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.1.IntroductionSimultaneous 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.@2.Theorem[1]: StackExchange: Given the equation A(qr−p)=B(pr+q), when can A=pr+q and B=qr−p be true?3.Reference

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