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It's easy to see that $(ab, bc, ca, d)$ is a Pythagorean quadruple. In order to find all solutions we use the primitive Pythagorean quadruples. That is

$$ \begin{array} aab= m^2+n^2-p^2-q^2 \\ bc=2(mq+np) \\ ca=2(nq-mp) \\ d=m^2+n^2+p^2+q^2 \end{array}$$

Where $m$, $n$, $p$, and $q$ are non-negative integers with greatest common divisor $1$ such that $m + n + p + q$ is odd.

One can easily conclude that

$$ \begin{array} aa^2= \frac{(m^2+n^2-p^2-q^2)(nq-mp)}{mq+np} \\ b^2= \frac{(m^2+n^2-p^2-q^2)(mq+np)}{nq-mp} \\ c^2=\frac{(mq+np)(nq-mp)}{m^2+n^2-p^2-q^2} \end{array}$$

How can we find all quadruples of $(m, n, p, q)$ that make above relations true?

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  • $\begingroup$ Ask a question on another... It is necessary to solve a system of nonlinear Diophantine equations. Write down the system of the first 3 equations. I think you can solve, but the formula of the parameterization will be very cumbersome. Why are you doing this??? What??? $\endgroup$ – individ Nov 6 '17 at 7:07
  • $\begingroup$ Although I think you can do without the solution of the system . We can solve this equation. $\endgroup$ – individ Nov 6 '17 at 7:11
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For the equation.

$$a^2b^2+c^2b^2+a^2c^2=d^2$$

The solution can be written directly.

$$b=t(tp^2+2kps+ts^2)$$

$$a=kt(p^2-s^2)$$

$$c=2kp(kp+ts)$$

$$d=kt((2k^2+t^2)p^4+6ktsp^3+2(k^2+t^2)p^2s^2+2ktps^3+t^2s^4)$$

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    $\begingroup$ Again! Some parametric forms without explanation! $\endgroup$ – Ghartal Nov 6 '17 at 18:56

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