4,9,13 according as k = 2,3,5,7,respectively. This contradicts x ≥ 2k . Thus k ≥ 11.For 11 ≤ k ≤ 31, we get x ≤ 71.


prime gaps

Finally for 11 ≤ k ≤ 113 and

x ≤ 71 for 11 ≤ k ≤ 31; x < 4k for 37 ≤ k ≤ 113

we use the following result.

Lemma 8.

Let X be a positive real number and k0 be a positive integer.Suppose that pi+1 − pi < k0 for any two consecutive primespi < pi+1 ≤ pπ(X)+1. Then

P(x(x − 1) · · · (x − k + 1)) > k

for 2k ≤ x ≤ X and k ≥ k0.


prime gaps

Proof.Let 2k ≤ x ≤ X . We may assume that none ofx , x − 1, · · · , x − k + 1 is a prime, since otherwise the resultfollows. Thus

pπ(x) < x − k + 1 < x < pπ(x)+1 ≤ pπ(X)+1.

Hence by our assumption, we have

k − 1 = x − (x − k + 1) < pπ(x)+1 − pπ(x) < k0,

which implies k − 1 < k0 − 1, a contradiction.


prime gaps

Computation gives:

pi+1 − pi <

{21 for pi+1 ≤ 457 = pπ(452)+1

8 for pi+1 ≤ 79 = pπ(73)+1.

Apply Lemma 8 as follows:For 23 ≤ k ≤ 31, take X = 452, k0 = 21;For 11 ≤ k ≤ 31, take X = 73, k0 = 10.This completes the proof.


Generalization of Sylvester’s Theorem

Recall Sylvester’s result:

P((n + 1) · · · (n + k)) > k for n ≥ k .

Moser: P((n + 1) · · · (n + k)) > 1110k

Hanson: P((n + 1) · · · (n + k)) > 1.5k unless(n, k) = (3,2), (8,2), (6,5).Faulkner: P((n + 1) · · · (n + k)) > 2k if n ≥ pπ(2k)+1.For k = 2, easy to see: P((n + 1)(n + 2)) > 4 except whenn = 1,2,7.



Laishram and Shorey: We have

P((n + 1) · · · (n + k)) > 1.95k for n ≥ k .

unless (n, k) belongs to an explicitly given finite set.Here 1.95 cannot be replaced by 2 asP((k + 1) · · · (k + k)) < 2k .Laishram and Shorey: We have

P((n + 1) · · · (n + k)) > 1.8k for n ≥ k .

unless n = k ∈ {3,4,5,8,11,13,14,18,63} or

(n, k) ∈ {(7,3), (5,4), (6,4), (14,13), (15,13)}.



Laishram and Shorey: We also have

P((n + 1) · · · (n + k)) > 2k for n ≥ max(

k + 13,279262

k).

and

P((n + 1) · · · (n + k)) > 1.97k for n ≥ k + 13.

Observe: 1.97 cannot be replaced by 2 since there arearbitrary long chains of consecutive composite positiveintegers.The assumption n ≥ 279

262k is necessary sinceP((n + 1) · · · (n + k)) ≤ 2× 262 when n = 278 andk = 262.


Generalization for arithmetic progression

Let n,d be positive coprime integers with d > 1 and put

∆(n,d , k) = n(n + d)(n + 2d) · · · (n + (k − 1)d).

Observe; P(∆(n,d ,2)) = 2 if and only if n = 1,d = 2r − 1with r > 1.Sylvester: ∆(n,d , k) > k if n ≥ d + k .Let d = 2. Then

∆(n,2, k) = n(n + 2)(n + 2d) · · · (n + 2(k − 1)).

Let n ≤ k . Then P(∆(n,2, k)) < 2k impliesP((n + k) · · · (n + 2k − 1)) ≤ 2k .This gives n + k − 1 < max(k + 13, 279

262k) orn < max(14, 17

262k + 1).We consider k > 2 and d > 2.



Langevin: P(∆(n,d , k)) > k if n > k .Shorey and Tijdeman: P(∆(n,d , k)) > k unless

(n,d , k) = (2,7,3) where P(2 · 9 · 16) = 3.

Laishram and Shorey: For k > 2 and d > 2, we have

P(n(n + d) · · · (n + (k − 1)d)) > 2k

unless (n,d , k) is given by

k = 3, n = 1,d = 4,7;

n = 2,d = 3,7,23,79;

n = 3,d = 61; n = 4,d = 23;

n = 5,d = 11; n = 18,d = 7;

k = 4, n = 1,d = 3,13; n = 3,d = 11;

k = 10, n = 1,d = 3.

This is best known result.Shanta Laishram Prime divisors in an arithmetic progression


Laishram and Shorey: P(∆(n,3, k)) > 3k when n > 3kand (n, k) 6= (125,2).Laishram and Shorey: P(∆(n,4, k)) > 4k when n > 4kand n 6= 21,45 when k = 2.Conjecture: Fix d . For n > dk and k > 1, we have

P(n(n + d) · · · (n + (k − 1)d)) > dk

except for finitely many n and k .This is open.



Though we have an elementary proof of Sylvester’s resultfor consecutive integers, the proof of other results needadvanced mathematical ideas from Prime number theory,linear forms in logarithms, Thue equations etc.However many of the ideas used in the elementary proofcan be generalised and are used in subsequent proofs.


Number of prime divisors

Let ω(m) denote the number of distinct prime divisors of m.For example, ω(10) = 2.The result of Sylvester is in fact

ω(n(n + 1) · · · (n + k − 1)) > π(k) if n > k .

ω(n(n + 1)) ≥ 3 except when n = 2p − 1, a MersennePrime or n = 22r

+ 1, a Fermat Prime or n ∈ {1,2,3,8}.In fact the famous Catalan Conjecture, now a theorem,states that x r = 9 = 32, ys = 8 = 23 are the only solutionsof

x r − ys = 1, with x > 1, y > 1, r > 1, s > 1.



Let k ≥ 3. Observe that

ω(n(n + 1) · · · (n + k − 1)) = π(2k) when n = k + 1.

If k + 1 is prime and 2k + 1 is composite, then we observethat

ω((k + 2) · · · (2k + 1)) = π(2k)− 1.

Let k + 1 be a prime of the form 3r + 2. Then2k + 1 = 3(2r + 1) is composite. Since there are infinitelymany primes of the form 3r + 2, we see that there areinfinitely many k for which k + 1 is prime and 2k + 1 iscomposite.Therefore ω(∆(n, k)) ≥ π(2k)− 1 is valid for infinitelymany k . Thus an inequality sharper thanω(∆(n, k)) ≥ π(2k)− 1 for n > k is not valid.



Saradha and Shorey:

ω(n(n + 1) · · · (n + k − 1)) ≥ π(k) + b13π(k)c+ 2 if n > k

unless (n, k) belongs to a finite set.Laishram and Shorey: For n > k , we have

ω(n(n + 1) · · · (n + k − 1)) ≥ π(k) + b34π(k)c − 1 + δ(k)

unless (n, k) belongs to a finite set and

δ(k) =

2 if 3 ≤ k ≤ 61 if 7 ≤ k ≤ 160 otherwise.



Laishram and Shorey: Let n > k . Then

ω(n(n + 1) · · · (n + k − 1))

≥min(π(k) + b3

4π(k)c − 1 + δ(k), π(2k)− 1

).

Laishram and Shorey: Let (n, k) 6= (6,4). Then

ω(n(n + 1) · · · (n + k − 1)) ≥ π(2k) for n >1712

k .

For n > k , we cannot have π(2k) or 2π(k) as a lowerbound.



Laishram and Shorey: Let ε > 0 and n > k . Then thereexists a computable number k0 depending only on ε suchthat for k ≥ k0, we have

ω(n(n + 1) · · · (n + k − 1)) ≥ (2− ε)π(k).


Grimm’s Conjecture

Suppose n + 1, · · · ,n + k are all composite numbers, thenthere are distinct primes Pj such that Pj |(n + j) for1 ≤ j ≤ k .This is open and considered quite difficult.This implies: If n,n + 1, · · · ,n + k − 1 are all composite,then ω(n(n + 1) · · · (n + k − 1)) ≥ k which is also open.Grimm’s Conjecture has been verified for all n ≤ 1.9× 1010

and for all k .


Schinzel’s Hypothesis H

Schinzel’s Hypothesis H:Let f1(x), · · · , fr (x) be irreducible non constant polynomials

with integer coefficients and the leading coefficientspositive. Assume that for every prime p, there is an integera such that the product f1(a) · · · fr (a) is not divisible by p.

Then there are infinitely many positive integers m such thatf1(m), · · · , fr (m) are all primes.

This is open. Considered quite difficult.Schinzel’s Hypothesis H implies thatω(∆(n,d , k)) = π(k), π(k) = 1 for k = 3 and k ∈ {4,5},respectively for infinitely many pairs (n,d).


Number of prime divisors in an AP

Shorey and Tijdeman: ω(∆(n,d , k)) ≥ π(k)

Moree: ω(∆(n,d , k)) > π(k) for k ≥ 4 and(n,d , k) 6= (1,2,5).Saradha, Shorey and Tijdeman: For k ≥ 6,

ω(∆(n,d , k)) >65π(k) + 1

unless (n,d , k) belongs to

{(1,2,6), (1,3,6), (1,2,7), (1,3,7), (1,4,7),

(2,3,7), (2,5,7), (3,2,7), (1,2,8), (1,2,11),

(1,3,11), (1,2,13), (3,2,13), (1,2,14)}.



Laishram and Shorey: For k ≥ 9

ω(∆(n,d , k)) ≥ π(2k)− ρ

unless n,d , k is given byn = 1, d = 3, k = 9,10,11,12,19,22,24,31;

n = 2, d = 3, k = 12; n = 4, d = 3, k = 9,10;

n = 2, d = 5, k = 9,10; n = 1, d = 7, k = 10.

where

ρ = ρ(d) =

{1 if d = 2,n ≤ k0 otherwise.

.



Laishram and Shorey: Let k ≥ 4. Then

ω(∆(n,d , k)) ≥ π(2k)− 1

except at (n,d , k) = (1,3,10).This proves a conjecture of Moree.


Applications

Binary Recurrence SequencesExponential Diophantine EquationsA weaker Conjecture: Fix d . For n > dk and k > 1, wehave

P(n(n + d) · · · (n + (k − 1)d)) >dk200

.

This conjecture implies a conjecture of Erdos which statesthat the equation

n(n + d) · · · (n + (k − 1)d) = y2

has no solution in positive integers n,d , k withgcd(n,d) = 1 and k ≥ 2.


Applications

Irreducibility of PolynomialsFor example, the Truncated Exponential Polynomials

En(x) = 1 + x +x2

2!+ · · ·+ xn

n!

are irreducible for all n ≥ 1.


Thank you


prime divisors in an arithmetic progression

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