# Find the real solutions for {(x^4-6x^2y^2+y^4=1),(4x^3y+4xy^3=1):}?

Sep 8, 2016

See explanation...

#### Explanation:

Subtract the second equation from the first to get:

$0 = {x}^{4} - 4 {x}^{3} y - 6 {x}^{2} {y}^{2} - 4 x {y}^{3} + {y}^{4}$

Divide through by ${x}^{2} {y}^{2}$ to get:

$0 = {\left(\frac{x}{y}\right)}^{2} - 4 \left(\frac{x}{y}\right) - 6 - 4 \left(\frac{y}{x}\right) + {\left(\frac{y}{x}\right)}^{2}$

$\textcolor{w h i t e}{0} = {\left(\frac{x}{y} + \frac{y}{x}\right)}^{2} - 4 \left(\frac{x}{y} + \frac{y}{x}\right) - 8$

$\textcolor{w h i t e}{0} = {\left(\frac{x}{y} + \frac{y}{x}\right)}^{2} - 4 \left(\frac{x}{y} + \frac{y}{x}\right) + 4 - 12$

$\textcolor{w h i t e}{0} = {\left(\left(\frac{x}{y} + \frac{y}{x}\right) - 2\right)}^{2} - {\left(2 \sqrt{3}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\frac{x}{y} + \frac{y}{x} - 2 - 3 \sqrt{2}\right) \left(\frac{x}{y} + \frac{y}{x} - 2 + 3 \sqrt{2}\right)$

From the two factors, multiplying each by $\left(\frac{x}{y}\right)$, we get two quadratic equations in $\left(\frac{x}{y}\right)$...

${\left(\frac{x}{y}\right)}^{2} + \left(- 2 - 3 \sqrt{2}\right) \left(\frac{x}{y}\right) + 1 = 0$

${\left(\frac{x}{y}\right)}^{2} + \left(- 2 + 3 \sqrt{2}\right) \left(\frac{x}{y}\right) + 1 = 0$

From these we get four values for $\left(\frac{x}{y}\right)$:

$\left(\frac{x}{y}\right) = 1 + \frac{3 \sqrt{2}}{2} \pm \frac{\sqrt{18 + 12 \sqrt{2}}}{2}$
(both positive)

$\left(\frac{x}{y}\right) = 1 - \frac{3 \sqrt{2}}{2} \pm \frac{\sqrt{18 - 12 \sqrt{2}}}{2}$
(both negative)

If $\left(\frac{x}{y}\right) = c$ then $x = c y$ and:

$1 = 4 {x}^{3} y + 4 x {y}^{3} = 4 c \left(1 + {c}^{2}\right) {x}^{4}$

Hence:

${x}^{4} = \frac{1}{4 c \left(1 + {c}^{2}\right)}$

In order to have Real solutions, we require $c > 0$

So we require:

$\left(\frac{x}{y}\right) = 1 + \frac{3 \sqrt{2}}{2} \pm \frac{\sqrt{18 - 12 \sqrt{2}}}{2}$

Due to the symmetry of the derivation in $x$ and $y$, these two values are reciprocals of one another.

So we need only consider one of them then allow $x$ and $y$ to be swapped in the final solution...

Let $c = \left(\frac{x}{y}\right) = 1 + \frac{3 \sqrt{2}}{2} + \frac{\sqrt{18 - 12 \sqrt{2}}}{2}$

Then:

$x = \pm \sqrt[4]{\frac{1}{4 c \left(1 + {c}^{2}\right)}} \text{ }$ and $\text{ } y = c x$

or:

$y = \pm \sqrt[4]{\frac{1}{4 c \left(1 + {c}^{2}\right)}} \text{ }$ and $\text{ } x = c y$