Problems & Puzzles:
Collection 20th
Coll.20th011.
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:

Numbers n such
that n concatenated with n+a gives the
product of two numbers which differ by b.

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=m1,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 p_{i} and p_{i+1} be consecutive primes where p_{i+1}
– p_{i} = 2d for some integer d. Then the product p_{i*}p_{i+1}
= (p_{i}+ d)^{2} – d^{2}.
Let m denote a block of digits which is concatenated with itself.
Then questions 1 and 2 can be considered in the equation
(p_{i}+ d)^{2} – d^{2} =
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
x^{2} = (pi+ d)^{2} == d^{2}
+^{ }1 (mod 10^n + 1)
(2a)
and
x^{2} = (pi+ d)^{2} == d^{2}
 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 xd are
consecutive primes. From here on, we look at only congruence (2a)
since the discussion easily adapts for congruence (2b)
Let a =
d^{2}
+^{ }1 and let (a/p_{j}) be the Legendre symbol
which shows that the congruence x^{2} == d^{2} +^{
}1 (mod p_{j}) is solvable for a prime p_{j}
whenever let (a/p_{j}) =1 and p_{j} divides 10^n +1.
Consulting a table of Factors of Q_{n} = 10^{n} + 1
^{1} and a table of
Quadratic Residues: (a/p) = +1, which show linear forms of primes
for which the congruence x^{2} == a (mod pj) is solvable,
one can ascertain which values of d^{2+}1and the
corresponding form of p_{j} must be tested.
For example: for n=8, we have 10^{8} +1 = 17*5882353. Since
both primes are of the form 8m +1, one needs to examine congruences
of the form
x^{2}
== d^{2} +^{ }1 (mod p) where
p = 17 and 5882353. Here d=1, 2 and 7 gives d^{2} + 1 =2, 5
and 50
respectively in congruence )(2a) and for d^{2} 1=1and 3 we
have d^{2}1 = 0 and 8 respectively in congruence (2b).
After combining the solutions generated using the Chinese Remainder
Theorem, only the
solution x ==
4999500(mod 10^{8}+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. 400403.
2.
Ibid. pp.438441.
***
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^k1 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^(d1) <= 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
***
