$fracq^p-1q-1$ squarefree? The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara$3^p-2^p$ squarefree?Are there other pseudo-random distributions like the prime-numbers?Is every primorial number squarefree?How many numbers $2^n-k$ are prime?Are mersenne numbers with prime exponent cube free?How did Euler disprove Mersenne's conjecture?There is at least one number with $k$ different prime factors in Mersenne sequence, right?Why do Mersenne primes only have prime inputs?On variations of Erdős squarefree conjecture: a conjecture involving the prime-counting functionIs it possible the density of Mersenne numbers in the primes gets arbitrarily close to $1$?$3^p-2^p$ squarefree?
How did passengers keep warm on sail ships?
Is this wall load bearing? Blueprints and photos attached
Is there a writing software that you can sort scenes like slides in PowerPoint?
Did the new image of black hole confirm the general theory of relativity?
How did the audience guess the pentatonic scale in Bobby McFerrin's presentation?
Is an up-to-date browser secure on an out-of-date OS?
Huge performance difference of the command find with and without using %M option to show permissions
ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?
What information about me do stores get via my credit card?
Accepted by European university, rejected by all American ones I applied to? Possible reasons?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Why can't devices on different VLANs, but on the same subnet, communicate?
Circular reasoning in L'Hopital's rule
Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?
Do I have Disadvantage attacking with an off-hand weapon?
How do spell lists change if the party levels up without taking a long rest?
Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?
Visa regaring travelling European country
Why not take a picture of a closer black hole?
1960s short story making fun of James Bond-style spy fiction
Loose spokes after only a few rides
Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?
What can I do if neighbor is blocking my solar panels intentionally?
"... to apply for a visa" or "... and applied for a visa"?
$fracq^p-1q-1$ squarefree?
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara$3^p-2^p$ squarefree?Are there other pseudo-random distributions like the prime-numbers?Is every primorial number squarefree?How many numbers $2^n-k$ are prime?Are mersenne numbers with prime exponent cube free?How did Euler disprove Mersenne's conjecture?There is at least one number with $k$ different prime factors in Mersenne sequence, right?Why do Mersenne primes only have prime inputs?On variations of Erdős squarefree conjecture: a conjecture involving the prime-counting functionIs it possible the density of Mersenne numbers in the primes gets arbitrarily close to $1$?$3^p-2^p$ squarefree?
$begingroup$
Is $fracq^p-1q-1$ always squarefree with $q,p$ prime and $p>2$ and $(q,p)=(3,5)$ excluded?
This is a follow up of $3^p-2^p$ squarefree?
I know the case $q=2$ (Mersenne) and $q=3$ are still open, but is there a similar/generalised conjecture for the other prime $q$?
number-theory prime-numbers prime-factorization mersenne-numbers
asked Mar 24 at 18:47
Collag3nCollag3n
784211
$endgroup$
add a comment |
$begingroup$
Is $fracq^p-1q-1$ always squarefree with $q,p$ prime and $p>2$ and $(q,p)=(3,5)$ excluded?
This is a follow up of $3^p-2^p$ squarefree?
I know the case $q=2$ (Mersenne) and $q=3$ are still open, but is there a similar/generalised conjecture for the other prime $q$?
number-theory prime-numbers prime-factorization mersenne-numbers
asked Mar 24 at 18:47
Collag3nCollag3n
784211
$endgroup$
$begingroup$
A prime number $r$ with the property $$r^2mid fracq^p-1q-1$$ must be a Wieferich-prime to base $q$, that is $$q^r-1equiv 1mod r^2$$ must hold.
$endgroup$
– Peter
Mar 24 at 19:06
$begingroup$
Yes, necessary, but not sufficient
$endgroup$
– Collag3n
Mar 24 at 19:14
$begingroup$
I am not actually sure whether it is even necessary, but if $r$ does not divide $q-1$, then surely.
$endgroup$
– Peter
Mar 24 at 19:15
$begingroup$
can't we apply the same reasoning as for the mersenne case? If $rmid fracq^p-1q-1$, than $p$ is the smallest exponent for which $rmid fracq^p-1q-1$. Since I look only at $p>2$, $rnmid (q-1)$
$endgroup$
– Collag3n
Mar 24 at 19:28
$begingroup$
For Mersenne numbers $ rmid q-1 $ is impossible because of $ q=2 $, so in this special case we actually have the necessary condition. Since the known Wieferich primes (to base $2$) can be ruled out and another Wieferich prime (to base $2$) must be huge, we can conclude that $2^p-1$ with prime $p$ is always squarefree with a high probability.
$endgroup$
– Peter
Mar 24 at 19:30
add a comment |
$begingroup$
Is $fracq^p-1q-1$ always squarefree with $q,p$ prime and $p>2$ and $(q,p)=(3,5)$ excluded?
This is a follow up of $3^p-2^p$ squarefree?
I know the case $q=2$ (Mersenne) and $q=3$ are still open, but is there a similar/generalised conjecture for the other prime $q$?
number-theory prime-numbers prime-factorization mersenne-numbers
asked Mar 24 at 18:47
Collag3nCollag3n
784211
$endgroup$
Is $fracq^p-1q-1$ always squarefree with $q,p$ prime and $p>2$ and $(q,p)=(3,5)$ excluded?
This is a follow up of $3^p-2^p$ squarefree?
I know the case $q=2$ (Mersenne) and $q=3$ are still open, but is there a similar/generalised conjecture for the other prime $q$?
number-theory prime-numbers prime-factorization mersenne-numbers
number-theory prime-numbers prime-factorization mersenne-numbers
asked Mar 24 at 18:47
Collag3nCollag3n
784211
asked Mar 24 at 18:47
Collag3nCollag3n
784211
asked Mar 24 at 18:47
Collag3nCollag3n
784211
asked Mar 24 at 18:47
Collag3nCollag3n
784211
asked Mar 24 at 18:47
Collag3nCollag3n
784211
784211
$begingroup$
A prime number $r$ with the property $$r^2mid fracq^p-1q-1$$ must be a Wieferich-prime to base $q$, that is $$q^r-1equiv 1mod r^2$$ must hold.
$endgroup$
– Peter
Mar 24 at 19:06
$begingroup$
Yes, necessary, but not sufficient
$endgroup$
– Collag3n
Mar 24 at 19:14
$begingroup$
I am not actually sure whether it is even necessary, but if $r$ does not divide $q-1$, then surely.
$endgroup$
– Peter
Mar 24 at 19:15
$begingroup$
can't we apply the same reasoning as for the mersenne case? If $rmid fracq^p-1q-1$, than $p$ is the smallest exponent for which $rmid fracq^p-1q-1$. Since I look only at $p>2$, $rnmid (q-1)$
$endgroup$
– Collag3n
Mar 24 at 19:28
$begingroup$
For Mersenne numbers $ rmid q-1 $ is impossible because of $ q=2 $, so in this special case we actually have the necessary condition. Since the known Wieferich primes (to base $2$) can be ruled out and another Wieferich prime (to base $2$) must be huge, we can conclude that $2^p-1$ with prime $p$ is always squarefree with a high probability.
$endgroup$
– Peter
Mar 24 at 19:30
add a comment |
$begingroup$
A prime number $r$ with the property $$r^2mid fracq^p-1q-1$$ must be a Wieferich-prime to base $q$, that is $$q^r-1equiv 1mod r^2$$ must hold.
$endgroup$
– Peter
Mar 24 at 19:06
$begingroup$
Yes, necessary, but not sufficient
$endgroup$
– Collag3n
Mar 24 at 19:14
$begingroup$
I am not actually sure whether it is even necessary, but if $r$ does not divide $q-1$, then surely.
$endgroup$
– Peter
Mar 24 at 19:15
$begingroup$
can't we apply the same reasoning as for the mersenne case? If $rmid fracq^p-1q-1$, than $p$ is the smallest exponent for which $rmid fracq^p-1q-1$. Since I look only at $p>2$, $rnmid (q-1)$
$endgroup$
– Collag3n
Mar 24 at 19:28
$begingroup$
For Mersenne numbers $ rmid q-1 $ is impossible because of $ q=2 $, so in this special case we actually have the necessary condition. Since the known Wieferich primes (to base $2$) can be ruled out and another Wieferich prime (to base $2$) must be huge, we can conclude that $2^p-1$ with prime $p$ is always squarefree with a high probability.
$endgroup$
– Peter
Mar 24 at 19:30
$begingroup$
A prime number $r$ with the property $$r^2mid fracq^p-1q-1$$ must be a Wieferich-prime to base $q$, that is $$q^r-1equiv 1mod r^2$$ must hold.
$endgroup$
– Peter
Mar 24 at 19:06
$begingroup$
A prime number $r$ with the property $$r^2mid fracq^p-1q-1$$ must be a Wieferich-prime to base $q$, that is $$q^r-1equiv 1mod r^2$$ must hold.
$endgroup$
– Peter
Mar 24 at 19:06
$begingroup$
Yes, necessary, but not sufficient
$endgroup$
– Collag3n
Mar 24 at 19:14
$begingroup$
Yes, necessary, but not sufficient
$endgroup$
– Collag3n
Mar 24 at 19:14
$begingroup$
I am not actually sure whether it is even necessary, but if $r$ does not divide $q-1$, then surely.
$endgroup$
– Peter
Mar 24 at 19:15
$begingroup$
I am not actually sure whether it is even necessary, but if $r$ does not divide $q-1$, then surely.
$endgroup$
– Peter
Mar 24 at 19:15
$begingroup$
can't we apply the same reasoning as for the mersenne case? If $rmid fracq^p-1q-1$, than $p$ is the smallest exponent for which $rmid fracq^p-1q-1$. Since I look only at $p>2$, $rnmid (q-1)$
$endgroup$
– Collag3n
Mar 24 at 19:28
$begingroup$
can't we apply the same reasoning as for the mersenne case? If $rmid fracq^p-1q-1$, than $p$ is the smallest exponent for which $rmid fracq^p-1q-1$. Since I look only at $p>2$, $rnmid (q-1)$
$endgroup$
– Collag3n
Mar 24 at 19:28
$begingroup$
For Mersenne numbers $ rmid q-1 $ is impossible because of $ q=2 $, so in this special case we actually have the necessary condition. Since the known Wieferich primes (to base $2$) can be ruled out and another Wieferich prime (to base $2$) must be huge, we can conclude that $2^p-1$ with prime $p$ is always squarefree with a high probability.
$endgroup$
– Peter
Mar 24 at 19:30
$begingroup$
For Mersenne numbers $ rmid q-1 $ is impossible because of $ q=2 $, so in this special case we actually have the necessary condition. Since the known Wieferich primes (to base $2$) can be ruled out and another Wieferich prime (to base $2$) must be huge, we can conclude that $2^p-1$ with prime $p$ is always squarefree with a high probability.
$endgroup$
– Peter
Mar 24 at 19:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The following PARI/GP-code finds examples :
? forprime(q=2,100,forprime(p=3,30,if(issquarefree((q^p-1)/(q-1))==0,print([q,p]
))))
[3, 5]
[53, 23]
[53, 29]
[67, 3]
[71, 23]
[79, 3]
?
So, for example $$frac53^23-153-1$$ is not squarfree.
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
$endgroup$
$begingroup$
Arff...indeed. was too concentrated on higher $p$ than higher $q$. I wonder if there are any pattern in the $q$ for which it would still be true
$endgroup$
– Collag3n
Mar 24 at 19:05
add a comment |
$begingroup$
You can construct counterexamples easily enough. For instance, say we want an example divisible by $29^2$. Noting that $varphi(29^2)$ is divisible by $7$ we first find an element of order $7$ $pmod 29^2$, I got $645$ with a little calculating. Then we find a prime $qequiv 645 pmod 29^2$, I think $22511$ is the least. Then we get $$frac 22511^7-122511-1equiv 0 pmod 29^2$$
answered Mar 24 at 19:12
lulululu
43.6k25081
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160871%2ffracqp-1q-1-squarefree%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The following PARI/GP-code finds examples :
? forprime(q=2,100,forprime(p=3,30,if(issquarefree((q^p-1)/(q-1))==0,print([q,p]
))))
[3, 5]
[53, 23]
[53, 29]
[67, 3]
[71, 23]
[79, 3]
?
So, for example $$frac53^23-153-1$$ is not squarfree.
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
$endgroup$
$begingroup$
Arff...indeed. was too concentrated on higher $p$ than higher $q$. I wonder if there are any pattern in the $q$ for which it would still be true
$endgroup$
– Collag3n
Mar 24 at 19:05
add a comment |
$begingroup$
The following PARI/GP-code finds examples :
? forprime(q=2,100,forprime(p=3,30,if(issquarefree((q^p-1)/(q-1))==0,print([q,p]
))))
[3, 5]
[53, 23]
[53, 29]
[67, 3]
[71, 23]
[79, 3]
?
So, for example $$frac53^23-153-1$$ is not squarfree.
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
$endgroup$
$begingroup$
Arff...indeed. was too concentrated on higher $p$ than higher $q$. I wonder if there are any pattern in the $q$ for which it would still be true
$endgroup$
– Collag3n
Mar 24 at 19:05
add a comment |
$begingroup$
The following PARI/GP-code finds examples :
? forprime(q=2,100,forprime(p=3,30,if(issquarefree((q^p-1)/(q-1))==0,print([q,p]
))))
[3, 5]
[53, 23]
[53, 29]
[67, 3]
[71, 23]
[79, 3]
?
So, for example $$frac53^23-153-1$$ is not squarfree.
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
$endgroup$
The following PARI/GP-code finds examples :
? forprime(q=2,100,forprime(p=3,30,if(issquarefree((q^p-1)/(q-1))==0,print([q,p]
))))
[3, 5]
[53, 23]
[53, 29]
[67, 3]
[71, 23]
[79, 3]
?
So, for example $$frac53^23-153-1$$ is not squarfree.
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
answered Mar 24 at 18:57
PeterPeter
49.1k1240138
49.1k1240138
$begingroup$
Arff...indeed. was too concentrated on higher $p$ than higher $q$. I wonder if there are any pattern in the $q$ for which it would still be true
$endgroup$
– Collag3n
Mar 24 at 19:05
add a comment |
$begingroup$
Arff...indeed. was too concentrated on higher $p$ than higher $q$. I wonder if there are any pattern in the $q$ for which it would still be true
$endgroup$
– Collag3n
Mar 24 at 19:05
$begingroup$
Arff...indeed. was too concentrated on higher $p$ than higher $q$. I wonder if there are any pattern in the $q$ for which it would still be true
$endgroup$
– Collag3n
Mar 24 at 19:05
$begingroup$
Arff...indeed. was too concentrated on higher $p$ than higher $q$. I wonder if there are any pattern in the $q$ for which it would still be true
$endgroup$
– Collag3n
Mar 24 at 19:05
add a comment |
$begingroup$
You can construct counterexamples easily enough. For instance, say we want an example divisible by $29^2$. Noting that $varphi(29^2)$ is divisible by $7$ we first find an element of order $7$ $pmod 29^2$, I got $645$ with a little calculating. Then we find a prime $qequiv 645 pmod 29^2$, I think $22511$ is the least. Then we get $$frac 22511^7-122511-1equiv 0 pmod 29^2$$
answered Mar 24 at 19:12
lulululu
43.6k25081
$endgroup$
add a comment |
$begingroup$
You can construct counterexamples easily enough. For instance, say we want an example divisible by $29^2$. Noting that $varphi(29^2)$ is divisible by $7$ we first find an element of order $7$ $pmod 29^2$, I got $645$ with a little calculating. Then we find a prime $qequiv 645 pmod 29^2$, I think $22511$ is the least. Then we get $$frac 22511^7-122511-1equiv 0 pmod 29^2$$
answered Mar 24 at 19:12
lulululu
43.6k25081
$endgroup$
add a comment |
$begingroup$
You can construct counterexamples easily enough. For instance, say we want an example divisible by $29^2$. Noting that $varphi(29^2)$ is divisible by $7$ we first find an element of order $7$ $pmod 29^2$, I got $645$ with a little calculating. Then we find a prime $qequiv 645 pmod 29^2$, I think $22511$ is the least. Then we get $$frac 22511^7-122511-1equiv 0 pmod 29^2$$
answered Mar 24 at 19:12
lulululu
43.6k25081
$endgroup$
You can construct counterexamples easily enough. For instance, say we want an example divisible by $29^2$. Noting that $varphi(29^2)$ is divisible by $7$ we first find an element of order $7$ $pmod 29^2$, I got $645$ with a little calculating. Then we find a prime $qequiv 645 pmod 29^2$, I think $22511$ is the least. Then we get $$frac 22511^7-122511-1equiv 0 pmod 29^2$$
answered Mar 24 at 19:12
lulululu
43.6k25081
answered Mar 24 at 19:12
lulululu
43.6k25081
answered Mar 24 at 19:12
lulululu
43.6k25081
answered Mar 24 at 19:12
lulululu
43.6k25081
43.6k25081
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160871%2ffracqp-1q-1-squarefree%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
A prime number $r$ with the property $$r^2mid fracq^p-1q-1$$ must be a Wieferich-prime to base $q$, that is $$q^r-1equiv 1mod r^2$$ must hold.
$endgroup$
– Peter
Mar 24 at 19:06
$begingroup$
Yes, necessary, but not sufficient
$endgroup$
– Collag3n
Mar 24 at 19:14
$begingroup$
I am not actually sure whether it is even necessary, but if $r$ does not divide $q-1$, then surely.
$endgroup$
– Peter
Mar 24 at 19:15
$begingroup$
can't we apply the same reasoning as for the mersenne case? If $rmid fracq^p-1q-1$, than $p$ is the smallest exponent for which $rmid fracq^p-1q-1$. Since I look only at $p>2$, $rnmid (q-1)$
$endgroup$
– Collag3n
Mar 24 at 19:28
$begingroup$
For Mersenne numbers $ rmid q-1 $ is impossible because of $ q=2 $, so in this special case we actually have the necessary condition. Since the known Wieferich primes (to base $2$) can be ruled out and another Wieferich prime (to base $2$) must be huge, we can conclude that $2^p-1$ with prime $p$ is always squarefree with a high probability.
$endgroup$
– Peter
Mar 24 at 19:30