Problems & Puzzles: Collection 20th

Coll.20th-013. 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.

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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.

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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(ps-kr)=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 Johns result

 

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On August 30, 2018, John Arioni wrote:

Here is the correct solution of the puzzle.
 

Coll.20th-013. 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 

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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. = {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.

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