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.

Question

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.

Answer 1

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?