1.Introduction

It's showed that simultaneous equation {A^2+B^2=C^2, Ap+Bq=Cr, Bp-Aq=C, p^2+q^2=r^2+1} has a solution on MathStackExchange website[1].

We show above simultaneous equation has a parametric solution.

2.Theorem
     
Simultaneous equation {A^2+B^2=C^2, Ap+Bq=Cr, Bp-Aq=C, p^2+q^2=r^2+1} has a parametric solution as follows.
    
(A,B,C,p,q,r)=(2n+1, 2(n+1)n, 2n^2+2n+1, 2n, 2n^2-1, 2n^2).
    
n is arbitrary.           

 
Proof.

A^2+B^2=C^2..........................................(1)

Ap+Bq=Cr.............................................(2)

Bp-Aq=C..............................................(3)

From equation (2) and (3), we obtain

A = pr-q
B = qr+p
C = p^2+q^2.

Substitute {A,B,C} to equation (1), and simplifying equation (1), we obtain

(p^2+q^2)(p^2-r^2-1+q^2)=0..........................(4)

Hence we find parametric solution of p^2+q^2=r^2+1.

Let (p,q,r)=(2n,2n^2-1,2n^2), we obtain a parametric solution for (A,B,C,p,q,r).

   
Q.E.D.　
       

3.Reference

[1]: StackExchange: Given the equation A(qr−p)=B(pr+q), when can A=pr+q and B=qr−p be true?