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I would appreciate an algebraic explanation to the system \begin{equation*} \left\{ \begin{array} ( (7t + 21x)(-7t+21x)(7t+7x)(7t-7x) = (2^{2} \cdot 49)^{2} \\ (8t+24x)(-8t+24x)(8t+8x)(8t-8x) = (2^{2} \cdot 64)^{2} \\ (9t+9x)(-3t+9x)(3t+9x)(9t-9x) = (2^{2} \cdot 27)^{2} \end{array} \right. \end{equation*} of equations in the variables $t$ and $x$ having solutions $x = \pm \sqrt{5/3}$ and $t = \pm\sqrt{3}$.

The system of equations is equivalent to \begin{equation*} (t + 3x)(-t+3x)(t+x)(t-x) = 2^{4} \end{equation*} and to \begin{equation*} (t^{2} - x^{2})(t^{2} - (3x)^{2}) = -2^{4} . \end{equation*} Why are the only solutions to this equation $x = \pm \sqrt{5/3}$ and $t = \pm\sqrt{3}$ ?

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    $\begingroup$ How about cancelling all those factors of $7$, $8$ and $9$? $\endgroup$ – Lord Shark the Unknown Oct 28 '18 at 14:01
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The equations are all the same, namely $$ (t^2-9x^2)(t^2-x^2)+16=0 $$ that becomes $$ t^4-10x^2t^2+9x^4+16=0 $$ This is a biquadratic in $t$: $$ t^2=5x^2\pm4\sqrt{x^4-1} $$ and requires $|x|\ge1$. For any $x$ with $|x|>1$ we get four distinct solutions for $t$. Just two for $|x|=1$, namely $t=\pm\sqrt{5}$.

If we consider the equation as a biquadratic in $x$, the discriminant is nonnegative for $t^4\ge9$, that is, $|t|\ge\sqrt{3}$ and we have $$ x^2=\frac{5t^2\pm4\sqrt{t^4-9}}{9} $$ For $t=\pm\sqrt{3}$ we have $$ x^2=\frac{15}{9}=\frac{5}{3} $$ that is, $x=\pm\sqrt{5/3}$, but this is certainly not the unique set of solutions.

For instance, with $x=\pm\sqrt{5/3}$, we get $$ t^2=\frac{25}{3}\pm4\sqrt{\frac{25}{9}-1}= \frac{25}{3}\pm\frac{16}{3} $$ so we get $t^2=3$ or $t^2=41/3$.

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  • $\begingroup$ There are infinitely many solutions. $\endgroup$ – A gal named Desire Oct 28 '18 at 18:28
  • $\begingroup$ Would you like to guess why Wolfram offers only $x = \pm\sqrt{5/3}$ and $t = \pm\sqrt{3}$ as solutions to $(t^{2} - x^{2})(t^{2} - (3x)^{2}) = -2^{4}$? $\endgroup$ – A gal named Desire Oct 28 '18 at 18:28
  • $\begingroup$ @AgalnamedDesire No, I wouldn't. ;-) $\endgroup$ – egreg Oct 28 '18 at 18:57
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$$\begin{equation*} \left\{ \begin{array} ( (7t + 21x)(-7t+21x)(7t+7x)(7t-7x) = (2^{2} \cdot 49)^{2} \\ (8t+24x)(-8t+24x)(8t+8x)(8t-8x) = (2^{2} \cdot 64)^{2} \\ (9t+9x)(-3t+9x)(3t+9x)(9t-9x) = (2^{2} \cdot 27)^{2} \end{array} \right. \end{equation*}$$ Cancel power of 7,8,3 to$$\begin{equation*} \left\{ \begin{array} ( (t + 3x)(-t+3x)(t+x)(t-x) = 16 \\ (t+3x)(-t+3x)(t+x)(t-x) =16\\ (t+x)(-t+3x)(t+3x)(t-x) =16 \end{array} \right. \end{equation*}$$ We get all identical equations.

It remains to solve $$(t+x)(-t+3x)(t+3x)(t-x) = 16$$

The given solutions satisfy this equation but I am not sure if that is the only set of solutions that we can get out of it.

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