# $\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$

I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$$ by reducing the system to a system of second degree.

What can I look for in such situations? What is the way to solve this kind of systems? The only thing I see here is that we can factor:

$$\begin{cases} x^2(x^2+y^2)+y^4=21 \\ x(x+y)+y^2=3 \end{cases}$$

Hint:

$$x^4+y^4+x^2y^2=(x^2+y^2)^2-(xy)^2=?$$

So, we know $$x^2+y^2,xy=9>0$$

So, $$x,y$$ will have the same sign

Hope you can take it from here using $$y=\dfrac9x$$

You can substitute $$u=x+y$$ and $$v=xy$$. Then $$x^2+xy+y^2=(x+y)^2-xy=u^2-v$$ and $$x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=(u^2-2v)^2-v^2$$ You obtain the system $$\begin{cases} u^4-4u^2v+3v^2=21\\ u^2-v=3 \end{cases}$$ From the second equation you get $$v=u^2-3$$. Substitute that in the first equation, and you'll immediately find $$u^2=1$$. After that, you can find $$u$$ and $$v$$, and I assume you can finish.

Multiply the first by $$x^2-y^2$$ and the second by $$x-y$$.$$x^6-y^6=21(x^2-y^2),\\x^3-y^3=3(x-y)$$ Then take the ratio $$x^3+y^3=7(x+y).$$ Adding the two above, $$2x^3=10x+4y$$ and using $$2y=x^3-5x$$,

$$4x^2+2x(x^3-5x)+(x^3-5x)^2-12=x^6-8x^4+19x^2-12=0.$$

By trial and error, $$x^2=1$$ are solutions and we factor as

$$(x^2-1)(x^4-7x^2+12)=(x^2-1)(x^2-3)(x^2-4).$$

• I would imagine that this is the intended approach. – Andrew Chin Jan 7 '20 at 18:18
• @AndrewChin: maybe, but lab's approach is nice. – Yves Daoust Jan 7 '20 at 18:22

I found $$9=(x^2+xy+y^2)^2=$$ $$=(x^4+x^2y^2+y^4)+2(x^2y^2+x^3y+xy^3)=$$ $$=21+2xy(xy+x^2+y^2)=21+2xy(3)=$$ $$=21+6xy$$ giving $$xy=-2.$$

Then $$(x+y)^2=(x^2+xy+y^2)+xy=3+xy=3-2=1.$$

Knowing $$(S,P)=(\pm 1,-2),$$ we can find all possible $$\{x,y\}$$ as the solutions of $$0=z^2-Sz+P=0=z^2\pm z-2.$$

• The idea was to square the 2nd Eq. and subtract it from the 1st Eq. just to see what happens. – DanielWainfleet Jan 7 '20 at 18:38