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Square-free integer

2

$begingroup$

Is it true that for any positive integer, the prime decomposition of N = 10^2n + 10^n + 1 is of the form N = pqrs... with distinct numbers p < q < r < s <...?

For the first values of n,hereafter the prime decompositions:

n = 1, N = 111 = 3*37

n = 2, N = 10101 = 3*7*13*37

n = 3, N = 1001001 = 3*333667

n = 4, N = 100010001 = 3*7*13*37*9901

n = 5, N = 10000100001 = 3*31*37*2906161

n = 6, N = 1000001000001 = 3*19*52579*333667

n = 7, N = 100000010000001 = 3*37*43*1933*10838689

n = 8, N = 10000000100000001 = 3*7*13*37*9901*99990001
etc....

number-theory

share|cite|improve this question

edited Mar 27 at 12:31

Oscar Lanzi

13.7k12136

asked Mar 26 at 9:26

Bernard VignesBernard Vignes

557

$endgroup$

add a comment | 

2

$begingroup$

Is it true that for any positive integer, the prime decomposition of N = 10^2n + 10^n + 1 is of the form N = pqrs... with distinct numbers p < q < r < s <...?

For the first values of n,hereafter the prime decompositions:

n = 1, N = 111 = 3*37

n = 2, N = 10101 = 3*7*13*37

n = 3, N = 1001001 = 3*333667

n = 4, N = 100010001 = 3*7*13*37*9901

n = 5, N = 10000100001 = 3*31*37*2906161

n = 6, N = 1000001000001 = 3*19*52579*333667

n = 7, N = 100000010000001 = 3*37*43*1933*10838689

n = 8, N = 10000000100000001 = 3*7*13*37*9901*99990001
etc....

number-theory

share|cite|improve this question

edited Mar 27 at 12:31

Oscar Lanzi

13.7k12136

asked Mar 26 at 9:26

Bernard VignesBernard Vignes

557

$endgroup$

add a comment | 

$begingroup$

Is it true that for any positive integer, the prime decomposition of N = 10^2n + 10^n + 1 is of the form N = pqrs... with distinct numbers p < q < r < s <...?

For the first values of n,hereafter the prime decompositions:

n = 1, N = 111 = 3*37

n = 2, N = 10101 = 3*7*13*37

n = 3, N = 1001001 = 3*333667

n = 4, N = 100010001 = 3*7*13*37*9901

n = 5, N = 10000100001 = 3*31*37*2906161

n = 6, N = 1000001000001 = 3*19*52579*333667

n = 7, N = 100000010000001 = 3*37*43*1933*10838689

n = 8, N = 10000000100000001 = 3*7*13*37*9901*99990001
etc....

number-theory

share|cite|improve this question

edited Mar 27 at 12:31

Oscar Lanzi

13.7k12136

asked Mar 26 at 9:26

Bernard VignesBernard Vignes

557

$endgroup$

share|cite|improve this question

edited Mar 27 at 12:31

Oscar Lanzi

13.7k12136

asked Mar 26 at 9:26

Bernard VignesBernard Vignes

557

share|cite|improve this question

edited Mar 27 at 12:31

Oscar Lanzi

13.7k12136

asked Mar 26 at 9:26

Bernard VignesBernard Vignes

557

edited Mar 27 at 12:31

Oscar Lanzi

13.7k12136

edited Mar 27 at 12:31

Oscar Lanzi

13.7k12136

asked Mar 26 at 9:26

Bernard VignesBernard Vignes

557

asked Mar 26 at 9:26

Bernard VignesBernard Vignes

557

2 Answers
2

active

oldest

votes

3

$begingroup$

No, for $n=14$ we have
$$
10^2n+10^n+1=3^1cdot 7^2cdot 13cdot 37cdot 43cdot 127cdot 1933cdots .
$$
So this number is not squarefree. In principle, we have seen this question here, a bit differently:

Prove that all numbers $10^n + 1$ are square free

This claim also was not true.

share|cite|improve this answer

edited Mar 26 at 13:03

Oscar Lanzi

13.7k12136

answered Mar 26 at 9:29

Dietrich BurdeDietrich Burde

82.1k649107

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @OscarLanzi Thank you, Oscar.
  $endgroup$
  – Dietrich Burde
  Mar 26 at 16:08
  

  
  
  
  

add a comment | 

2

$begingroup$

We set out to find a counterexample that is divisible by $49$, using a method similar to that demonstrated here.

First solve $x^2+x+1equiv 0bmod 49$. Thereby $xin18,30bmod 49$, both of these residues being units. Now we must put in $x=10^n$ and determine whether some whole number $n$ gives $xin18,30bmod 49$.

The residue $3$ is a primitive root $bmod 7$. Then among the units $bmod 49$ the only nonprimitive residue $equiv 3bmod 7$ is $3^7equiv 31bmod 49$. Since $10$ is different from this solutions to $10^nin18,30 bmod 49$ must exist and with them, so do multiples of $49$ matching $10^2n+10^n+1$.

share|cite|improve this answer

edited Mar 26 at 12:32

answered Mar 26 at 10:18

Oscar LanziOscar Lanzi

13.7k12136

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks for your explanations.
  $endgroup$
  – Bernard Vignes
  Mar 26 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May I say that if I consider N = 3^(2*3^n) + 3^(3^n) + 1, N is not square-free for some integers n?
  $endgroup$
  – Bernard Vignes
  Mar 26 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ask a new question, please.
  $endgroup$
  – Oscar Lanzi
  Mar 26 at 12:33
  

  
  
  
  

add a comment | 

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3

$begingroup$

No, for $n=14$ we have
$$
10^2n+10^n+1=3^1cdot 7^2cdot 13cdot 37cdot 43cdot 127cdot 1933cdots .
$$
So this number is not squarefree. In principle, we have seen this question here, a bit differently:

Prove that all numbers $10^n + 1$ are square free

This claim also was not true.

share|cite|improve this answer

edited Mar 26 at 13:03

Oscar Lanzi

13.7k12136

answered Mar 26 at 9:29

Dietrich BurdeDietrich Burde

82.1k649107

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @OscarLanzi Thank you, Oscar.
  $endgroup$
  – Dietrich Burde
  Mar 26 at 16:08
  

  
  
  
  

add a comment | 

3

$begingroup$

No, for $n=14$ we have
$$
10^2n+10^n+1=3^1cdot 7^2cdot 13cdot 37cdot 43cdot 127cdot 1933cdots .
$$
So this number is not squarefree. In principle, we have seen this question here, a bit differently:

Prove that all numbers $10^n + 1$ are square free

This claim also was not true.

share|cite|improve this answer

edited Mar 26 at 13:03

Oscar Lanzi

13.7k12136

answered Mar 26 at 9:29

Dietrich BurdeDietrich Burde

82.1k649107

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @OscarLanzi Thank you, Oscar.
  $endgroup$
  – Dietrich Burde
  Mar 26 at 16:08
  

  
  
  
  

add a comment | 

$begingroup$

No, for $n=14$ we have
$$
10^2n+10^n+1=3^1cdot 7^2cdot 13cdot 37cdot 43cdot 127cdot 1933cdots .
$$
So this number is not squarefree. In principle, we have seen this question here, a bit differently:

Prove that all numbers $10^n + 1$ are square free

This claim also was not true.

share|cite|improve this answer

edited Mar 26 at 13:03

Oscar Lanzi

13.7k12136

answered Mar 26 at 9:29

Dietrich BurdeDietrich Burde

82.1k649107

$endgroup$

share|cite|improve this answer

edited Mar 26 at 13:03

Oscar Lanzi

13.7k12136

answered Mar 26 at 9:29

Dietrich BurdeDietrich Burde

82.1k649107

edited Mar 26 at 13:03

Oscar Lanzi

13.7k12136

edited Mar 26 at 13:03

Oscar Lanzi

13.7k12136

answered Mar 26 at 9:29

Dietrich BurdeDietrich Burde

82.1k649107

answered Mar 26 at 9:29

Dietrich BurdeDietrich Burde

82.1k649107

2

$begingroup$

We set out to find a counterexample that is divisible by $49$, using a method similar to that demonstrated here.

First solve $x^2+x+1equiv 0bmod 49$. Thereby $xin18,30bmod 49$, both of these residues being units. Now we must put in $x=10^n$ and determine whether some whole number $n$ gives $xin18,30bmod 49$.

The residue $3$ is a primitive root $bmod 7$. Then among the units $bmod 49$ the only nonprimitive residue $equiv 3bmod 7$ is $3^7equiv 31bmod 49$. Since $10$ is different from this solutions to $10^nin18,30 bmod 49$ must exist and with them, so do multiples of $49$ matching $10^2n+10^n+1$.

share|cite|improve this answer

edited Mar 26 at 12:32

answered Mar 26 at 10:18

Oscar LanziOscar Lanzi

13.7k12136

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks for your explanations.
  $endgroup$
  – Bernard Vignes
  Mar 26 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May I say that if I consider N = 3^(2*3^n) + 3^(3^n) + 1, N is not square-free for some integers n?
  $endgroup$
  – Bernard Vignes
  Mar 26 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ask a new question, please.
  $endgroup$
  – Oscar Lanzi
  Mar 26 at 12:33
  

  
  
  
  

add a comment | 

2

$begingroup$

We set out to find a counterexample that is divisible by $49$, using a method similar to that demonstrated here.

First solve $x^2+x+1equiv 0bmod 49$. Thereby $xin18,30bmod 49$, both of these residues being units. Now we must put in $x=10^n$ and determine whether some whole number $n$ gives $xin18,30bmod 49$.

The residue $3$ is a primitive root $bmod 7$. Then among the units $bmod 49$ the only nonprimitive residue $equiv 3bmod 7$ is $3^7equiv 31bmod 49$. Since $10$ is different from this solutions to $10^nin18,30 bmod 49$ must exist and with them, so do multiples of $49$ matching $10^2n+10^n+1$.

share|cite|improve this answer

edited Mar 26 at 12:32

answered Mar 26 at 10:18

Oscar LanziOscar Lanzi

13.7k12136

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks for your explanations.
  $endgroup$
  – Bernard Vignes
  Mar 26 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May I say that if I consider N = 3^(2*3^n) + 3^(3^n) + 1, N is not square-free for some integers n?
  $endgroup$
  – Bernard Vignes
  Mar 26 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ask a new question, please.
  $endgroup$
  – Oscar Lanzi
  Mar 26 at 12:33
  

  
  
  
  

add a comment | 

$begingroup$

We set out to find a counterexample that is divisible by $49$, using a method similar to that demonstrated here.

First solve $x^2+x+1equiv 0bmod 49$. Thereby $xin18,30bmod 49$, both of these residues being units. Now we must put in $x=10^n$ and determine whether some whole number $n$ gives $xin18,30bmod 49$.

The residue $3$ is a primitive root $bmod 7$. Then among the units $bmod 49$ the only nonprimitive residue $equiv 3bmod 7$ is $3^7equiv 31bmod 49$. Since $10$ is different from this solutions to $10^nin18,30 bmod 49$ must exist and with them, so do multiples of $49$ matching $10^2n+10^n+1$.

share|cite|improve this answer

edited Mar 26 at 12:32

answered Mar 26 at 10:18

Oscar LanziOscar Lanzi

13.7k12136

$endgroup$

share|cite|improve this answer

edited Mar 26 at 12:32

answered Mar 26 at 10:18

Oscar LanziOscar Lanzi

13.7k12136

answered Mar 26 at 10:18

Oscar LanziOscar Lanzi

13.7k12136

answered Mar 26 at 10:18

Oscar LanziOscar Lanzi

13.7k12136