Is it possible for $a^2bc+2, ab^2c+2, abc^2+2$ to all be perfect squares? [on hold] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prime Numbers And Perfect Squaresperfect squares possible?Sum of two numbers is equal to hcf +lcmShow that if $x,y,z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square iff $xy +1, yz +1, zx+1$ are all perfect squares.Find all possible sum of digits of perfect squaresThere are two irreducible rational numbers with denominators 600 and 700. Find the minimal possible value of the denominator of their sum.An interesting number theory question for math contest training $x^2+x=2y^3$ (finding integer solutions)Prove that for all positive integers $n>1, (n^2+1)-n$ is not a perfect square.Prove that for all integers $r, s$ and $t$, that $gcd(gcd(r, s), t) = gcd(r, gcd(s, t))$.Prove that there are no positive integer solutions for $a,b,c,d$ if $a^2-b=c^2, b^2-a=d^2$
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Is it possible for $a^2bc+2, ab^2c+2, abc^2+2$ to all be perfect squares? [on hold]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prime Numbers And Perfect Squaresperfect squares possible?Sum of two numbers is equal to hcf +lcmShow that if $x,y,z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square iff $xy +1, yz +1, zx+1$ are all perfect squares.Find all possible sum of digits of perfect squaresThere are two irreducible rational numbers with denominators 600 and 700. Find the minimal possible value of the denominator of their sum.An interesting number theory question for math contest training $x^2+x=2y^3$ (finding integer solutions)Prove that for all positive integers $n>1, (n^2+1)-n$ is not a perfect square.Prove that for all integers $r, s$ and $t$, that $gcd(gcd(r, s), t) = gcd(r, gcd(s, t))$.Prove that there are no positive integer solutions for $a,b,c,d$ if $a^2-b=c^2, b^2-a=d^2$
$begingroup$
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
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
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
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
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
edited Apr 2 at 19:22
Rick Almeida
1989
asked Apr 2 at 8:35
user587054user587054
59111
$endgroup$
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
elementary-number-theory
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
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
edited Apr 2 at 19:22
Rick Almeida
1989
1989
asked Apr 2 at 8:35
user587054user587054
59111
asked Apr 2 at 8:35
user587054user587054
59111
asked Apr 2 at 8:35
user587054user587054
59111
59111
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
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
1
$begingroup$
Brute force says: Not if $a,b,c leq 3000$.
$endgroup$
– PierreCarre
Apr 2 at 9:41
add a comment |
1
$begingroup$
Brute force says: Not if $a,b,c leq 3000$.
$endgroup$
– PierreCarre
Apr 2 at 9:41
1
1
$begingroup$
Brute force says: Not if $a,b,c leq 3000$.
$endgroup$
– PierreCarre
Apr 2 at 9:41
$begingroup$
Brute force says: Not if $a,b,c leq 3000$.
$endgroup$
– PierreCarre
Apr 2 at 9:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Apr 2 at 10:42
Rick AlmeidaRick Almeida
1989
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered Apr 2 at 10:42
Rick AlmeidaRick Almeida
1989
$endgroup$
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.
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
answered Apr 2 at 10:42
Rick AlmeidaRick Almeida
1989
1989
add a comment |
add a comment |
1
$begingroup$
Brute force says: Not if $a,b,c leq 3000$.
$endgroup$
– PierreCarre
Apr 2 at 9:41