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. How to prove that system 
$$
a^2+b^2=2x^2+1, \\
c^2+d^2=2y^2+1, \\
a\cdot c-b \cdot d=1
$$
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
$$(2x^2+1)(2у^2+1)=(а^2+b^2)(c^2+d^2)=(аd+bc)^2+(аc-bd)^2=(аd+bc)^2+1$$
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
$$2х^2+1=а^2+b^2; 2у^2+1=c^2+d^2$$. Besides 
$$1=m^2+n^2-p^2-q^2=(m-q)(m+q)-(p-n)(p+n)=аc-bd$$
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.

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
$$\begin{cases}a=2XZ\\b=Z^2-X^2-Y^2\\x^2=2YZ\\x^2+1=X^2+Y^2+Z^2\end{cases}$$
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?
