1.Introduction

We show a parametric solution for { x^2+A=u^2, x^2-A=v^2 }

2.Theorem
 
There is a parametric solution for simultaneous equation { x^2+A=u^2, x^2-A=v^2 },

      A = -4k(k-1)(k+1)
      
      x = k^2+1

      u = k^2-2k-1

      v = k^2+2k-1
    
      k: arbitrary


     
Proof.

x^2+a=u^2......................................................(1)

x^2-a=v^2......................................................(2)

From above equation, we obtain u^2+v^2-2x^2=0..................(3).
Since equation (3) has a rational solution (u,v,x)=(1,-1,1), we can obtain a parametric solution below.

(x,u,v)=( -(k^2+1)/((k-1)(k+1)), (k^2-1-2k)/((k-1)(k+1)), -(k^2-1+2k)/((k-1)(k+1)) )
k is arbitrary.

Then a=u^2-x^2=-4k/((k-1)(k+1)).

Finally we obtain a parametric solution.

(A,x,u,v)=( -4k(k-1)(k+1), k^2+1, k^2-2k-1, k^2+2k-1 )
 
Q.E.D.