# Algebraically solving a system of equations in two variables

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 $$t = \pm\sqrt{3}$$ ?

• How about cancelling all those factors of $7$, $8$ and $9$? – Lord Shark the Unknown Oct 28 '18 at 14:01

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 $$t^2=41/3$$.

• There are infinitely many solutions. – A gal named Desire Oct 28 '18 at 18:28
• Would you like to guess why Wolfram offers only $x = \pm\sqrt{5/3}$ and $t = \pm\sqrt{3}$ as solutions to $(t^{2} - x^{2})(t^{2} - (3x)^{2}) = -2^{4}$? – A gal named Desire Oct 28 '18 at 18:28
• @AgalnamedDesire No, I wouldn't. ;-) – egreg Oct 28 '18 at 18:57

$$\begin{equation*} \left\{ \begin{array} ( (7t + 21x)(-7t+21x)(7t+7x)(7t-7x) = (2^{2} \cdot 49)^{2} \\ (8t+24x)(-8t+24x)(8t+8x)(8t-8x) = (2^{2} \cdot 64)^{2} \\ (9t+9x)(-3t+9x)(3t+9x)(9t-9x) = (2^{2} \cdot 27)^{2} \end{array} \right. \end{equation*}$$ 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.