Problems & Puzzles: Puzzles

Puzzle 976. m/d+/-d primes.

Emmanuel Vantieghem sent the following nice puzzle:

I was wondering which natural numbers  m  have the following property :

"m  has at least one divisor  d  such that  m/d - d  and  m/d + d  are both primes, d < Sqrt(m)"  (***).

The first such  m  are quickly found :

4, 6, 10, 12, 18, 24, 28, 30, 36, 40, 42, 48, 60, 66, 70, 72, 78, 84, 88, 90, 102, 108, 120, 126, 130, 132, 138, 150, 162, 168, 180, 192, 198

For instance, 42  is in this sequence because its divisors are : 1, 2, 3, 6, 7, 14, 21, 42
and  42/2 - 2 = 19, 42/2 + 2 = 23  are both prime.

Clearly, such an  m  cannot be odd.

A bit less clear but sure true is the following statement:

"there are infinitely many  m  with the property (***)"

Q1 prove that statement.

Other questions are:

Q2. find very big  m  with the property  (***).

Q3. find numbers with the property (***) with as great as possible number of such divisors.

 

Contributions came from Fausto Morales, Fred Schneider, Paul Cleary, Metin Sariyar, Adam Stinchcombe, Gennady Gusev, Paolo Lava, Jeff Heleen, Simon Cavegn, Oscar Volpatti, Emmanuel Vantieghem

All of them wrote on the week 26-31 October, 2019

***

Fausto wrote

Q1:
The result on prime gaps by Yitang Zhang (*) implies that there is a (currently unknown) constant k < N = 7*(10^7) appearing infinitely many times as gap between two consecutive primes.
 
Consider any pair of consecutive primes {p(n), p(n+1)}, with p(n) > k/2 and  p(n+1) - p(n) = k.

Then the integer
 
m = (k/2) * [p(n) + p(n+1)]/2 
 
meets the condition with d = k/2 (< sqrt(m)):
 
(i)     (m/d) - d = p(n)
(ii)    (m/d) + d = p(n+1)
 
Since there are infinitely many such choices for m, the proposition is true. 
 
Moreover, d can be further constrained to be less than c*m for any positive constant c, and the proposition still holds.
(*)
Bounded gaps between primes
Yitang Zhang
Annals of Mathematics
SECOND SERIES, Vol. 179, No. 3 (May, 2014), pp. 1121-1174

Q2:
 
With M51= 2^82589933 - 1 (the largest known prime I am aware of), the approach described in Q1 comes in handy to produce a large example as well:
 
Letting
 
m = [(M51)^2 - 9] / 4    and
d = [(M51) - 3] / 2m/d + d = M51
m/d - d = 3

Q3


This looks like an extraordinary problem to me, probably worth throwing some heavy hardware at - along with some clever search strategy...

The following case is just a boutique example (with small m and 9 valid divisors):
 
m = 2310

d           m/d - d         m/d + d
 
1             2309              2311
5               457                467
14             151                179
21               89                131
22               83                137
30               47                107
33               37                103
35               31                101
42               13                  97

***

Fred wrote

Q1:

For every pair of primes (p, p + 2d) with d > 0,  m= d * (p+d) has a solution for d

m/d -d = p + d - d = p
m/d +d = p + d + d = p + 2d
 
As there is an infinite number of primes, there is an infinite number of prime pairs.
 
So, there are an infinite number of m.
 
Q2) The largest known twin primes are : 2996863034895 ∑ 21290000 +/- 1
 
So m = 2996863034895 ∑ 21290000
is a large solution

Source: https://en.wikipedia.org/wiki/Twin_prime
 
Q3) Looking at this table for divisor-rich numbers (http://oeis.org/A002201/list):
 
6064949221531200 has 89 prime pair solutions.  Here are the associated divisors:
[2025,3553,4807,7475,9425,10925,12673,44051,46400,67925,127075,234175,375193,390775,395675
,469557,567567,580203,978112,1000384,1311552,1323328,1397825,1425600,1550016,1653696,
1767744,1847104,1884025,2091663,2203200,2273600,2647323,2977975,3333275,3454451,3568131,
3741309,4673669,5048813,5143775,5757475,6111721,7793344,8227648,8976825,9434048,9863451,
10621975,10759232,13344669,13873275,13873600,15505425,16385600,17223941,18179392,19250287,
21523887,21772881,22496576,22980672,23297344,24012352,24235200,24545025,25329325,26734400,
28112832,29120553,29450304,29668275,30948075,31011904,31128867,32011200,41860800,45004491,
48859200,51264576,52055003,53566016,56635200,59246275,63892800,68493425,70018325,70747425,
73920275]

***

Paul wrote

My first attempt at Q3 is the number

 

2438560915062710206517929282865911302.  This has 17082 pairs of solutions.

 

...

 

I donít think I can get much bigger than this for Q3.  As it was it used almost 32GB of memory.  The number is

 

177209197509194632020570614172936200654501805139182, with 1115615 pairs of solutions.

 

Here is a link to my Dropbox of the file with the solutions.  Caution it is a 145 MB file.

 

https://www.dropbox.com/s/g7ze4mpi2pfa5zy/puzzle%20976%201.txt?dl=0

 

 ***

Metin wrote

 

Q1:
Proof :

It is known that there are infinitely many primes.So let p1 and p2 with p2>p1 be two  primes.Then n=(p1+p2)/2 is the arithmetic mean of these two primes and m=n*(p2-n)=n*(n-p1) (and it is obvious that here d=p2-n=n-p1<sqrt(m) because d<n) has this property and here d=(n-p1)=(p2-n) is the divisor of m such that (m/d)+d=p2 and (m/d)-d=p1 .So there are infinitely many numbers m with this property. 
An example :

for p1=31, p2=37 , n= (31+37)/2=34, m=34*(37-34)=34*(34-31)=102 and

102/3+3=37, 102/3-3=31.      

Q2: 
By taking the largest known two primes we can find a large m with this property such as:
P2:2^82589933 - 1
P1:2^77232917 - 1

m=((p1+p2)/2)*(p2-((p1+p2)/2))=((p1+p2)/2)*(((p1+p2)/2)-p1)

Q3: 
I found an example with "34" of its 192 divisors, 1021020 has this property. But I think this is a very small example ( which I was able to find with my home pc)

***

Adam wrote

It is probably true due to the twin prime conjecture: if p and p+2 are two twin primes then (p+1)/1-1 and (p+1)/2+1 are of the required form.

It is definitely true by (all variables below are positive):

Let    1 + 2*n1, 1 + 2*n2 be two primes with n2>n1.  Write 1+2*n1=m/d-d and 1+2*n2=m/d+d and solve for d=n2-n1.  Next solve for m=(1+2*n1+d)*d.  Clearly d is a divisor of m but also m=(1+2*n1+d)*d>d*d=d^2 so sqrt(m)>d.

Example: 31 = 1 + 2*15 and 73=1 +2*36, write d=21 and m=(31+21)*21 = 1092.  Then 1092/21-21 = 31 and 1092/21+21 = 73 and approximately,  sqrt(1092) = 33.05>d .

Big m can be achieved with big d, say 1+2*n1=3 and 1+2*n2 the largest known prime.

Many divisors similarly can be achieved by a lot of divisors of d.  For a given large k, use Dirichlet's theorem to find infinitely many primes of the form 1+k!*n and then, given two of them, d is divisible by k!/2 (at least).

...

Say the prime pair {p1,p2} speak for m if there is a d<sqrt(m) such that {p1,p2}={m/d-d,m/d+d}.  The most spoken for m would have the most numerous prime pairs.  In a search up to m=4,596,527, the first m that have k prime pairs speaking for m are (excluding the beginning record holders) the following (k,m) pairs:  (26,510510), (30,746130), (33,870870), and (34,1021020).

For instance, the prime pairs that speak for 1021020 are {{255251, 255259}, {15773, 15643}, {19687, 19583}, {12071, 12239}, {13183, 13337}, {14947, 15083}, {92809, 92831}, {51071, 51031}, {29207, 29137}, {12097, 11927}, {9829, 9619}, {8699, 8461}, {7867, 7603}, {6997, 7283}, {6389, 6701}, {4801, 5209}, {1669, 2621}, {1789, 2699}, {2029, 181}, {1213, 2357}, {1259, 2381}, {2267, 3037}, {5647, 5273}, {4861, 4421}, {3343, 2663}, {2207, 887}, {2441, 3169}, {2011, 2851}, {3217, 2503}, {2039, 271}, {157, 2027}, {3467, 4013}, {2203, 877}, {617, 2113}}.


Note: typo in the original message, not (p+1)/2+1 but rather  (p+1)/1+1.

...

It is probably true due to the twin prime conjecture: if p and p+2 are two twin primes then (p+1)/1-1 and (p+1)/1+1 are of the required form.

***

Gennady wrote

Let primes p1=m/d-d and p2=m/d+d.

Then p2-p1=2*d, d=(p2-p1)/2;

p1+p2=2*m/d=4*m/(p2-p1);

And m=(p1+p2)/(p2-p1)/4=(p2^2-p1^2)/4; note that divisor (p2-p1)/2 is less than sqrt(m).

Let p1=3, then m=(p2^2-9)/4; The number of primes is infinitely, so number of m is infinitely too.

Q1 is proved.

Q2. m maybe arbitrary large. For proved primes p1 and p2 we take p1=3 and p2 = 51st mersenne prime 2^p - 1 (the largest known prime), where p=82589931.

Then m=(2^(2*p)-2^(p+1)+1-9)/4=2^(2*p-2)-2^p-2=2^165179860-2^82589931-2.

...

There is no answer to Q3 question. I.e. if you say number with N such divisors I say number with N1>N such divisors.
 
For example, number m=29#=2*3*5*7*11*13*17*19*23*29=6469693230 yields 106 pairs of prime with divisors of m:

 

3, 17, 57, 145, 209, 406, 418, 442, 462, 546, 609, 667, 759, 770, 782, 805, 874, 1015, 1131, 1254, 1330, 1463, 1870, 2233, 2346, 2431, 2717, 2730, 3135, 3230, 3289, 3315, 3451, 3570, 3705, 3770, 4147, 4370, 4485, 4830, 4862, 5278, 5423, 5474, 5681, 5865, 6270, 6279, 6555, 6670, 6783, 7293, 7854, 7917, 10010, 10166, 11339, 13398, 13585, 14007, 14326, 15249, 15470, 15834, 16302, 16422, 17342, 18734, 19227, 19285, 20010, 20930, 23023, 24310, 28014, 28101, 29393, 33649, 34034, 35581, 37145, 37961, 38019, 39585, 40755, 41055, 42427, 43355, 46410, 51870, 54230, 56202, 57057, 58058, 62491, 62985, 64090, 66990, 67298, 67830, 68034, 70035, 74613, 75922, 79373, 79534.

23# -> 56;

31# -> 199;

37# -> 328;

41# -> 484;

43# -> 835;

47# -> 1702;

53# -> 2354

and, I think, so on but I cannot prove it.

***

Paolo wrote

I just searched for the minimum numbers with property (***) that admit n distinct divisors.

For m<= 10^7 we have:

n

m

1

4

2

18

3

30

4

60

5

390

6

210

7

462

8

1932

9

2310

10

5460

11

2730

12

13860

13

59052

14

117390

15

51870

16

43890

17

46410

18

78540

19

30030

20

221340

21

570570

22

131670

23

1385670

24

1345890

25

1591590

26

510510

27

1141140

28

1272810

29

1610070

30

746130

31

1845690


 
and n=33  m=870870,  n=34  m=1021020,  n=43  m=9699690

***

Jeff wrote

For puzzle 976 Q3 the table below shows the smallest m which has k
factors that give solutions.
 k        m
 1        4               = 2^2
 2        6               = 2 ◊ 3
 3        78              = 2 ◊ 3 ◊ 13
 4        30              = 2 ◊ 3 ◊ 5
 5        390             = 2 ◊ 3 ◊ 5 ◊ 13
 6        330             = 2 ◊ 3 ◊ 5 ◊ 11
 7        210             = 2 ◊ 3 ◊ 5 ◊ 7
 8        462             = 2 ◊ 3 ◊ 7 ◊ 11
 9        2310            = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11
10        5460            = 2^2 ◊ 3 ◊ 5 ◊ 7 ◊ 13
11        2730            = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 13
12        13860           = 2^2 ◊ 3^2 ◊ 5 ◊ 7 ◊ 11
13        23562           = 2 ◊ 3^2 ◊ 7 ◊ 11 ◊ 17
14        117390          = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 13 ◊ 43
15        51870           = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 13 ◊ 19
16        85470           = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 37
17        43890           = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 19
18        53130           = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 23
19        30030           = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13
20        221340          = 2^2 ◊ 3 ◊ 5 ◊ 7 ◊ 17 ◊ 31
21        570570          = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 19
22        131670          = 2 ◊ 3^2 ◊ 5 ◊ 7 ◊ 11 ◊ 19
23        1385670         = 2 ◊ 3 ◊ 5 ◊ 11 ◊ 13 ◊ 17 ◊ 19
24        1345890         = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 13 ◊ 17 ◊ 29
25        1591590         = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 53
26        690690          = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 23
27        510510          = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17
28        1272810         = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 19 ◊ 29
29        1610070         = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 17 ◊ 41
30        746130          = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 17 ◊ 19
31        1845690         = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 17 ◊ 47
32        26996970        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 29 ◊ 31
33        870870          = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 29
34        1021020         = 2^2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17
35        14804790        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17 ◊ 29
36        15825810        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17 ◊ 31
37        29274630        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 19 ◊ 23 ◊ 29
38        19399380        = 2^2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17 ◊ 19
39        11741730        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17 ◊ 23
40        51381330        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 29 ◊ 59
41        16546530        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 19 ◊ 29
42        26246220        = 2^2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 19 ◊ 23
43        9699690         = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17 ◊ 19
44        21637770        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 17 ◊ 19 ◊ 29
45        37447410        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 29 ◊ 43
46        40750710        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 23 ◊ 59
47        none <10^8
48        35225190        = 2 ◊ 3^2 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 17 ◊ 23
49        21111090        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 19 ◊ 37
50        23393370        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 19 ◊ 41
51        56258202        = 2 ◊ 3 ◊ 7 ◊ 11 ◊ 13 ◊ 17 ◊ 19 ◊ 29
52        81363282        = 2 ◊ 3 ◊ 7 ◊ 11 ◊ 13 ◊ 19 ◊ 23 ◊ 31
53-58     none <10^8
59        34804770        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 13 ◊ 19 ◊ 61
60-61     none <10^8
62        17160990        = 2 ◊ 3 ◊ 5 ◊ 7 ◊ 11 ◊ 17 ◊ 19 ◊ 23
and no more <10^8

***

Simon wrote:

Q1:
First we find two primes p1 and p2, where p1 < p2, (not too far apart for the d < sqrt(m) condition). Then we can construct m and d: d = (p2-p1)/2, m = (p2-d)*d = (p1+d)*d. Short outline for a proof: With the upper bound for prime gaps small enough to fullfill d < sqrt(m), and with the construction of m and d, it follows infinitly many m and d can be constructed from infinitly many primes.
 
Q2:
Largest probable prime solution I found with the computer:
m = 1694*10^11585 + 45116302
d = 1694
p1 = 10^11585 + 24939
p2 = 10^11585 + 28327
 
Largest cousin probable primes (that differ by 4) found on internet:
m = (474435381*2^98394 − 3) * 2
d = 2
p1 = 474435381*2^98394 − 5
p2 = 474435381*2^98394 − 1
 
Q3:
The highest m where I could calculate amount(d): m = 23984823528925228172706521638692258396210 amount(d) = 1706980 (m is the multiplication of the first primes up to 103)
I noticed: When m = multiplication of the first n primes, the amount of such divisors d is big, but not maximal. (There may be smaller m with larger amount of such divisors.) However it is an easy way to construct m with many such divisors d. I assume the amount of such divisors d goes to infinity when n goes to infinity. Is this assumption true? Is it proofable?

***

Oscar wrote:

Dear Carlos,

 
for any pair of odd primes (p,q) with p>q, we can find an integer m and a proper divisor d of m such that  m/d + d = p  and  m/d - d = q.

 
Proof.
Defining c = m/d, we obtain the linear system:
c+d = p,
c-d = q;
with solution:
c = (p+q)/2,
d = (p-q)/2,
m = c*d = (p^2 - q^2)/4.
As p and q are both odd, the numbers c,d,m are positive integers, as claimed.

 
Q1.
Emmanuel Vantieghem's statement easily follows: keeping q fixed and letting p vary, we obtain infinitely many distinct numbers m.

 
Q2.
Let p and q be the largest known primes:
p = 2^82589933 - 1,
q = 2^77232917 - 1,
c = 2^82589932 + 2^77232916 - 1,
d = 2^82589932 - 2^77232916,
m = 2^165179864 - 2^154465832 - 2^82589932 + 2^77232916,
where m has 49724094 digits.

 
Q3.
As a first step, I found the numbers m < 1e7 with a record-breaking number of good divisors (gd).
Such list includes many primorials:

 
gd   m
====
  1   4
----------------
  2   18 = 2*9
 
-------------------------
  3   30 = 2*3*5 = 5#
 
  4   60 = 4*3*5
 
-------------------------------
  6     210 = 2*3*5*7 = 7#
 
  7     462 = 2*3*7*11
 
  8   1932 = 4*3*7*23
--------------------------------------
  9     2310 = 2*3*5*7*11 = 11#
11     2730 = 2*3*5*7*13
 
12   13860 = 4*9*5*7*11
-------------------------------------------
19     30030 = 2*3*5*7*11*13 = 13#
 
22   131670 = 2*9*5*7*11*19
-------------------------------------------------
26     510510 = 2*3*5*7*11*13*17 = 17#
30     746130 = 2*3*5*7*11*17*19
33     870870 = 2*3*5*7*11*13*29
 
34   1021020 = 4*3*5*7*11*13*17
-----------------------------------------------------
43   9699690 = 2*3*5*7*11*13*17*19 = 19#

 
For a given integer m, consider a good decomposition m = c*d.
If some prime divides both c and d, it is a proper divisor of  c+d  and  c-d  too.
So, c and d must be coprime: if some prime power f^n divides m, it divides either c or d, but not both.
If the factorization of m contains k distinct primes, there can be at most 2^(k-1) good divisors (not 2^k, because d must also be smaller than c).
As k-th primorial is the smallest number with k distinct prime factors, it has at least twice as many candidate good divisors as any previous number; moreover, the numbers  c+d  and  c-d  are coprime with the first k primes and are slightly more likely to be prime.

 
Let f(k) be the actual number of good divisors for k-th primorial.
I computed f(k) for k <=29:

 
p(k)#   f(k)
======
    2#   0
    3#   1
    5#   3
    7#   6
  11#   9
  13#   19
  17#   26
  19#   43
  23#   56
  29#   106
  31#   199
  37#   328
  41#   484
  43#   835
  47#   1702
  53#   2354
  59#   4642
  61#   7474
  67#   13970
 
  71#   22248
  73#   42334
 
  79#   77377
  83#   152843
  89#   300338
  97#   528407
101#   857005
103#   1706980
107#   2923940
 
109#   5595865

 
f(k) is always smaller than 2^(k-1), but it seems to grow exponentially.
A simple estimate is H*(2^k)/(k^2), with a constant H about 9.

 

***

Emmanuel wrote:

Q1.
Let  m  have the property (***).
Put  p = d + m/d
       q = -d + m/d.
It is very easy to solve  d  and  m/d  from these two equations :
       d = (p -q)/2  and  m/d = (p +q)/2.
So, if we take  p, q  as big as we want we will find that the number
       m = (p^2 - q^2)/4
has the propety (***) and is as big as we want.  Thus, there are infinitely many such numbers.

Q2.
For  p  I take  2^82589933 - 1 (actually the biggest known prime)  and for  q  I take 3.
Then,
       M = (p^2 - q^2)/4 = 2*(2^82589932 + 1)*(2^82589931 - 1).
It is a number of  49723094  digits.
I think that's a record that will stand til a new Mersenne prime is found.

Q3.
Let's denote by  L(m)  the number of prime pairs of the form  (-d + m/d, d + m/d).
The greatest value I found for  L(m)  is when  m = 89# (the 24th primorial)  and is  300338.
My PC has not enough memory to compute the L-value of higher primorials but I conjecture
that  L(p_n #)  is unbounded.

By the way, L(M) = 1  when  M  is the number of  Q2.

***


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