MHB How Do You Solve for \(x^4 + y^4\) Given These Equations?

[anemone]

Here is this week's POTW:

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Let $x$ and $y$ be real numbers satisfying both equations below:

\[x^4 + 8y = 4(x^3 - 1) - 16 \sqrt{3}\] \[y^4 + 8x = 4(y^3 - 1) + 16 \sqrt{3}.\]

Find $x^4 + y^4$.

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[anemone]

Congratulations to the following members for their correct solution:

1. Opalg
2. Olinguito
3. lfdahl

Solution from Opalg:

Spoiler

Write the equations as \[(\textcolor{blue}{1})\qquad y = \tfrac18(-x^4 + 4x^3 - 4 - 16\sqrt3),\]
\[(\textcolor{green}{2})\qquad x = \tfrac18(-y^4 + 4y^3 - 4 + 16\sqrt3).\]
From the graph (below) it looks as though these two curves are separated by the line \[(\textcolor{orange}{3})\qquad x-y = 2\sqrt3.\]
In fact, if we write $(\textcolor{orange}{3})$ as $y = x-2\sqrt3$, and compare that with $(\textcolor{blue}{1})$, we see that \[(x-2\sqrt3) - \tfrac18(-x^4 + 4x^3 - 4 - 16\sqrt3) = \tfrac18(x^4 - 4x^3 + 8x + 4) = \tfrac18(x^2 - 2x - 2)^2 \geqslant 0.\] It follows that the graph of $(\textcolor{blue}{1})$ is always below the line $(\textcolor{orange}{3})$, and touches it only when $x^2 - 2x - 2 = 0$, or $x = 1\pm\sqrt3.$

Similarly, comparing $(\textcolor{orange}{3})$ with $(\textcolor{green}{2})$, we see that \[(y+2\sqrt3) - \tfrac18(-y^4 + 4y^3 - 4 + 16\sqrt3) = \tfrac18(y^4 - 4y^3 + 8y + 4) = \tfrac18(y^2 - 2y - 2)^2 \geqslant 0.\] It follows that the graph of $(\textcolor{green}{2})$ is always to the left of the line $(\textcolor{orange}{3})$, and touches it only when $y = 1\pm\sqrt3.$

So the curves $(\textcolor{blue}{1})$ and $(\textcolor{green}{2})$ can only meet when $x = 1\pm\sqrt3$ and $y = 1\pm\sqrt3$, and in fact that only happens at the point $(x,y) = (1+\sqrt3,1-\sqrt3).$ At that point, $x^4 + y^4 = (28+16\sqrt3) + (28-16\sqrt3) = 56.$

[DESMOS]advanced: {"version":5,"graph":{"squareAxes":false,"viewport":{"xmin":-2.0490314769975804,"ymin":-4.849999999999998,"xmax":7.95096852300242,"ymax":5.150000000000002}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"x^{4\\ }+8y\\ =\\ 4\\left(x^3-1\\right)-16\\sqrt{3}","style":"SOLID"},{"type":"expression","id":"2","color":"#388c46","latex":"y^{4\\ }+8x\\ =\\ 4\\left(y^3-1\\right)+16\\sqrt{3}","style":"SOLID"},{"type":"expression","id":"3","color":"#fa7e19","latex":"x-y\\ =\\ 2\\sqrt{3}","style":"SOLID"}]}}[/DESMOS]

Alternate solution from Olinguito:

Spoiler

Adding the two equations gives
$$\left(x^4-4x^3+8x+4\right)+\left(y^4-4y^3+8y+4\right)\ =\ 0$$

$\implies\ \left(x^2-2x-2\right)^2+\left(y^2-2y-2\right)^2\ =\ 0.$

So $x,y$ are roots of the quadratic $t^2-2t-2=0$; $\therefore\ x,y\ =\ 1\pm\sqrt3$. Of the four possible solutions, the only one that satisfies the original set of equations is $x=1+\sqrt3,\,y=1-\sqrt3$. Hence:

$$x^4+y^4=(1+\sqrt3)^4+(1-\sqrt3)^4\ = \boxed{56}.$$