Given the equation $A(qr-p)=B(pr+q)$, when can $A=pr+q$ and $B=qr-p$ be true?

Question

The following equations are known: Given Equations \begin{align} A^2+B^2&=C^2\\ Ap+Bq&=Cr\\ p^2+q^2&=r^2+1\\ A&\neq B\\ Aq-Bp&=C \end{align} $A(qr-p)=B(pr+q)$ comes from the equations given above.

For this, I tried constructing a proof involving the prime factorization of $A(qr-p)$ and $B(pr+q)$ (which are just equal) and showed that since $A$ is not equal to $B$, then $A$ is equal to $pr+q$ and vice versa. However, I realized that I can make a lot of counterexamples for this (like $6 \cdot 4=8 \cdot 3$).

I really appreciate any form of hints or help that you could give. Thank you very much in advance!

Answer 1

Mr.Seiji Tomita has given parametric solution to the simultaneous equations shown below which was posted by "OP" The link to his web site is article # 284 & his web address is given below:

www.maroon.dti.ne.jp

And select "Computational number Theory"

$\begin{align} A^2+B^2&=C^2\\ Ap+Bq&=Cr\\ p^2+q^2&=r^2+1\\ A&\neq B\\ Bp-Aq&=C \end{align}$

So, the answer is yes to the query by "OP" And $A=(pr-q)$ & $B=(qr+p)$

But "OP" equation has a typo & required a sign change

Parametric solution given by Seiji Tomita gives another numerical solution to the above system of equations:

$(A,B,C)=((2n),(2n^2+2n),(2n^2+2n+1))$

$(p,q,r) =((2n),(2n^2-1),(2n^2))$

So, for $n=3$ we get: $(A,B,C)=(7,24,25)$ & $(p,q,r)=(6,17,18)$

Answer 2

Since you are tagging your post with "diophantine equations", it is to assume that we are dealing with integers.

We have that $$ \left\{ \matrix{ Ax = By \hfill \cr \gcd (x,y) = 1 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ A = ny \hfill \cr B = nx \hfill \cr} \right. $$

Then it is clear how to proceed if $gcd(x,y) \ne 1$.

Answer 3

Above equation shown below:

$\begin{align} A^2+B^2&=C^2\\ Ap+Bq&=Cr\\ p^2+q^2&=r^2+1\\ A&\neq B\\ Bp-Aq&=C \end{align}$

"OP" had made a typo in the last equality which needed a sign change:

So, now the above simultaneous equations are satisfied by:

$(A,B,C)=(4,3,5)$ & $(p,q,r)=(7,4,8)$