Problems & Puzzles: Collection 20th

Coll.20th-011. One more as 43*47=2021

On April 6, 2018, Giovanni Resta wrote:

One may notice that the product of the two consecutive
primes 43 and 47 is equal to the concatenation of 2 consecutive integers:

43 * 47 = 2021

Q1 Find another pair of consecutive primes whose product is the
concatenation of two consecutive integers in ascending order.

Q2
Similarly, find a pair of consecutive primes whose product is a concatenation of two consecutive integers in descending order.

Jan van Delden wrote on Sep 12, 2018:

No solution I’m afraid, but a few remarks.

In Sloane encyclopedia integer sequences can be found relating to questions of the following two relating questions:

1. Numbers n such that n concatenated with n+a gives the product of two numbers which differ by b.
2. Numbers n such that n concatenated with n+a is a square. (or b=0).

We are looking for solutions of the first type where a equals +/-1, b is even and n, n+/-b are prime.

Several of our puzzler’s, but mainly Patrick de Geest and Giovanni Resta, have contributed these sequences.

Integer sequences A030465 and A054214 solve type b. for a=1 and a=-1 respectively.

Trying to solve Q2 for twin primes p,q:

p.q = n(10^d+1) – 1 gives:

m^2=n(10^d+1)  (using p=m-1,q=m+1)

So a solution to type b. with even m and m+/-1 prime would be nice.

Checking the even solutions of A102567 will not give our sought solutions.

I studied solutions to pq=n(10^d+1)+r with |r|>0  and |r| as close to 1 as possible.

My best solution is (r=-4):

81818181817,81818181821,6694214876166942148757

Maybe Giovanni could share some of his thoughts on the matter?

***

All I [CR] can say at this moment is that Giovanni Resta positively sent one solution for each question. Let's see if he wants to share some other thoughts on the matter.

***

On April 6, 2019, Ken Wilke wrote:

I have obtained a “potential solution” not previously mentioned on your website. I say potential because initial zeroes are involved (and are not specifically excluded by the statement of the problem.

The potential solution is 4999493*4999507= 0024995000249951.

Methodology:

Let pi and pi+1 be consecutive primes where pi+1 – pi = 2d for some integer d. Then the product pi*pi+1 = (pi+ d)2 – d2.

Let m denote a block of digits which is concatenated with itself. Then questions 1 and 2 can be considered in the equation

(pi+ d)2 – d2 = m(10^n + 1) ± 1                                                 (1)

where n is a positive integer and the upper(lower) sign corresponds to ascending(descending) order of concatenation.

Then equation (1) implies the corresponding congruences

x2 =  (pi+ d)2 == d2 + 1 (mod 10^n + 1)                                    (2a)

and                    x2 =  (pi+ d)2 == d2 - 1 (mod 10^n + 1)                                    (2 b)

Then for given values of d and n, any solution x for either congruence one can easily check whether both x + d and x-d are consecutive primes. From here on, we look at only congruence (2a) since the discussion easily adapts for congruence (2b)

Let a = d2 + 1 and let (a/pj) be the Legendre symbol which shows that the congruence x2 == d2 + 1 (mod pj) is solvable for a prime pj whenever let (a/pj) =1 and pj divides 10^n +1. Consulting a table of Factors of Qn = 10n + 1 1 and a table of Quadratic Residues: (a/p) = +1, which show linear forms of primes for which the congruence x2 == a (mod pj) is solvable, one can ascertain which values of d2+1and the corresponding form of pj must be tested.

For example: for n=8, we have 108 +1 = 17*5882353. Since both primes are of the form 8m +1, one needs to examine congruences of the form x2 == d2 + 1 (mod p) where p = 17 and 5882353. Here d=1, 2 and 7 gives d2 + 1 =2, 5 and 50

respectively in congruence )(2a) and for d2- 1=1and 3 we have d2-1 = 0 and 8 respectively in congruence (2b). After combining the solutions generated using the Chinese Remainder Theorem, only the solution                                              x == 4999500(mod 108+1) yield the consecutive primes 4999493 and 4999507, the product of which is 24995000249951. By adding two initial zeroes, we have a new solution to question 1.

1.       Hans Riesel,Prime Numbers and Computer Methods for Factorization, Birkhauser , (1985), pp. 400-403.

2.       Ibid. pp.438-441.

***

Giovanni Resta wrote on April 11, 2019:

The solutions I found when I submitted the puzzle were

(a)
891077215721081784886888257701070827 *
891077215721081784886888257701070829 =
794018604377235322848433897872605582 ||
794018604377235322848433897872605583

(b) 4803478892324963 * 4803478892324969 =
2307340946901148 || 2307340946901147

where || stands for concatenation.

Note that in case (a) the two primes are twin primes.

The approach was simple. I show case (a) since (b) uses the same method.

Say that p and q = p+g (g is the gap between the two primes)
are the two consecutive primes and I want their
product to be the concatenation of x and x+1. We can assume that x and
x+1 have the same number of digits (because otherwise one should be
10^k-1 and the other 10^k, and these cases can be checked separately).

Assuming that x and x+1 have d digits, the concatenation of x and x+1 is
simply  x*10^d + (x+1).

So we have a Diophantine equation of the form  p*(p+g) = x*10^d + (x+1).

If we fix g and d, it is a quadratic Diophantine equation with the
further constraint 10^(d-1) <= x < 10^d because x must have d digits,
otherwise x*10^d + (x+1) is not the concatenation of x and x+1.

This kind of equation can be solved with Mathematica or Maple (and
probably with other softwares).

For example, if g=4 and d=2 the equation become p(p+4) = 100x+x+1. This
equation has infinite solutions, but with the constraint 10<=x<100 has
only two  (p=54 x=31) and (p=43, x=20). What I have to do now is to
check if p in the solution is prime and if p+g is the next prime. This
is true for p=43, because p+4=47 is the next prime.

So what I did was to compute the solution of a bunch of equations for
fixed values of d and g. Note that g (the gap between the two primes)
can be reasonably bounded once d is fixed, because p will be roughly of
the same size of x, so around 10^d.

In general it is conjectured that the maximal gap between two primes
grows proportionally to the square of log(p). So for each length d I
established a finite range for g.

When d grows the value of (log 10^d)^2 becomes quite large, and also
solving the Diophantine equations become slower. Being a little
impatient, I fixed an upper bound of g<4000 for the gaps up to d=50, and
then an upper bound g<1000 for 50<d<=63.

With these constraints, I did not find any other solution for (a). For
(b) I stopped earlier at d=58 but I found another solution:

p=8286379552872432369913616581715872482998493372751
q=p+228

and p*q =

6866408609426233220587132313827017515323918295230 ||
6866408609426233220587132313827017515323918295229

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