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System of Diophantine equations $a^2+b^2=2x^2+1,c^2+d^2=2y^2+1,ac-bd=1$ has no natural solutions.
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How to prove that system
$$
a^2+b^2=2x^2+1, \
c^2+d^2=2y^2+1, \
acdot c-b cdot d=1
$$
has no natural solutions?
It can be proved that system equal to the equation
$$(2х^2+1)(2у^2+1)=4z^2+1$$
In one direction
from Fibonacci identity
$$(2x^2+1)(2у^2+1)=(ð^2+b^2)(c^2+d^2)=(ðd+bc)^2+(ðc-bd)^2=(ðd+bc)^2+1$$
and using Pythagorean quadruple $$(1; 2z; х^2-у^2; 1+х^2+у^2)$$
$$1+(2z)^2+(x^2-у^2)^2=(1+х^2+у^2)^2$$
$$1=m^2+n^2-p^2-q^2; 1+x^2+у^2=m^2+n^2+p^2+q^2$$
$$х^2-у^2$$ equal $$2(mq+np)$$ or $$2(nq-mp)$$
Let
$$х^2-у^2=2(mq+np)$$
We get
$$2х^2+1=(m+q)^2+(n+p)^2; 2у^2+1=(m-q)^2+(n-p)^2$$
and
$$2х^2+1=ð^2+b^2; 2у^2+1=c^2+d^2$$. Besides
$$1=m^2+n^2-p^2-q^2=(m-q)(m+q)-(p-n)(p+n)=ðc-bd$$
and we get system again.
For equation $$(2х^2+1)(2у^2+1)=z^2+1$$
I have tried rewrite it
$$4x^2y^2+2x^2+2y^2=z^2$$
Let $$z = 2xy+k$$
$$4z^2=4x^2y^2+4xyk+k^2 $$
$$4x^2y^2+2x^2+2y^2 = 4x^2y^2+4xyk+k^2$$
$$2x^2+2y^2 =4xyk+k^2$$
We may assume $$k=2m$$
$$2x^2+2y^2 =8xym+4m^2$$
$$x^2+y^2= 4xym +2m^2$$
$$x^2+y^2-4xym =2m^2$$
$$(x-2my)^2+y^2-4m^2y^2 =2m^2$$
$$(x-2my)^2-y^2(4m^2-1) =2m^2$$
And have tried standard techniques like Legendre theorem,but it didn't help. .
I have seen this problem in one social math community on the russian site like facebook.
number-theory diophantine-equations
asked Jul 14 at 14:51
Anton Sorokovskiy
746
add a comment |Â
up vote
3
down vote
favorite
How to prove that system
$$
a^2+b^2=2x^2+1, \
c^2+d^2=2y^2+1, \
acdot c-b cdot d=1
$$
has no natural solutions?
It can be proved that system equal to the equation
$$(2х^2+1)(2у^2+1)=4z^2+1$$
In one direction
from Fibonacci identity
$$(2x^2+1)(2у^2+1)=(ð^2+b^2)(c^2+d^2)=(ðd+bc)^2+(ðc-bd)^2=(ðd+bc)^2+1$$
and using Pythagorean quadruple $$(1; 2z; х^2-у^2; 1+х^2+у^2)$$
$$1+(2z)^2+(x^2-у^2)^2=(1+х^2+у^2)^2$$
$$1=m^2+n^2-p^2-q^2; 1+x^2+у^2=m^2+n^2+p^2+q^2$$
$$х^2-у^2$$ equal $$2(mq+np)$$ or $$2(nq-mp)$$
Let
$$х^2-у^2=2(mq+np)$$
We get
$$2х^2+1=(m+q)^2+(n+p)^2; 2у^2+1=(m-q)^2+(n-p)^2$$
and
$$2х^2+1=ð^2+b^2; 2у^2+1=c^2+d^2$$. Besides
$$1=m^2+n^2-p^2-q^2=(m-q)(m+q)-(p-n)(p+n)=ðc-bd$$
and we get system again.
For equation $$(2х^2+1)(2у^2+1)=z^2+1$$
I have tried rewrite it
$$4x^2y^2+2x^2+2y^2=z^2$$
Let $$z = 2xy+k$$
$$4z^2=4x^2y^2+4xyk+k^2 $$
$$4x^2y^2+2x^2+2y^2 = 4x^2y^2+4xyk+k^2$$
$$2x^2+2y^2 =4xyk+k^2$$
We may assume $$k=2m$$
$$2x^2+2y^2 =8xym+4m^2$$
$$x^2+y^2= 4xym +2m^2$$
$$x^2+y^2-4xym =2m^2$$
$$(x-2my)^2+y^2-4m^2y^2 =2m^2$$
$$(x-2my)^2-y^2(4m^2-1) =2m^2$$
And have tried standard techniques like Legendre theorem,but it didn't help. .
I have seen this problem in one social math community on the russian site like facebook.
number-theory diophantine-equations
asked Jul 14 at 14:51
Anton Sorokovskiy
746
3
What have you tried? The third equation tells us something straight off does it not.
– Daniel Buck
Jul 14 at 15:12
Y have tried divisibility by 2 and 3. Looks like this system equal to (2x^2+1)*(2y^2+1)=z^2+1
– Anton Sorokovskiy
Jul 14 at 15:16
1
That's great. You should put your observations in your question with a description of how you arrived at them.
– Daniel Buck
Jul 14 at 15:19
If third equation was ab-cd=1 then it's easy to prove that system has no solutions.
– Anton Sorokovskiy
Jul 14 at 15:19
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How to prove that system
$$
a^2+b^2=2x^2+1, \
c^2+d^2=2y^2+1, \
acdot c-b cdot d=1
$$
has no natural solutions?
It can be proved that system equal to the equation
$$(2х^2+1)(2у^2+1)=4z^2+1$$
In one direction
from Fibonacci identity
$$(2x^2+1)(2у^2+1)=(ð^2+b^2)(c^2+d^2)=(ðd+bc)^2+(ðc-bd)^2=(ðd+bc)^2+1$$
and using Pythagorean quadruple $$(1; 2z; х^2-у^2; 1+х^2+у^2)$$
$$1+(2z)^2+(x^2-у^2)^2=(1+х^2+у^2)^2$$
$$1=m^2+n^2-p^2-q^2; 1+x^2+у^2=m^2+n^2+p^2+q^2$$
$$х^2-у^2$$ equal $$2(mq+np)$$ or $$2(nq-mp)$$
Let
$$х^2-у^2=2(mq+np)$$
We get
$$2х^2+1=(m+q)^2+(n+p)^2; 2у^2+1=(m-q)^2+(n-p)^2$$
and
$$2х^2+1=ð^2+b^2; 2у^2+1=c^2+d^2$$. Besides
$$1=m^2+n^2-p^2-q^2=(m-q)(m+q)-(p-n)(p+n)=ðc-bd$$
and we get system again.
For equation $$(2х^2+1)(2у^2+1)=z^2+1$$
I have tried rewrite it
$$4x^2y^2+2x^2+2y^2=z^2$$
Let $$z = 2xy+k$$
$$4z^2=4x^2y^2+4xyk+k^2 $$
$$4x^2y^2+2x^2+2y^2 = 4x^2y^2+4xyk+k^2$$
$$2x^2+2y^2 =4xyk+k^2$$
We may assume $$k=2m$$
$$2x^2+2y^2 =8xym+4m^2$$
$$x^2+y^2= 4xym +2m^2$$
$$x^2+y^2-4xym =2m^2$$
$$(x-2my)^2+y^2-4m^2y^2 =2m^2$$
$$(x-2my)^2-y^2(4m^2-1) =2m^2$$
And have tried standard techniques like Legendre theorem,but it didn't help. .
I have seen this problem in one social math community on the russian site like facebook.
number-theory diophantine-equations
asked Jul 14 at 14:51
Anton Sorokovskiy
746
How to prove that system
$$
a^2+b^2=2x^2+1, \
c^2+d^2=2y^2+1, \
acdot c-b cdot d=1
$$
has no natural solutions?
It can be proved that system equal to the equation
$$(2х^2+1)(2у^2+1)=4z^2+1$$
In one direction
from Fibonacci identity
$$(2x^2+1)(2у^2+1)=(ð^2+b^2)(c^2+d^2)=(ðd+bc)^2+(ðc-bd)^2=(ðd+bc)^2+1$$
and using Pythagorean quadruple $$(1; 2z; х^2-у^2; 1+х^2+у^2)$$
$$1+(2z)^2+(x^2-у^2)^2=(1+х^2+у^2)^2$$
$$1=m^2+n^2-p^2-q^2; 1+x^2+у^2=m^2+n^2+p^2+q^2$$
$$х^2-у^2$$ equal $$2(mq+np)$$ or $$2(nq-mp)$$
Let
$$х^2-у^2=2(mq+np)$$
We get
$$2х^2+1=(m+q)^2+(n+p)^2; 2у^2+1=(m-q)^2+(n-p)^2$$
and
$$2х^2+1=ð^2+b^2; 2у^2+1=c^2+d^2$$. Besides
$$1=m^2+n^2-p^2-q^2=(m-q)(m+q)-(p-n)(p+n)=ðc-bd$$
and we get system again.
For equation $$(2х^2+1)(2у^2+1)=z^2+1$$
I have tried rewrite it
$$4x^2y^2+2x^2+2y^2=z^2$$
Let $$z = 2xy+k$$
$$4z^2=4x^2y^2+4xyk+k^2 $$
$$4x^2y^2+2x^2+2y^2 = 4x^2y^2+4xyk+k^2$$
$$2x^2+2y^2 =4xyk+k^2$$
We may assume $$k=2m$$
$$2x^2+2y^2 =8xym+4m^2$$
$$x^2+y^2= 4xym +2m^2$$
$$x^2+y^2-4xym =2m^2$$
$$(x-2my)^2+y^2-4m^2y^2 =2m^2$$
$$(x-2my)^2-y^2(4m^2-1) =2m^2$$
And have tried standard techniques like Legendre theorem,but it didn't help. .
I have seen this problem in one social math community on the russian site like facebook.
number-theory diophantine-equations
asked Jul 14 at 14:51
Anton Sorokovskiy
746
edited Jul 16 at 18:05
asked Jul 14 at 14:51
Anton Sorokovskiy
746
asked Jul 14 at 14:51
Anton Sorokovskiy
746
asked Jul 14 at 14:51
Anton Sorokovskiy
746
746
3
What have you tried? The third equation tells us something straight off does it not.
– Daniel Buck
Jul 14 at 15:12
Y have tried divisibility by 2 and 3. Looks like this system equal to (2x^2+1)*(2y^2+1)=z^2+1
– Anton Sorokovskiy
Jul 14 at 15:16
1
That's great. You should put your observations in your question with a description of how you arrived at them.
– Daniel Buck
Jul 14 at 15:19
If third equation was ab-cd=1 then it's easy to prove that system has no solutions.
– Anton Sorokovskiy
Jul 14 at 15:19
add a comment |Â
3
What have you tried? The third equation tells us something straight off does it not.
– Daniel Buck
Jul 14 at 15:12
Y have tried divisibility by 2 and 3. Looks like this system equal to (2x^2+1)*(2y^2+1)=z^2+1
– Anton Sorokovskiy
Jul 14 at 15:16
1
That's great. You should put your observations in your question with a description of how you arrived at them.
– Daniel Buck
Jul 14 at 15:19
If third equation was ab-cd=1 then it's easy to prove that system has no solutions.
– Anton Sorokovskiy
Jul 14 at 15:19
3
3
What have you tried? The third equation tells us something straight off does it not.
– Daniel Buck
Jul 14 at 15:12
What have you tried? The third equation tells us something straight off does it not.
– Daniel Buck
Jul 14 at 15:12
Y have tried divisibility by 2 and 3. Looks like this system equal to (2x^2+1)*(2y^2+1)=z^2+1
– Anton Sorokovskiy
Jul 14 at 15:16
Y have tried divisibility by 2 and 3. Looks like this system equal to (2x^2+1)*(2y^2+1)=z^2+1
– Anton Sorokovskiy
Jul 14 at 15:16
1
1
That's great. You should put your observations in your question with a description of how you arrived at them.
– Daniel Buck
Jul 14 at 15:19
That's great. You should put your observations in your question with a description of how you arrived at them.
– Daniel Buck
Jul 14 at 15:19
If third equation was ab-cd=1 then it's easy to prove that system has no solutions.
– Anton Sorokovskiy
Jul 14 at 15:19
If third equation was ab-cd=1 then it's easy to prove that system has no solutions.
– Anton Sorokovskiy
Jul 14 at 15:19
add a comment |Â
1 Answer
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COMMENT.-It is clear that variables $a, b$ have distinct parity; let $a$ be even and $b$ odd.
It is known that the general solution of the equation
$$x^2+y^2+z^2=w^2$$ is given by the identity $$(2XZ)^2+(2YZ)^2+(Z^2-X^2-Y^2)^2=(X^2+Y^2+Z^2)^2$$ It follows because of the first equation is equivalent to $$a^2+b^2+x^4=(x^2+1)^2$$ we have
$$begincasesa=2XZ\b=Z^2-X^2-Y^2\x^2=2YZ\x^2+1=X^2+Y^2+Z^2endcases$$
Then the three parameters $X,Y,Z$ are related by $$(Y-Z)^2+X^2=1$$
Similar reasoning with the second equation.
I have no time to try to get the end of the proof in case this remark is useful. Can you do it?
answered Jul 14 at 21:23
Piquito
17.4k31234
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
COMMENT.-It is clear that variables $a, b$ have distinct parity; let $a$ be even and $b$ odd.
It is known that the general solution of the equation
$$x^2+y^2+z^2=w^2$$ is given by the identity $$(2XZ)^2+(2YZ)^2+(Z^2-X^2-Y^2)^2=(X^2+Y^2+Z^2)^2$$ It follows because of the first equation is equivalent to $$a^2+b^2+x^4=(x^2+1)^2$$ we have
$$begincasesa=2XZ\b=Z^2-X^2-Y^2\x^2=2YZ\x^2+1=X^2+Y^2+Z^2endcases$$
Then the three parameters $X,Y,Z$ are related by $$(Y-Z)^2+X^2=1$$
Similar reasoning with the second equation.
I have no time to try to get the end of the proof in case this remark is useful. Can you do it?
answered Jul 14 at 21:23
Piquito
17.4k31234
add a comment |Â
up vote
1
down vote
COMMENT.-It is clear that variables $a, b$ have distinct parity; let $a$ be even and $b$ odd.
It is known that the general solution of the equation
$$x^2+y^2+z^2=w^2$$ is given by the identity $$(2XZ)^2+(2YZ)^2+(Z^2-X^2-Y^2)^2=(X^2+Y^2+Z^2)^2$$ It follows because of the first equation is equivalent to $$a^2+b^2+x^4=(x^2+1)^2$$ we have
$$begincasesa=2XZ\b=Z^2-X^2-Y^2\x^2=2YZ\x^2+1=X^2+Y^2+Z^2endcases$$
Then the three parameters $X,Y,Z$ are related by $$(Y-Z)^2+X^2=1$$
Similar reasoning with the second equation.
I have no time to try to get the end of the proof in case this remark is useful. Can you do it?
answered Jul 14 at 21:23
Piquito
17.4k31234
add a comment |Â
up vote
1
down vote
up vote
1
down vote
COMMENT.-It is clear that variables $a, b$ have distinct parity; let $a$ be even and $b$ odd.
It is known that the general solution of the equation
$$x^2+y^2+z^2=w^2$$ is given by the identity $$(2XZ)^2+(2YZ)^2+(Z^2-X^2-Y^2)^2=(X^2+Y^2+Z^2)^2$$ It follows because of the first equation is equivalent to $$a^2+b^2+x^4=(x^2+1)^2$$ we have
$$begincasesa=2XZ\b=Z^2-X^2-Y^2\x^2=2YZ\x^2+1=X^2+Y^2+Z^2endcases$$
Then the three parameters $X,Y,Z$ are related by $$(Y-Z)^2+X^2=1$$
Similar reasoning with the second equation.
I have no time to try to get the end of the proof in case this remark is useful. Can you do it?
answered Jul 14 at 21:23
Piquito
17.4k31234
COMMENT.-It is clear that variables $a, b$ have distinct parity; let $a$ be even and $b$ odd.
It is known that the general solution of the equation
$$x^2+y^2+z^2=w^2$$ is given by the identity $$(2XZ)^2+(2YZ)^2+(Z^2-X^2-Y^2)^2=(X^2+Y^2+Z^2)^2$$ It follows because of the first equation is equivalent to $$a^2+b^2+x^4=(x^2+1)^2$$ we have
$$begincasesa=2XZ\b=Z^2-X^2-Y^2\x^2=2YZ\x^2+1=X^2+Y^2+Z^2endcases$$
Then the three parameters $X,Y,Z$ are related by $$(Y-Z)^2+X^2=1$$
Similar reasoning with the second equation.
I have no time to try to get the end of the proof in case this remark is useful. Can you do it?
answered Jul 14 at 21:23
Piquito
17.4k31234
answered Jul 14 at 21:23
Piquito
17.4k31234
answered Jul 14 at 21:23
Piquito
17.4k31234
answered Jul 14 at 21:23
Piquito
17.4k31234
17.4k31234
add a comment |Â
add a comment |Â
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3
What have you tried? The third equation tells us something straight off does it not.
– Daniel Buck
Jul 14 at 15:12
Y have tried divisibility by 2 and 3. Looks like this system equal to (2x^2+1)*(2y^2+1)=z^2+1
– Anton Sorokovskiy
Jul 14 at 15:16
1
That's great. You should put your observations in your question with a description of how you arrived at them.
– Daniel Buck
Jul 14 at 15:19
If third equation was ab-cd=1 then it's easy to prove that system has no solutions.
– Anton Sorokovskiy
Jul 14 at 15:19