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Why do all primes $n>3$ satisfy $,309mid 20^n-13^n-7^n$

6

4

$begingroup$

Solve the following... $309|(20^n-13^n-7^n)$ in $mathbb{Z}^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following... Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!

elementary-number-theory

share|cite|improve this question

edited 8 hours ago

Bill Dubuque

214k29198660

asked 11 hours ago

Shamim AkhtarShamim Akhtar

469

New contributor

Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it false that it generates all the primes?
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
  $endgroup$
  – Mark Fischler
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are no even numbers ..
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It seems like $n equiv pm1 pmod 6$.
  $endgroup$
  – md2perpe
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah right but why do only specific composites arrive?
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  

 | 
show 2 more comments

6

4

$begingroup$

Solve the following... $309|(20^n-13^n-7^n)$ in $mathbb{Z}^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following... Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!

elementary-number-theory

share|cite|improve this question

edited 8 hours ago

Bill Dubuque

214k29198660

asked 11 hours ago

Shamim AkhtarShamim Akhtar

469

New contributor

Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it false that it generates all the primes?
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is easy to picture why there might be a solution for all prime $n$. However, the interesting part of the pattern is that $n=5cdot 5$ and $n=5cdot 7$ both appear, yet neither $n=3cdot 3$ nor $n=5cdot 5$ appear. Which odd composites admit solutions?
  $endgroup$
  – Mark Fischler
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are no even numbers ..
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It seems like $n equiv pm1 pmod 6$.
  $endgroup$
  – md2perpe
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah right but why do only specific composites arrive?
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Solve the following... $309|(20^n-13^n-7^n)$ in $mathbb{Z}^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following... Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!

elementary-number-theory

share|cite|improve this question

edited 8 hours ago

Bill Dubuque

214k29198660

asked 11 hours ago

Shamim AkhtarShamim Akhtar

469

New contributor

Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

Bill Dubuque

214k29198660

asked 11 hours ago

Shamim AkhtarShamim Akhtar

469

New contributor

Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 8 hours ago

Bill Dubuque

214k29198660

asked 11 hours ago

Shamim AkhtarShamim Akhtar

469

New contributor

Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 8 hours ago

Bill Dubuque

214k29198660

edited 8 hours ago

Bill Dubuque

214k29198660

asked 11 hours ago

Shamim AkhtarShamim Akhtar

469

New contributor

Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 11 hours ago

Shamim AkhtarShamim Akhtar

469

3 Answers
3

active

oldest

votes

11

$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^{n+6} equiv 13^{n+6} + 7^{n+6} bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.

All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Robert IsraelRobert Israel

333k23223484

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 really cool.
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
  $endgroup$
  – Bill Dubuque
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At mathoverflow.net/questions/328782/… there is a proof that in general $(a^2-ab+b^2)$ divides $a^{6k+r} -b^{6k+r} -(a-b)^{6k+r}$ for $r = 1, 5$. The question at hand is a particular case with $a=20, b=13$ where $(a^2-ab+b^2)=309$
  $endgroup$
  – Mark Fischler
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
  $endgroup$
  – Bill Dubuque
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^{th}$ prime without having to do lots more computation. Quite cool, though.

share|cite|improve this answer

answered 11 hours ago

Michael StachowskyMichael Stachowsky

1,409418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain why this works?
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tapped the show more but wolframalpha doesnt show more
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid I canot
  $endgroup$
  – Michael Stachowsky
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
  $endgroup$
  – Michael Stachowsky
  10 hours ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

A trivial observation is that it does not generate all the odd primes: $3$ is missing.

However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.

This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.

But that fact is at least as interesting as your observation, and I don't know how to prove it.

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Mark FischlerMark Fischler

34.6k12552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can you verify?
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Mathematica is very fast in determining residues mod 309.
  $endgroup$
  – Mark Fischler
  10 hours ago
  

  
  
  
  

add a comment | 

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11

$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^{n+6} equiv 13^{n+6} + 7^{n+6} bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.

All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Robert IsraelRobert Israel

333k23223484

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 really cool.
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
  $endgroup$
  – Bill Dubuque
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At mathoverflow.net/questions/328782/… there is a proof that in general $(a^2-ab+b^2)$ divides $a^{6k+r} -b^{6k+r} -(a-b)^{6k+r}$ for $r = 1, 5$. The question at hand is a particular case with $a=20, b=13$ where $(a^2-ab+b^2)=309$
  $endgroup$
  – Mark Fischler
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
  $endgroup$
  – Bill Dubuque
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

11

$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^{n+6} equiv 13^{n+6} + 7^{n+6} bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.

All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Robert IsraelRobert Israel

333k23223484

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 really cool.
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ShamimAkhtar If you had entered your sequence $,5,7,11,13,ldots, 65$ into OEIS you'd obtain a unique match: A007310 = Numbers congruent to $1$ or $5$ mod $6$. Once you know that the rest is easy. That's the first method one should try on problems like this (when the pattern is not already clear from prior knowledge).
  $endgroup$
  – Bill Dubuque
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At mathoverflow.net/questions/328782/… there is a proof that in general $(a^2-ab+b^2)$ divides $a^{6k+r} -b^{6k+r} -(a-b)^{6k+r}$ for $r = 1, 5$. The question at hand is a particular case with $a=20, b=13$ where $(a^2-ab+b^2)=309$
  $endgroup$
  – Mark Fischler
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
  $endgroup$
  – Bill Dubuque
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, so $20^n equiv 13^n + 7^n mod 309$ if and only if $20^{n+6} equiv 13^{n+6} + 7^{n+6} bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n equiv 13^n + 7^n bmod 309$ if and only if $n equiv 1$ or $5 bmod 6$.

All primes except $2$ and $3$ are congruent to $1$ or $5 bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Robert IsraelRobert Israel

333k23223484

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Robert IsraelRobert Israel

333k23223484

answered 10 hours ago

Robert IsraelRobert Israel

333k23223484

answered 10 hours ago

Robert IsraelRobert Israel

333k23223484

1

$begingroup$

So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^{th}$ prime without having to do lots more computation. Quite cool, though.

share|cite|improve this answer

answered 11 hours ago

Michael StachowskyMichael Stachowsky

1,409418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain why this works?
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tapped the show more but wolframalpha doesnt show more
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid I canot
  $endgroup$
  – Michael Stachowsky
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
  $endgroup$
  – Michael Stachowsky
  10 hours ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^{th}$ prime without having to do lots more computation. Quite cool, though.

share|cite|improve this answer

answered 11 hours ago

Michael StachowskyMichael Stachowsky

1,409418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain why this works?
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tapped the show more but wolframalpha doesnt show more
  $endgroup$
  – Shamim Akhtar
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid I canot
  $endgroup$
  – Michael Stachowsky
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's a perfectly acceptable question, although it's probably much more difficult to answer than we think. As a result, you're probably either not going to get an answer or get a very strange one. I could be wrong, number theory isn't my thing.
  $endgroup$
  – Michael Stachowsky
  10 hours ago
  

  
  
  
  

 | 
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So far you have only shown evidence that some of the primes are generated here, not all of them. It's interesting how many there are, though. There are some polynomials that generate an exorbitant number of primes. Here are some examples. You will likely have to do a lot more work than just exploring the idea numerically to prove that all primes are generated. Even if you do, these functions aren't super interesting because they produce non-primes as well. You won't know whether a number that is generated is prime or not until you test it for primality, so it's not like it can be used to generate the $n^{th}$ prime without having to do lots more computation. Quite cool, though.

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answered 11 hours ago

Michael StachowskyMichael Stachowsky

1,409418

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share|cite|improve this answer

answered 11 hours ago

Michael StachowskyMichael Stachowsky

1,409418

answered 11 hours ago

Michael StachowskyMichael Stachowsky

1,409418

answered 11 hours ago

Michael StachowskyMichael Stachowsky

1,409418

1

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A trivial observation is that it does not generate all the odd primes: $3$ is missing.

However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.

This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.

But that fact is at least as interesting as your observation, and I don't know how to prove it.

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edited 10 hours ago

answered 10 hours ago

Mark FischlerMark Fischler

34.6k12552

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can you verify?
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Mathematica is very fast in determining residues mod 309.
  $endgroup$
  – Mark Fischler
  10 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

A trivial observation is that it does not generate all the odd primes: $3$ is missing.

However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.

This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.

But that fact is at least as interesting as your observation, and I don't know how to prove it.

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Mark FischlerMark Fischler

34.6k12552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can you verify?
  $endgroup$
  – Shamim Akhtar
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Mathematica is very fast in determining residues mod 309.
  $endgroup$
  – Mark Fischler
  10 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

A trivial observation is that it does not generate all the odd primes: $3$ is missing.

However, I can verify that all primes up to $59359$ (other than $2$ and $3$) are represented.

This has a lot do to with the fact that $(a^2-ab+^2)$ is a apparently factor of $a^n-b^n-(a-b)^n$ for all prime $n>3$. Here, $a=20, b=13$.

But that fact is at least as interesting as your observation, and I don't know how to prove it.

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Mark FischlerMark Fischler

34.6k12552

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Mark FischlerMark Fischler

34.6k12552

answered 10 hours ago

Mark FischlerMark Fischler

34.6k12552

answered 10 hours ago

Mark FischlerMark Fischler

34.6k12552