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Solve for real x and y:

$x(x^2-3y^2)=2$

$y(3x^2-y^2)=11$

My attempt: I got $(x-y)^3=13$ but this doesn't always hold, I got a solution $(2,1)$.

How to proceed?

Tapi
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2 Answers2

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The system can be written as \begin{align*} x^3-3xy^2&=2\\ 3x^2y-y^3&=11 \end{align*}

Then \begin{align*} x^3-3xy^2+(3x^2y-y^3)i&=x^3+3ix^2y-3xy^2-iy^3=2+11i\\ (x+iy)^3&=2+11i\quad (\star) \end{align*} but $$2+11i=8+12i-6-i=2^3+3(2^2)(i)+3(2)(i^2)+i^3=(2+i)^3$$ Now, there are three cubic roots of $2+11i$: $2+i,\;(2+i)\omega\;\text{and}\;(2+i)\omega^2$ where $\omega=\frac{-1+i\sqrt 3}{2}$.

From equation $(\star)$ we get

a) $x+iy=2+i\qquad \implies\qquad (x,y)=(2,1)$

b) $x+iy=(2+i)\omega=(2+i)\left(\frac{-1+i\sqrt 3}2\right)=-1-\frac{\sqrt 3}2+\left(-\frac12+\sqrt 3\right)i$, then $(x,y)=\left(-1-\frac{\sqrt 3}2,\;-\frac12+\sqrt 3\right)$

c) $x+iy=(2+i)\omega^2=(2+i)\left(\frac{-1-i\sqrt 3}2\right)=-1+\frac{\sqrt 3}2+\left(-\frac12-\sqrt 3\right)i$, then $(x,y)=\left(-1+\frac{\sqrt 3}2,\;-\frac12-\sqrt 3\right)$

Ángel Mario Gallegos
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NOTE: You made a mistake in getting $(x-y)^3=13$. Once you add the two equations, the left side is not $(x-y)^3$.

Idea:

\begin{align*} \frac{x(x^2-3y^2)}{y(3x^2-y^2)}&=\frac{2}{11}. \end{align*} Let $y=mx$, then we have \begin{align*} \frac{(1-3m^2)}{m(3-m^2)}&=\frac{2}{11}. \end{align*} This gives $$2m^3-33m^2-6m+11=0 \implies (2m-1)(m^2-16m-11)=0.$$ One of the solutions for this is $m=1/2$, now others can be found as well and they are $8 \pm 5\sqrt{3}$.

Anurag A
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  • The substitution is really clever, what is the intuition behind it? – Tapi Apr 01 '20 at 14:36
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    @Tapi we want intersection of two curves in a plane. So if a solutions exists, then it will be located on some line passing ($y=mx$) through the origin. – Anurag A Apr 02 '20 at 09:32