Dgdrxrt

Universal covering space of the real projective line?

I can't produce songs

Mounting TV on a weird wall that has some material between the drywall and stud

Why is a lens darker than other ones when applying the same settings?

NERDTreeMenu Remapping

After Sam didn't return home in the end, were he and Al still friends?

What initially awakened the Balrog?

What is the chair depicted in Cesare Maccari's 1889 painting "Cicerone denuncia Catilina"?

Show current row "win streak"

How does light 'choose' between wave and particle behaviour?

Found this skink in my tomato plant bucket. Is he trapped? Or could he leave if he wanted?

Tannaka duality for semisimple groups

One-one communication

Nose gear failure in single prop aircraft: belly landing or nose-gear up landing?

What is the difference between CTSS and ITS?

Random body shuffle every night—can we still function?

The Nth Gryphon Number

retrieve food groups from food item list

In musical terms, what properties are varied by the human voice to produce different words / syllables?

How can a team of shapeshifters communicate?

Printing attributes of selection in ArcPy?

Test print coming out spongy

Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?

Flight departed from the gate 5 min before scheduled departure time. Refund options

Is it possible for $a^2bc+2, ab^2c+2, abc^2+2$ to all be perfect squares? [on hold]

1

2

$begingroup$

This question in my opinion is challenging. I tried to put $gcd(a, b, c)=d$ and $a=dx, b=dy, c=dz$ but failed. Any help would be appreciated.

elementary-number-theory

share|cite|improve this question

edited Apr 2 at 19:22

Rick Almeida

1989

asked Apr 2 at 8:35

user587054user587054

59111

$endgroup$

put on hold as off-topic by user21820, Saad, Brahadeesh, Cesareo, Paul Frost Apr 17 at 15:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Brahadeesh, Cesareo, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Brute force says: Not if $a,b,c leq 3000$.
  $endgroup$
  – PierreCarre
  Apr 2 at 9:41
  

  
  
  
  

add a comment | 

1

2

$begingroup$

This question in my opinion is challenging. I tried to put $gcd(a, b, c)=d$ and $a=dx, b=dy, c=dz$ but failed. Any help would be appreciated.

elementary-number-theory

share|cite|improve this question

edited Apr 2 at 19:22

Rick Almeida

1989

asked Apr 2 at 8:35

user587054user587054

59111

$endgroup$

put on hold as off-topic by user21820, Saad, Brahadeesh, Cesareo, Paul Frost Apr 17 at 15:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Brahadeesh, Cesareo, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Brute force says: Not if $a,b,c leq 3000$.
  $endgroup$
  – PierreCarre
  Apr 2 at 9:41
  

  
  
  
  

add a comment | 

$begingroup$

This question in my opinion is challenging. I tried to put $gcd(a, b, c)=d$ and $a=dx, b=dy, c=dz$ but failed. Any help would be appreciated.

elementary-number-theory

share|cite|improve this question

edited Apr 2 at 19:22

Rick Almeida

1989

asked Apr 2 at 8:35

user587054user587054

59111

$endgroup$

share|cite|improve this question

edited Apr 2 at 19:22

Rick Almeida

1989

asked Apr 2 at 8:35

user587054user587054

59111

share|cite|improve this question

edited Apr 2 at 19:22

Rick Almeida

1989

asked Apr 2 at 8:35

user587054user587054

59111

edited Apr 2 at 19:22

Rick Almeida

1989

edited Apr 2 at 19:22

Rick Almeida

1989

asked Apr 2 at 8:35

user587054user587054

59111

asked Apr 2 at 8:35

user587054user587054

59111

1 Answer
1

active

oldest

votes

3

$begingroup$

First, observe that if any of $a, b, c$ is even, then one of the terms cannot be a square. WLOG take $a=2n$, then $4n^2bcequiv -2pmod4$ has no solution. So all of them must be odd.

Observe that only 1 is a quadratic residue mod 8.

$a^2bc+2equiv 1pmod8Longrightarrow a^2bcequiv (1)bc=bcequiv 7pmod8$

So we have the system

$$
abequiv 7pmod8quad acequiv 7pmod8quad bcequiv 7pmod8
$$

This implies $bequiv cpmod8$, hence from the last equation $b^2equiv 7pmod8$, but 7 is not a quadratic residue mode 8.

So there is no solution.

share|cite|improve this answer

answered Apr 2 at 10:42

Rick AlmeidaRick Almeida

1989

$endgroup$

add a comment | 

3

$begingroup$

First, observe that if any of $a, b, c$ is even, then one of the terms cannot be a square. WLOG take $a=2n$, then $4n^2bcequiv -2pmod4$ has no solution. So all of them must be odd.

Observe that only 1 is a quadratic residue mod 8.

$a^2bc+2equiv 1pmod8Longrightarrow a^2bcequiv (1)bc=bcequiv 7pmod8$

So we have the system

$$
abequiv 7pmod8quad acequiv 7pmod8quad bcequiv 7pmod8
$$

This implies $bequiv cpmod8$, hence from the last equation $b^2equiv 7pmod8$, but 7 is not a quadratic residue mode 8.

So there is no solution.

share|cite|improve this answer

answered Apr 2 at 10:42

Rick AlmeidaRick Almeida

1989

$endgroup$

add a comment | 

3

$begingroup$

First, observe that if any of $a, b, c$ is even, then one of the terms cannot be a square. WLOG take $a=2n$, then $4n^2bcequiv -2pmod4$ has no solution. So all of them must be odd.

Observe that only 1 is a quadratic residue mod 8.

$a^2bc+2equiv 1pmod8Longrightarrow a^2bcequiv (1)bc=bcequiv 7pmod8$

So we have the system

$$
abequiv 7pmod8quad acequiv 7pmod8quad bcequiv 7pmod8
$$

This implies $bequiv cpmod8$, hence from the last equation $b^2equiv 7pmod8$, but 7 is not a quadratic residue mode 8.

So there is no solution.

share|cite|improve this answer

answered Apr 2 at 10:42

Rick AlmeidaRick Almeida

1989

$endgroup$

add a comment | 

$begingroup$

First, observe that if any of $a, b, c$ is even, then one of the terms cannot be a square. WLOG take $a=2n$, then $4n^2bcequiv -2pmod4$ has no solution. So all of them must be odd.

Observe that only 1 is a quadratic residue mod 8.

$a^2bc+2equiv 1pmod8Longrightarrow a^2bcequiv (1)bc=bcequiv 7pmod8$

So we have the system

$$
abequiv 7pmod8quad acequiv 7pmod8quad bcequiv 7pmod8
$$

This implies $bequiv cpmod8$, hence from the last equation $b^2equiv 7pmod8$, but 7 is not a quadratic residue mode 8.

So there is no solution.

share|cite|improve this answer

answered Apr 2 at 10:42

Rick AlmeidaRick Almeida

1989

$endgroup$

share|cite|improve this answer

answered Apr 2 at 10:42

Rick AlmeidaRick Almeida

1989

answered Apr 2 at 10:42

Rick AlmeidaRick Almeida

1989

answered Apr 2 at 10:42

Rick AlmeidaRick Almeida

1989