Problems & Puzzles:
Collection 20th
Coll.20th013.
The set {a+m,a+2m,...} contains infinitely...
On April 10, 2018,
Emmanuel Vantieghem,
wrote:
We suppose you know that, when a and m are two
coprime numbers, then the set { a+m, a+2m, a+3m, ... } contains
infinitely many primes.
Q1. Does that set also contains
infinitely many semiprimes ?
Q2. Does it contains infinitely many
square free numbers with three, four, ...... prime factors ?
Prove your
statement or give counterexamples.
Contribution came from John Arioni.
***
John wrote on August 20, 2018:
Q1
Let a = p*q, where p is
a prime number and q is any number such that a and m
are coprime. Obviously if q=1 then a=p is
prime, differently q and m are coprime.
Let's consider the set B, subset of A =
{ a+m, a+2m, a+3m,
... }:
B ={ a+pm, a+2pm, a+3pm,
... } = { p(q+m), p(q+2m), p(q+3m)
... }
It's plain to see that the subset B contains
one semiprime p*p' for each p' of the infinitely many primes in the set C =
{ (q+m), (q+2m), (q+3m,)
... }.
Q2
After Q1 we can state that a set like C =
{ (q+m), (q+2m), (q+3m,)
... }, as well as the set A = { a+m, a+2m,a+3m,
... }, contains infinitely many semiprimes, then B,
subset of A , (B = { p(q+m), p(q+2m), p(q+3m)
... }) contains infinitely many integers of the type p*p'*p'', that is
square free numbers with 3 prime factors. This proves that C,
as well as A, contains infinitely many integers of the
type p*p'*p'' , then we can state that the set B = {p(q+m), p(q+2m), p(q+3m)
... } contains infinitely many primes of the type p*p'*p''*p''' ….. and
so on.
***
On August 25, 2018, Jan van Delden wrote:
I read the contribution by John. Very elegant, where the
argument by Dirichlet is only used in the last step.
It limits a to pq and for m such that (pq,m)=1, with p
prime and q not necessarily prime.
Which means that the situation a=1 is not covered. In
this case one has (a,m)=1, for all m>1.
All one needs to do is to find a prime p and a number q
with (q,m)=1, such that pq=1 mod m, for m>1. [*]
This congruence relation pq=1 mod m has a solution if and
only if (p,m) divides 1.
We therefore choose a prime p such that (p,m)=1, or in other words p
does not divide m.
Suppose (q,m)=d>1, so m=dr and q=ds for some r,s>0.
We have pds=1+kdr for some k or d(pskr)=1 with d>1, a contradiction.
So we have (q,m)=1 as required.
[*]
Consider the set O = {1+m,1+2m,1+3m,…}
We can find p,q such that pq=1+km for some k.
Consider A = { 1+(k+1)m,1+(k+2)m,..} = {pq+1m,pq+2m,pq+3m,..}
subset of O and apply John’s result
***
On August 30, 2018, John Arioni wrote:
Here is the correct solution of the
puzzle.
Coll.20th013. The set {a+m,a+2m,...} contains
infinitely...
Q1
Let a = p*q, where p is
a prime number and q is any number such that a and m
are coprime. Obviously if q=1 thena=p is
prime, differently q and m are coprime.
Let's consider the set B, subset of A =
{ a+m, a+2m, a+3m,
... }:
B ={ a+pm, a+2pm, a+3pm,
... } = { p(q+m), p(q+2m), p(q+3m)
... }
It's plain to see that the subset B contains
one semiprime p*p' for each p' of the infinitely many primes of the set C =
{ (q+m), (q+2m), (q+3m,)
... } that are greater than p. This proves that the semiprimes in A are
infinitely many.
Q2
After Q1 we can state that a set like C =
{ (q+m), (q+2m), (q+3m,)
... }, as well as the set A = { a+m, a+2m,a+3m,
... }, contains infininitely many semiprimes p*p', then B,
subset of A , (B = { p(q+m), p(q+2m), p(q+3m)
... }) contains one integer of the type p*p'*p'' for each p” of the
infinitely many primes of the set C that are greater
than p and p' (p'>p).
This proves that also C, as well as A, contains
infinitely many integers of the type p*p'*p'' that are square free, then
we can state that the set B = {p(q+m), p(q+2m), p(q+3m)
... } contains infinitely many primes of the type p*p'*p''*p''' where
p''' is one of the infinitely many primes of the set C that
are greater than p, p' and p”(p">p'>p).
This proves that also C, as well as A,
contains infinitely many integers of the type p*p'*p''*p'' that are
square free
***
On Sep 12, 2018 John Arioni wrote:
Jan's remark is absolutely correct and his solution
suggested me the following simple one for the case a=1,
i.e. A = {1+m, 1+2m, 1+3m,
... }.
Let's consider the set D= { 1+(m+1)m, 1+(m+2)m, 1+(m+3)m,
... }, subset of A = { 1+m, 1+2m, 1+3m,
... }, and rewrite D as follows:
D= {(1+m^2)+m,
(1+m^2)+2m, (1+m^2)+3m, ... }
If we put (1+m^2)=a,
then:
D ={a+m, a+2m, a+3m,
... }, where obviously (a,m)=1.
***
