Why do all primes $n>3$ satisfy $,309mid 20^n-13^n-7^n$Prime factor of $A=14^7+14^2+1$Number of divisors excluding set of primes SCalculating $e^3x pmod27$Help understanding number theory definitionWhat's wrong with this proof of the infinity of primes?Unclear on why Meissel's approach to counting primes worksIf $a^n+n^bmid c^n+n^d$ for every $n$ then $c=a^k$ and $d=kb$ .What does x equivalent to 2 mod 15 mean?Olympiad problem: Prove that there are infinitely many primes which divide the sequence $ S(k) _kin mathbbN$Primes of Atlantis: a definition and a problemWhat is an irreducible prime?
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Why do all primes $n>3$ satisfy $,309mid 20^n-13^n-7^n$
Prime factor of $A=14^7+14^2+1$Number of divisors excluding set of primes SCalculating $e^3x pmod27$Help understanding number theory definitionWhat's wrong with this proof of the infinity of primes?Unclear on why Meissel's approach to counting primes worksIf $a^n+n^bmid c^n+n^d$ for every $n$ then $c=a^k$ and $d=kb$ .What does x equivalent to 2 mod 15 mean?Olympiad problem: Prove that there are infinitely many primes which divide the sequence $ S(k) _kin mathbbN$Primes of Atlantis: a definition and a problemWhat is an irreducible prime?
$begingroup$
Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
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
edited Apr 23 at 19:31
Bill Dubuque
214k29198660
asked Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
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.
$endgroup$
|
show 2 more comments
$begingroup$
Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
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
edited Apr 23 at 19:31
Bill Dubuque
214k29198660
asked Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
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
Apr 23 at 16:33
$begingroup$
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
Apr 23 at 16:38
$begingroup$
There are no even numbers ..
$endgroup$
– Shamim Akhtar
Apr 23 at 16:39
3
$begingroup$
It seems like $n equiv pm1 pmod 6$.
$endgroup$
– md2perpe
Apr 23 at 16:46
$begingroup$
Yeah right but why do only specific composites arrive?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:47
|
show 2 more comments
$begingroup$
Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
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
edited Apr 23 at 19:31
Bill Dubuque
214k29198660
asked Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
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.
$endgroup$
Solve the following... $309|(20^n-13^n-7^n)$ in $mathbbZ^+$.
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
elementary-number-theory
edited Apr 23 at 19:31
Bill Dubuque
214k29198660
asked Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
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 Apr 23 at 19:31
Bill Dubuque
214k29198660
asked Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
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 Apr 23 at 19:31
Bill Dubuque
214k29198660
edited Apr 23 at 19:31
Bill Dubuque
214k29198660
edited Apr 23 at 19:31
Bill Dubuque
214k29198660
214k29198660
asked Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
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 Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
asked Apr 23 at 16:31
Shamim AkhtarShamim Akhtar
13911
13911
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.
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.
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Is it false that it generates all the primes?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:33
$begingroup$
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
Apr 23 at 16:38
$begingroup$
There are no even numbers ..
$endgroup$
– Shamim Akhtar
Apr 23 at 16:39
3
$begingroup$
It seems like $n equiv pm1 pmod 6$.
$endgroup$
– md2perpe
Apr 23 at 16:46
$begingroup$
Yeah right but why do only specific composites arrive?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:47
|
show 2 more comments
$begingroup$
Is it false that it generates all the primes?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:33
$begingroup$
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
Apr 23 at 16:38
$begingroup$
There are no even numbers ..
$endgroup$
– Shamim Akhtar
Apr 23 at 16:39
3
$begingroup$
It seems like $n equiv pm1 pmod 6$.
$endgroup$
– md2perpe
Apr 23 at 16:46
$begingroup$
Yeah right but why do only specific composites arrive?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:47
$begingroup$
Is it false that it generates all the primes?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:33
$begingroup$
Is it false that it generates all the primes?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:33
$begingroup$
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
Apr 23 at 16:38
$begingroup$
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
Apr 23 at 16:38
$begingroup$
There are no even numbers ..
$endgroup$
– Shamim Akhtar
Apr 23 at 16:39
$begingroup$
There are no even numbers ..
$endgroup$
– Shamim Akhtar
Apr 23 at 16:39
3
3
$begingroup$
It seems like $n equiv pm1 pmod 6$.
$endgroup$
– md2perpe
Apr 23 at 16:46
$begingroup$
It seems like $n equiv pm1 pmod 6$.
$endgroup$
– md2perpe
Apr 23 at 16:46
$begingroup$
Yeah right but why do only specific composites arrive?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:47
$begingroup$
Yeah right but why do only specific composites arrive?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:47
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, 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 $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
$endgroup$
$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
Apr 23 at 17:12
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
Apr 23 at 19:53
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
Apr 23 at 22:04
$begingroup$
@Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
$endgroup$
– Bill Dubuque
Apr 23 at 22:38
add a comment |
$begingroup$
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.
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
$endgroup$
$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:36
$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
Apr 23 at 16:37
1
$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
Apr 23 at 16:40
$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
Apr 23 at 16:41
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
Apr 23 at 16:46
|
show 1 more 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.
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
$endgroup$
$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
Apr 23 at 17:06
1
$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
Apr 23 at 17:11
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, 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 $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
$endgroup$
$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
Apr 23 at 17:12
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
Apr 23 at 19:53
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
Apr 23 at 22:04
$begingroup$
@Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
$endgroup$
– Bill Dubuque
Apr 23 at 22:38
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 $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
$endgroup$
$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
Apr 23 at 17:12
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
Apr 23 at 19:53
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
Apr 23 at 22:04
$begingroup$
@Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
$endgroup$
– Bill Dubuque
Apr 23 at 22:38
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 $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
$endgroup$
$20^6 equiv 13^6 equiv 7^6 equiv 229 bmod 309$, 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 $25 = 5^2$, $35 = 5 cdot 7$, $49 = 7 cdot 7$, $55 = 5 cdot 11$, ...
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
edited Apr 23 at 17:14
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
answered Apr 23 at 17:10
Robert IsraelRobert Israel
333k23223484
333k23223484
$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
Apr 23 at 17:12
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
Apr 23 at 19:53
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
Apr 23 at 22:04
$begingroup$
@Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
$endgroup$
– Bill Dubuque
Apr 23 at 22:38
add a comment |
$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
Apr 23 at 17:12
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
Apr 23 at 19:53
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
Apr 23 at 22:04
$begingroup$
@Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
$endgroup$
– Bill Dubuque
Apr 23 at 22:38
$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
Apr 23 at 17:12
$begingroup$
+1 really cool.
$endgroup$
– Shamim Akhtar
Apr 23 at 17:12
2
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
Apr 23 at 19:53
$begingroup$
@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).
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– Bill Dubuque
Apr 23 at 19:53
1
1
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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 $(a^2-ab+b^2)=309$
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– Mark Fischler
Apr 23 at 22:04
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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 $(a^2-ab+b^2)=309$
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– Mark Fischler
Apr 23 at 22:04
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@Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
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– Bill Dubuque
Apr 23 at 22:38
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@Mark We can also proceed analogously as I do here. (iirc this specific case is done in some other question here).
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– Bill Dubuque
Apr 23 at 22:38
add a comment |
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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.
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
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Can you explain why this works?
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– Shamim Akhtar
Apr 23 at 16:36
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I tapped the show more but wolframalpha doesnt show more
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– Shamim Akhtar
Apr 23 at 16:37
1
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I'm afraid I canot
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– Michael Stachowsky
Apr 23 at 16:40
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Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
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– Shamim Akhtar
Apr 23 at 16:41
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.
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– Michael Stachowsky
Apr 23 at 16:46
|
show 1 more comment
$begingroup$
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.
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
$endgroup$
$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:36
$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
Apr 23 at 16:37
1
$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
Apr 23 at 16:40
$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
Apr 23 at 16:41
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
Apr 23 at 16:46
|
show 1 more comment
$begingroup$
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.
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
$endgroup$
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.
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
answered Apr 23 at 16:36
Michael StachowskyMichael Stachowsky
1,409418
1,409418
$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:36
$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
Apr 23 at 16:37
1
$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
Apr 23 at 16:40
$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
Apr 23 at 16:41
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
Apr 23 at 16:46
|
show 1 more comment
$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:36
$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
Apr 23 at 16:37
1
$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
Apr 23 at 16:40
$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
Apr 23 at 16:41
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
Apr 23 at 16:46
$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:36
$begingroup$
Can you explain why this works?
$endgroup$
– Shamim Akhtar
Apr 23 at 16:36
$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
Apr 23 at 16:37
$begingroup$
I tapped the show more but wolframalpha doesnt show more
$endgroup$
– Shamim Akhtar
Apr 23 at 16:37
1
1
$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
Apr 23 at 16:40
$begingroup$
I'm afraid I canot
$endgroup$
– Michael Stachowsky
Apr 23 at 16:40
$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
Apr 23 at 16:41
$begingroup$
Is this a standard question in mse? I too am afraid if any moderator marks this as off topic
$endgroup$
– Shamim Akhtar
Apr 23 at 16:41
1
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
Apr 23 at 16:46
$begingroup$
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
Apr 23 at 16:46
|
show 1 more 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.
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
$endgroup$
$begingroup$
How can you verify?
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– Shamim Akhtar
Apr 23 at 17:06
1
$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
Apr 23 at 17:11
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.
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
$endgroup$
$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
Apr 23 at 17:06
1
$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
Apr 23 at 17:11
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.
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
$endgroup$
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.
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
edited Apr 23 at 17:16
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
answered Apr 23 at 16:59
Mark FischlerMark Fischler
34.7k12552
34.7k12552
$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
Apr 23 at 17:06
1
$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
Apr 23 at 17:11
add a comment |
$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
Apr 23 at 17:06
1
$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
Apr 23 at 17:11
$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
Apr 23 at 17:06
$begingroup$
How can you verify?
$endgroup$
– Shamim Akhtar
Apr 23 at 17:06
1
1
$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
Apr 23 at 17:11
$begingroup$
Mathematica is very fast in determining residues mod 309.
$endgroup$
– Mark Fischler
Apr 23 at 17:11
add a comment |
Shamim Akhtar is a new contributor. Be nice, and check out our Code of Conduct.
Shamim Akhtar is a new contributor. Be nice, and check out our Code of Conduct.
Shamim Akhtar is a new contributor. Be nice, and check out our Code of Conduct.
Shamim Akhtar is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Is it false that it generates all the primes?
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– Shamim Akhtar
Apr 23 at 16:33
$begingroup$
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?
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– Mark Fischler
Apr 23 at 16:38
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There are no even numbers ..
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– Shamim Akhtar
Apr 23 at 16:39
3
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It seems like $n equiv pm1 pmod 6$.
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– md2perpe
Apr 23 at 16:46
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Yeah right but why do only specific composites arrive?
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– Shamim Akhtar
Apr 23 at 16:47