For prime $p$ do we have $p^3+p^2+p+1=n^2$ infinitely often? [duplicate]

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This question already has an answer here:

• 
  Find p is the prime number which $dfracp+12$ and $dfracp^2+12$ both are square number.
  
  1 answer
  
  

This is a question to ponder about the occurrence of prime $p$ giving
$p^3 + p^2 + p +1=n^2$ as is true with $7$ giving $400=20^2$. Do you
think this will ever happen again?

prime-numbers

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edited Jul 17 at 18:47

Marcus M

8,1731847

asked Jul 17 at 18:45

J. M. Bergot

35118

marked as duplicate by greedoid, Xander Henderson, Parcly Taxel, Trần Thúc Minh Trí, Claude Leibovici Jul 18 at 9:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Factor, $(p+1)(p^2+1)$ should be a square. For $p > 2$ the $gcd$ of the factors is $2$, so $p+1 = 2k^2$ and $p^2+1 = 2m^2$. From $p^2 - 2m^2 = - 1$, you get a quick recurrence for candidates. Which of those are primes such that $(p+1)/2$ is a square …?
  – Daniel Fischer♦
  Jul 17 at 18:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  posed first by Fermat in 1657. See volume I of Dickson's History, pages 54-58. I think $7$ turns out to be the only prime with $sigma(x^3) = y^2$
  – Will Jagy
  Jul 17 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does anyone know up to what p was this examined? One also has 50 as twice 5 squared minus 1 is seven squared. Does this ever happen again?
  – J. M. Bergot
  Jul 17 at 19:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One also has 18^2 + 18 + 1 = 343=7^3. Can another composite like 18 produce a prime cubed?
  – J. M. Bergot
  Jul 17 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  books.google.com/…
  – Will Jagy
  Jul 17 at 19:09
  

  
  
  
  

 | 
show 3 more comments

up vote
6
down vote

favorite

2

This question already has an answer here:

• 
  Find p is the prime number which $dfracp+12$ and $dfracp^2+12$ both are square number.
  
  1 answer
  
  

This is a question to ponder about the occurrence of prime $p$ giving
$p^3 + p^2 + p +1=n^2$ as is true with $7$ giving $400=20^2$. Do you
think this will ever happen again?

prime-numbers

share|cite|improve this question

edited Jul 17 at 18:47

Marcus M

8,1731847

asked Jul 17 at 18:45

J. M. Bergot

35118

marked as duplicate by greedoid, Xander Henderson, Parcly Taxel, Trần Thúc Minh Trí, Claude Leibovici Jul 18 at 9:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Factor, $(p+1)(p^2+1)$ should be a square. For $p > 2$ the $gcd$ of the factors is $2$, so $p+1 = 2k^2$ and $p^2+1 = 2m^2$. From $p^2 - 2m^2 = - 1$, you get a quick recurrence for candidates. Which of those are primes such that $(p+1)/2$ is a square …?
  – Daniel Fischer♦
  Jul 17 at 18:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  posed first by Fermat in 1657. See volume I of Dickson's History, pages 54-58. I think $7$ turns out to be the only prime with $sigma(x^3) = y^2$
  – Will Jagy
  Jul 17 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does anyone know up to what p was this examined? One also has 50 as twice 5 squared minus 1 is seven squared. Does this ever happen again?
  – J. M. Bergot
  Jul 17 at 19:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One also has 18^2 + 18 + 1 = 343=7^3. Can another composite like 18 produce a prime cubed?
  – J. M. Bergot
  Jul 17 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  books.google.com/…
  – Will Jagy
  Jul 17 at 19:09
  

  
  
  
  

 | 
show 3 more comments

up vote
6
down vote

favorite

2

up vote
6
down vote

favorite

2

2

This question already has an answer here:

• 
  Find p is the prime number which $dfracp+12$ and $dfracp^2+12$ both are square number.
  
  1 answer
  
  

This is a question to ponder about the occurrence of prime $p$ giving
$p^3 + p^2 + p +1=n^2$ as is true with $7$ giving $400=20^2$. Do you
think this will ever happen again?

prime-numbers

share|cite|improve this question

edited Jul 17 at 18:47

Marcus M

8,1731847

asked Jul 17 at 18:45

J. M. Bergot

35118

This question already has an answer here:

• 
  Find p is the prime number which $dfracp+12$ and $dfracp^2+12$ both are square number.
  
  1 answer
  
  

This is a question to ponder about the occurrence of prime $p$ giving
$p^3 + p^2 + p +1=n^2$ as is true with $7$ giving $400=20^2$. Do you
think this will ever happen again?

This question already has an answer here:

• 
  Find p is the prime number which $dfracp+12$ and $dfracp^2+12$ both are square number.
  
  1 answer
  
  

prime-numbers

share|cite|improve this question

edited Jul 17 at 18:47

Marcus M

8,1731847

asked Jul 17 at 18:45

J. M. Bergot

35118

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 18:47

Marcus M

8,1731847

edited Jul 17 at 18:47

Marcus M

8,1731847

edited Jul 17 at 18:47

Marcus M

8,1731847

8,1731847

asked Jul 17 at 18:45

J. M. Bergot

35118

asked Jul 17 at 18:45

J. M. Bergot

35118

asked Jul 17 at 18:45

J. M. Bergot

35118

35118

marked as duplicate by greedoid, Xander Henderson, Parcly Taxel, Trần Thúc Minh Trí, Claude Leibovici Jul 18 at 9:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by greedoid, Xander Henderson, Parcly Taxel, Trần Thúc Minh Trí, Claude Leibovici Jul 18 at 9:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Factor, $(p+1)(p^2+1)$ should be a square. For $p > 2$ the $gcd$ of the factors is $2$, so $p+1 = 2k^2$ and $p^2+1 = 2m^2$. From $p^2 - 2m^2 = - 1$, you get a quick recurrence for candidates. Which of those are primes such that $(p+1)/2$ is a square …?
  – Daniel Fischer♦
  Jul 17 at 18:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  posed first by Fermat in 1657. See volume I of Dickson's History, pages 54-58. I think $7$ turns out to be the only prime with $sigma(x^3) = y^2$
  – Will Jagy
  Jul 17 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does anyone know up to what p was this examined? One also has 50 as twice 5 squared minus 1 is seven squared. Does this ever happen again?
  – J. M. Bergot
  Jul 17 at 19:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One also has 18^2 + 18 + 1 = 343=7^3. Can another composite like 18 produce a prime cubed?
  – J. M. Bergot
  Jul 17 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  books.google.com/…
  – Will Jagy
  Jul 17 at 19:09
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Factor, $(p+1)(p^2+1)$ should be a square. For $p > 2$ the $gcd$ of the factors is $2$, so $p+1 = 2k^2$ and $p^2+1 = 2m^2$. From $p^2 - 2m^2 = - 1$, you get a quick recurrence for candidates. Which of those are primes such that $(p+1)/2$ is a square …?
  – Daniel Fischer♦
  Jul 17 at 18:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  posed first by Fermat in 1657. See volume I of Dickson's History, pages 54-58. I think $7$ turns out to be the only prime with $sigma(x^3) = y^2$
  – Will Jagy
  Jul 17 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does anyone know up to what p was this examined? One also has 50 as twice 5 squared minus 1 is seven squared. Does this ever happen again?
  – J. M. Bergot
  Jul 17 at 19:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One also has 18^2 + 18 + 1 = 343=7^3. Can another composite like 18 produce a prime cubed?
  – J. M. Bergot
  Jul 17 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  books.google.com/…
  – Will Jagy
  Jul 17 at 19:09
  

  
  
  
  

1

1

Factor, $(p+1)(p^2+1)$ should be a square. For $p > 2$ the $gcd$ of the factors is $2$, so $p+1 = 2k^2$ and $p^2+1 = 2m^2$. From $p^2 - 2m^2 = - 1$, you get a quick recurrence for candidates. Which of those are primes such that $(p+1)/2$ is a square …?
– Daniel Fischer♦
Jul 17 at 18:58

Factor, $(p+1)(p^2+1)$ should be a square. For $p > 2$ the $gcd$ of the factors is $2$, so $p+1 = 2k^2$ and $p^2+1 = 2m^2$. From $p^2 - 2m^2 = - 1$, you get a quick recurrence for candidates. Which of those are primes such that $(p+1)/2$ is a square …?
– Daniel Fischer♦
Jul 17 at 18:58

1

1

posed first by Fermat in 1657. See volume I of Dickson's History, pages 54-58. I think $7$ turns out to be the only prime with $sigma(x^3) = y^2$
– Will Jagy
Jul 17 at 19:02

posed first by Fermat in 1657. See volume I of Dickson's History, pages 54-58. I think $7$ turns out to be the only prime with $sigma(x^3) = y^2$
– Will Jagy
Jul 17 at 19:02

Does anyone know up to what p was this examined? One also has 50 as twice 5 squared minus 1 is seven squared. Does this ever happen again?
– J. M. Bergot
Jul 17 at 19:07

Does anyone know up to what p was this examined? One also has 50 as twice 5 squared minus 1 is seven squared. Does this ever happen again?
– J. M. Bergot
Jul 17 at 19:07

One also has 18^2 + 18 + 1 = 343=7^3. Can another composite like 18 produce a prime cubed?
– J. M. Bergot
Jul 17 at 19:09

One also has 18^2 + 18 + 1 = 343=7^3. Can another composite like 18 produce a prime cubed?
– J. M. Bergot
Jul 17 at 19:09

books.google.com/…
– Will Jagy
Jul 17 at 19:09

books.google.com/…
– Will Jagy
Jul 17 at 19:09

 | 
show 3 more comments

1 Answer
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There are no more positive prime solution other than $p = 7$.

First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.

Let's say $p$ is an odd prime such that

$$p^3 + p^2 + p + 1 = n^2quadiffquad(p^2+1)(p+1) = n^2$$

Since $gcd(p+1,p^2+1) = gcd(p+1,2) = 2$, we find for some $k, m > 0$,

$$begincases
p^2 + 1 &= 2m^2\
p + 1 &= 2k^2\
endcases quadiffquad begincases m = sqrtfracp^2+12\k = sqrtfracp+12endcases$$

Subtracting the two equations on the left, we obtain

$$p(p-1) = 2(m+k)(m-k)$$

Since $0 < m - k < m = sqrtfracp^2+12 < sqrtfracp^2+p^22 = p$, both $2$ and $m-k$ are coprime to $p$.

Since $p$ is a prime, we obtain $p | (m+k)$.

Since $m+k < 2m < 2p$, this forces $p = m + k$.
As a result,

$$beginalign
& 2(p^4 + m^4 + k^4) - (p^2 + m^2 + k^2)^2 = (p+m+k)(p+m-k)(p-m+k)(p-m-k) = 0\
iff &
p^4 - 6p^3 - 7p^2 = (p-7)(p+1)p^2 = 0\
implies & p = 7endalign
$$

In short, the equation

$$p^3 + p^2 + p + 1 = n^2$$
has one and only one positive prime solution. Namely, $p = 7$.

share|cite|improve this answer

answered Jul 17 at 19:26

achille hui

91k5127246

add a comment | 

1 Answer
1

active

oldest

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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
10
down vote

There are no more positive prime solution other than $p = 7$.

First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.

Let's say $p$ is an odd prime such that

$$p^3 + p^2 + p + 1 = n^2quadiffquad(p^2+1)(p+1) = n^2$$

Since $gcd(p+1,p^2+1) = gcd(p+1,2) = 2$, we find for some $k, m > 0$,

$$begincases
p^2 + 1 &= 2m^2\
p + 1 &= 2k^2\
endcases quadiffquad begincases m = sqrtfracp^2+12\k = sqrtfracp+12endcases$$

Subtracting the two equations on the left, we obtain

$$p(p-1) = 2(m+k)(m-k)$$

Since $0 < m - k < m = sqrtfracp^2+12 < sqrtfracp^2+p^22 = p$, both $2$ and $m-k$ are coprime to $p$.

Since $p$ is a prime, we obtain $p | (m+k)$.

Since $m+k < 2m < 2p$, this forces $p = m + k$.
As a result,

$$beginalign
& 2(p^4 + m^4 + k^4) - (p^2 + m^2 + k^2)^2 = (p+m+k)(p+m-k)(p-m+k)(p-m-k) = 0\
iff &
p^4 - 6p^3 - 7p^2 = (p-7)(p+1)p^2 = 0\
implies & p = 7endalign
$$

In short, the equation

$$p^3 + p^2 + p + 1 = n^2$$
has one and only one positive prime solution. Namely, $p = 7$.

share|cite|improve this answer

answered Jul 17 at 19:26

achille hui

91k5127246

add a comment | 

up vote
10
down vote

There are no more positive prime solution other than $p = 7$.

First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.

Let's say $p$ is an odd prime such that

$$p^3 + p^2 + p + 1 = n^2quadiffquad(p^2+1)(p+1) = n^2$$

Since $gcd(p+1,p^2+1) = gcd(p+1,2) = 2$, we find for some $k, m > 0$,

$$begincases
p^2 + 1 &= 2m^2\
p + 1 &= 2k^2\
endcases quadiffquad begincases m = sqrtfracp^2+12\k = sqrtfracp+12endcases$$

Subtracting the two equations on the left, we obtain

$$p(p-1) = 2(m+k)(m-k)$$

Since $0 < m - k < m = sqrtfracp^2+12 < sqrtfracp^2+p^22 = p$, both $2$ and $m-k$ are coprime to $p$.

Since $p$ is a prime, we obtain $p | (m+k)$.

Since $m+k < 2m < 2p$, this forces $p = m + k$.
As a result,

$$beginalign
& 2(p^4 + m^4 + k^4) - (p^2 + m^2 + k^2)^2 = (p+m+k)(p+m-k)(p-m+k)(p-m-k) = 0\
iff &
p^4 - 6p^3 - 7p^2 = (p-7)(p+1)p^2 = 0\
implies & p = 7endalign
$$

In short, the equation

$$p^3 + p^2 + p + 1 = n^2$$
has one and only one positive prime solution. Namely, $p = 7$.

share|cite|improve this answer

answered Jul 17 at 19:26

achille hui

91k5127246

add a comment | 

up vote
10
down vote

up vote
10
down vote

There are no more positive prime solution other than $p = 7$.

First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.

Let's say $p$ is an odd prime such that

$$p^3 + p^2 + p + 1 = n^2quadiffquad(p^2+1)(p+1) = n^2$$

Since $gcd(p+1,p^2+1) = gcd(p+1,2) = 2$, we find for some $k, m > 0$,

$$begincases
p^2 + 1 &= 2m^2\
p + 1 &= 2k^2\
endcases quadiffquad begincases m = sqrtfracp^2+12\k = sqrtfracp+12endcases$$

Subtracting the two equations on the left, we obtain

$$p(p-1) = 2(m+k)(m-k)$$

Since $0 < m - k < m = sqrtfracp^2+12 < sqrtfracp^2+p^22 = p$, both $2$ and $m-k$ are coprime to $p$.

Since $p$ is a prime, we obtain $p | (m+k)$.

Since $m+k < 2m < 2p$, this forces $p = m + k$.
As a result,

$$beginalign
& 2(p^4 + m^4 + k^4) - (p^2 + m^2 + k^2)^2 = (p+m+k)(p+m-k)(p-m+k)(p-m-k) = 0\
iff &
p^4 - 6p^3 - 7p^2 = (p-7)(p+1)p^2 = 0\
implies & p = 7endalign
$$

In short, the equation

$$p^3 + p^2 + p + 1 = n^2$$
has one and only one positive prime solution. Namely, $p = 7$.

share|cite|improve this answer

answered Jul 17 at 19:26

achille hui

91k5127246

There are no more positive prime solution other than $p = 7$.

First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.

Let's say $p$ is an odd prime such that

$$p^3 + p^2 + p + 1 = n^2quadiffquad(p^2+1)(p+1) = n^2$$

Since $gcd(p+1,p^2+1) = gcd(p+1,2) = 2$, we find for some $k, m > 0$,

$$begincases
p^2 + 1 &= 2m^2\
p + 1 &= 2k^2\
endcases quadiffquad begincases m = sqrtfracp^2+12\k = sqrtfracp+12endcases$$

Subtracting the two equations on the left, we obtain

$$p(p-1) = 2(m+k)(m-k)$$

Since $0 < m - k < m = sqrtfracp^2+12 < sqrtfracp^2+p^22 = p$, both $2$ and $m-k$ are coprime to $p$.

Since $p$ is a prime, we obtain $p | (m+k)$.

Since $m+k < 2m < 2p$, this forces $p = m + k$.
As a result,

$$beginalign
& 2(p^4 + m^4 + k^4) - (p^2 + m^2 + k^2)^2 = (p+m+k)(p+m-k)(p-m+k)(p-m-k) = 0\
iff &
p^4 - 6p^3 - 7p^2 = (p-7)(p+1)p^2 = 0\
implies & p = 7endalign
$$

In short, the equation

$$p^3 + p^2 + p + 1 = n^2$$
has one and only one positive prime solution. Namely, $p = 7$.

share|cite|improve this answer

answered Jul 17 at 19:26

achille hui

91k5127246

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 19:26

achille hui

91k5127246

answered Jul 17 at 19:26

achille hui

91k5127246

answered Jul 17 at 19:26

achille hui

91k5127246

91k5127246

add a comment | 

add a comment |