Prove the equation $left(2x^2+1right)left(2y^2+1right)=4z^2+1$ has no solution in the positive integers
$begingroup$
Prove the equation
$$left(2x^2+1right)left(2y^2+1right)=4z^2+1$$
has no solution in the positive integers
My work:
1) I have the usually problem
$$left(nx^2+1right)left(my^2+1right)=(m+n)z^2+1$$
in the positive integers. Initially I use case $gcd(m,n)=1$
2) Let $m=n=2$. It is this case. I need to prove that $(x^2-y^2)^2+(x^2+y^2) $ is not perfect square for any $x,y$
number-theory elementary-number-theory diophantine-equations integers
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
$endgroup$
|
show 3 more comments
$begingroup$
Prove the equation
$$left(2x^2+1right)left(2y^2+1right)=4z^2+1$$
has no solution in the positive integers
My work:
1) I have the usually problem
$$left(nx^2+1right)left(my^2+1right)=(m+n)z^2+1$$
in the positive integers. Initially I use case $gcd(m,n)=1$
2) Let $m=n=2$. It is this case. I need to prove that $(x^2-y^2)^2+(x^2+y^2) $ is not perfect square for any $x,y$
number-theory elementary-number-theory diophantine-equations integers
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
$endgroup$
2
$begingroup$
Seems to be hard. I tried infinite descent, but I failed. Also considering the prime factors of $2x^2+1$,$2y^2+1$ and $4z^2+1$ led to nowhere. I am curious whether someone can solve this ...
$endgroup$
– Peter
Jul 10 '18 at 17:28
$begingroup$
If it's any help, it's easy to prove the impossibility of the special case $x = y$.
$endgroup$
– Connor Harris
Jul 10 '18 at 17:42
1
$begingroup$
Could you clarify the relevance, in (2), of the expression $(x^2 - y^2)^2 + (x^2 + y^2)$, which equals $x^4 - 2x^2y^2 + y^4 + x^2 + y^2$? With its 4th powers, it seems unrelated to your initial problem statement.
$endgroup$
– Adam Bailey
Jul 14 '18 at 12:30
1
$begingroup$
In the case of $(2)$ (equation in the title) it can be shown that $x,y,z$ must be divisible by $3$, hence there is no solution in positive integers when $$36m^2n^2+2m^2+2n^2$$ cannot be a positive perfect square. But this apparently does not help much.
$endgroup$
– Peter
Jul 15 '18 at 13:42
3
$begingroup$
Found similar question: Math Overflow Link
$endgroup$
– cvogt8
Jul 26 '18 at 7:59
|
show 3 more comments
$begingroup$
Prove the equation
$$left(2x^2+1right)left(2y^2+1right)=4z^2+1$$
has no solution in the positive integers
My work:
1) I have the usually problem
$$left(nx^2+1right)left(my^2+1right)=(m+n)z^2+1$$
in the positive integers. Initially I use case $gcd(m,n)=1$
2) Let $m=n=2$. It is this case. I need to prove that $(x^2-y^2)^2+(x^2+y^2) $ is not perfect square for any $x,y$
number-theory elementary-number-theory diophantine-equations integers
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
$endgroup$
Prove the equation
$$left(2x^2+1right)left(2y^2+1right)=4z^2+1$$
has no solution in the positive integers
My work:
1) I have the usually problem
$$left(nx^2+1right)left(my^2+1right)=(m+n)z^2+1$$
in the positive integers. Initially I use case $gcd(m,n)=1$
2) Let $m=n=2$. It is this case. I need to prove that $(x^2-y^2)^2+(x^2+y^2) $ is not perfect square for any $x,y$
number-theory elementary-number-theory diophantine-equations integers
number-theory elementary-number-theory diophantine-equations integers
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
asked Jul 10 '18 at 11:53
Roman83Roman83
14.4k31956
14.4k31956
2
$begingroup$
Seems to be hard. I tried infinite descent, but I failed. Also considering the prime factors of $2x^2+1$,$2y^2+1$ and $4z^2+1$ led to nowhere. I am curious whether someone can solve this ...
$endgroup$
– Peter
Jul 10 '18 at 17:28
$begingroup$
If it's any help, it's easy to prove the impossibility of the special case $x = y$.
$endgroup$
– Connor Harris
Jul 10 '18 at 17:42
1
$begingroup$
Could you clarify the relevance, in (2), of the expression $(x^2 - y^2)^2 + (x^2 + y^2)$, which equals $x^4 - 2x^2y^2 + y^4 + x^2 + y^2$? With its 4th powers, it seems unrelated to your initial problem statement.
$endgroup$
– Adam Bailey
Jul 14 '18 at 12:30
1
$begingroup$
In the case of $(2)$ (equation in the title) it can be shown that $x,y,z$ must be divisible by $3$, hence there is no solution in positive integers when $$36m^2n^2+2m^2+2n^2$$ cannot be a positive perfect square. But this apparently does not help much.
$endgroup$
– Peter
Jul 15 '18 at 13:42
3
$begingroup$
Found similar question: Math Overflow Link
$endgroup$
– cvogt8
Jul 26 '18 at 7:59
|
show 3 more comments
2
$begingroup$
Seems to be hard. I tried infinite descent, but I failed. Also considering the prime factors of $2x^2+1$,$2y^2+1$ and $4z^2+1$ led to nowhere. I am curious whether someone can solve this ...
$endgroup$
– Peter
Jul 10 '18 at 17:28
$begingroup$
If it's any help, it's easy to prove the impossibility of the special case $x = y$.
$endgroup$
– Connor Harris
Jul 10 '18 at 17:42
1
$begingroup$
Could you clarify the relevance, in (2), of the expression $(x^2 - y^2)^2 + (x^2 + y^2)$, which equals $x^4 - 2x^2y^2 + y^4 + x^2 + y^2$? With its 4th powers, it seems unrelated to your initial problem statement.
$endgroup$
– Adam Bailey
Jul 14 '18 at 12:30
1
$begingroup$
In the case of $(2)$ (equation in the title) it can be shown that $x,y,z$ must be divisible by $3$, hence there is no solution in positive integers when $$36m^2n^2+2m^2+2n^2$$ cannot be a positive perfect square. But this apparently does not help much.
$endgroup$
– Peter
Jul 15 '18 at 13:42
3
$begingroup$
Found similar question: Math Overflow Link
$endgroup$
– cvogt8
Jul 26 '18 at 7:59
2
2
$begingroup$
Seems to be hard. I tried infinite descent, but I failed. Also considering the prime factors of $2x^2+1$,$2y^2+1$ and $4z^2+1$ led to nowhere. I am curious whether someone can solve this ...
$endgroup$
– Peter
Jul 10 '18 at 17:28
$begingroup$
Seems to be hard. I tried infinite descent, but I failed. Also considering the prime factors of $2x^2+1$,$2y^2+1$ and $4z^2+1$ led to nowhere. I am curious whether someone can solve this ...
$endgroup$
– Peter
Jul 10 '18 at 17:28
$begingroup$
If it's any help, it's easy to prove the impossibility of the special case $x = y$.
$endgroup$
– Connor Harris
Jul 10 '18 at 17:42
$begingroup$
If it's any help, it's easy to prove the impossibility of the special case $x = y$.
$endgroup$
– Connor Harris
Jul 10 '18 at 17:42
1
1
$begingroup$
Could you clarify the relevance, in (2), of the expression $(x^2 - y^2)^2 + (x^2 + y^2)$, which equals $x^4 - 2x^2y^2 + y^4 + x^2 + y^2$? With its 4th powers, it seems unrelated to your initial problem statement.
$endgroup$
– Adam Bailey
Jul 14 '18 at 12:30
$begingroup$
Could you clarify the relevance, in (2), of the expression $(x^2 - y^2)^2 + (x^2 + y^2)$, which equals $x^4 - 2x^2y^2 + y^4 + x^2 + y^2$? With its 4th powers, it seems unrelated to your initial problem statement.
$endgroup$
– Adam Bailey
Jul 14 '18 at 12:30
1
1
$begingroup$
In the case of $(2)$ (equation in the title) it can be shown that $x,y,z$ must be divisible by $3$, hence there is no solution in positive integers when $$36m^2n^2+2m^2+2n^2$$ cannot be a positive perfect square. But this apparently does not help much.
$endgroup$
– Peter
Jul 15 '18 at 13:42
$begingroup$
In the case of $(2)$ (equation in the title) it can be shown that $x,y,z$ must be divisible by $3$, hence there is no solution in positive integers when $$36m^2n^2+2m^2+2n^2$$ cannot be a positive perfect square. But this apparently does not help much.
$endgroup$
– Peter
Jul 15 '18 at 13:42
3
3
$begingroup$
Found similar question: Math Overflow Link
$endgroup$
– cvogt8
Jul 26 '18 at 7:59
$begingroup$
Found similar question: Math Overflow Link
$endgroup$
– cvogt8
Jul 26 '18 at 7:59
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
As it says in the comments this is answered on Math Overflow. The only solution is indeed the trivial one: $(0,0,0)$. This can be found in Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations.
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
$endgroup$
2
$begingroup$
My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote?
$endgroup$
– Mason
Dec 13 '18 at 16:39
1
$begingroup$
You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 4 at 16:07
add a comment |
$begingroup$
The question is concerned only with $x,y,zge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $mod x$.
$1cdot (2y^2+1)equiv 4z^2+1mod x$.
Let $zequiv amod x$. Then $2y^2+1=4(nx+a)^2+1$.
Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2Rightarrow 2=frac{y^2}{(nx+a)^2} $ which would make $sqrt{2}$ rational.
Hence there are no positive integers that satisfy the equation.
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
$endgroup$
2
$begingroup$
There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$
$endgroup$
– jjagmath
Dec 12 '18 at 18:45
$begingroup$
I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $mod x dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this.
$endgroup$
– Mason
Dec 17 '18 at 20:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As it says in the comments this is answered on Math Overflow. The only solution is indeed the trivial one: $(0,0,0)$. This can be found in Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations.
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
$endgroup$
2
$begingroup$
My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote?
$endgroup$
– Mason
Dec 13 '18 at 16:39
1
$begingroup$
You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 4 at 16:07
add a comment |
$begingroup$
As it says in the comments this is answered on Math Overflow. The only solution is indeed the trivial one: $(0,0,0)$. This can be found in Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations.
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
$endgroup$
2
$begingroup$
My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote?
$endgroup$
– Mason
Dec 13 '18 at 16:39
1
$begingroup$
You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 4 at 16:07
add a comment |
$begingroup$
As it says in the comments this is answered on Math Overflow. The only solution is indeed the trivial one: $(0,0,0)$. This can be found in Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations.
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
$endgroup$
As it says in the comments this is answered on Math Overflow. The only solution is indeed the trivial one: $(0,0,0)$. This can be found in Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations.
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
answered Dec 12 '18 at 1:01
MasonMason
1,7791630
1,7791630
2
$begingroup$
My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote?
$endgroup$
– Mason
Dec 13 '18 at 16:39
1
$begingroup$
You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 4 at 16:07
add a comment |
2
$begingroup$
My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote?
$endgroup$
– Mason
Dec 13 '18 at 16:39
1
$begingroup$
You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 4 at 16:07
2
2
$begingroup$
My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote?
$endgroup$
– Mason
Dec 13 '18 at 16:39
$begingroup$
My main reason for posting this answer was to remove it from the list of unanswered questions. This was answered quite a while back in the comments. I am a little curious about the delete vote... I would expect this answer to maintain no upvotes and no down votes because it's mostly just site maintenance but why the delete vote?
$endgroup$
– Mason
Dec 13 '18 at 16:39
1
1
$begingroup$
You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 4 at 16:07
$begingroup$
You have to ping the reviewer who voted to delete your answer in the LQR so that he knows you're asking him.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 4 at 16:07
add a comment |
$begingroup$
The question is concerned only with $x,y,zge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $mod x$.
$1cdot (2y^2+1)equiv 4z^2+1mod x$.
Let $zequiv amod x$. Then $2y^2+1=4(nx+a)^2+1$.
Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2Rightarrow 2=frac{y^2}{(nx+a)^2} $ which would make $sqrt{2}$ rational.
Hence there are no positive integers that satisfy the equation.
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
$endgroup$
2
$begingroup$
There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$
$endgroup$
– jjagmath
Dec 12 '18 at 18:45
$begingroup$
I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $mod x dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this.
$endgroup$
– Mason
Dec 17 '18 at 20:42
add a comment |
$begingroup$
The question is concerned only with $x,y,zge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $mod x$.
$1cdot (2y^2+1)equiv 4z^2+1mod x$.
Let $zequiv amod x$. Then $2y^2+1=4(nx+a)^2+1$.
Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2Rightarrow 2=frac{y^2}{(nx+a)^2} $ which would make $sqrt{2}$ rational.
Hence there are no positive integers that satisfy the equation.
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
$endgroup$
2
$begingroup$
There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$
$endgroup$
– jjagmath
Dec 12 '18 at 18:45
$begingroup$
I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $mod x dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this.
$endgroup$
– Mason
Dec 17 '18 at 20:42
add a comment |
$begingroup$
The question is concerned only with $x,y,zge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $mod x$.
$1cdot (2y^2+1)equiv 4z^2+1mod x$.
Let $zequiv amod x$. Then $2y^2+1=4(nx+a)^2+1$.
Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2Rightarrow 2=frac{y^2}{(nx+a)^2} $ which would make $sqrt{2}$ rational.
Hence there are no positive integers that satisfy the equation.
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
$endgroup$
The question is concerned only with $x,y,zge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $mod x$.
$1cdot (2y^2+1)equiv 4z^2+1mod x$.
Let $zequiv amod x$. Then $2y^2+1=4(nx+a)^2+1$.
Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2Rightarrow 2=frac{y^2}{(nx+a)^2} $ which would make $sqrt{2}$ rational.
Hence there are no positive integers that satisfy the equation.
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
edited Dec 12 '18 at 17:26
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
answered Dec 12 '18 at 17:15
Keith BackmanKeith Backman
1,5441812
1,5441812
2
$begingroup$
There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$
$endgroup$
– jjagmath
Dec 12 '18 at 18:45
$begingroup$
I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $mod x dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this.
$endgroup$
– Mason
Dec 17 '18 at 20:42
add a comment |
2
$begingroup$
There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$
$endgroup$
– jjagmath
Dec 12 '18 at 18:45
$begingroup$
I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $mod x dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this.
$endgroup$
– Mason
Dec 17 '18 at 20:42
2
2
$begingroup$
There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$
$endgroup$
– jjagmath
Dec 12 '18 at 18:45
$begingroup$
There's a mistake in your proof, you can't conclude that $2y^2+1=4(nx+a)^2+1$
$endgroup$
– jjagmath
Dec 12 '18 at 18:45
$begingroup$
I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $mod x dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this.
$endgroup$
– Mason
Dec 17 '18 at 20:42
$begingroup$
I think from your argument you can conclude that $2y^2+1=mx+4(nx+a)^2+1$ because you started by saying consider it $mod x dots$ Anyway. I don't think this one will have an easily contained answer given that the Kashihara paper I reference takes several pages to do this.
$endgroup$
– Mason
Dec 17 '18 at 20:42
add a comment |
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2
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Seems to be hard. I tried infinite descent, but I failed. Also considering the prime factors of $2x^2+1$,$2y^2+1$ and $4z^2+1$ led to nowhere. I am curious whether someone can solve this ...
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– Peter
Jul 10 '18 at 17:28
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If it's any help, it's easy to prove the impossibility of the special case $x = y$.
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– Connor Harris
Jul 10 '18 at 17:42
1
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Could you clarify the relevance, in (2), of the expression $(x^2 - y^2)^2 + (x^2 + y^2)$, which equals $x^4 - 2x^2y^2 + y^4 + x^2 + y^2$? With its 4th powers, it seems unrelated to your initial problem statement.
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– Adam Bailey
Jul 14 '18 at 12:30
1
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In the case of $(2)$ (equation in the title) it can be shown that $x,y,z$ must be divisible by $3$, hence there is no solution in positive integers when $$36m^2n^2+2m^2+2n^2$$ cannot be a positive perfect square. But this apparently does not help much.
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– Peter
Jul 15 '18 at 13:42
3
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Found similar question: Math Overflow Link
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– cvogt8
Jul 26 '18 at 7:59