System of equations with radicals: $2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}$ and $2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2}$

Question

Solve the system of equations (in $\mathbb R$): $$\begin{matrix} 2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2} \\ 2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2} \end{matrix}.$$

This is from an older question, which was closed and deleted because of lack of context. (Here is link for users who can see deleted questions.) I will post my solution below - so I hope this time the question will not be closed for the lack of effort.

I found the system not too easy and somewhat interesting. (Of course I might have missed some straightforward way to the solution.) I'd be glad to see some other methods to solve it.

Answer 1

Without loss of generality, we can assume that $x,y\ge0$. (Notice that all exponents are even. We can then add signs to get the remaining solutions.)

So our equations are changed to

$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}y\tag{1}$$ $$2\sqrt[4]{\frac{y^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{2}$$

If we subtract the two equations, we get (1)-(2): $$2\left(\sqrt[4]{\frac{x^4}{3}+4}-\sqrt[4]{\frac{y^4}{3}+4}\right)=y-x.\tag{3}$$ Notice that the function $x\mapsto\sqrt[4]{\frac{x^4}{3}+4}$ is increasing.

So for $x>y$ the LHS is positive and the RHS is negative. Similarly, for $y<x$ the signs of the two expressions are opposite.

So we can only find a real solution for $x=y$, which gives us $$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{4}$$

From this we get $$16\left(\frac{x^4}{3}+4\right)=\left(1+\sqrt{\frac{3}{2}}x\right)^4\tag{5}$$

This is a quartic equation. In theory, this can be done by hand, but it will be very probably quite messy. WolframAlpha returns this. (One of the solution, according to WA, is $\sqrt6$.)

Answer 2

As you say, you will have solution corresponding to $x=\pm y$. Now, if you expand the last equation, you have (if $x \geq 0$),$$-\frac{37 x^4}{12}+3 \sqrt{6} x^3+9 x^2+2 \sqrt{6} x-63=0$$ To get rid of the $\sqrt{6}$'s, define $x=\sqrt{6} z$ and the equation becomes $$-111 z^4+108 z^3+54 z^2+12 z-63=0$$ By inspection $z=1$ is a solution. Making the long division let you with $$-111 z^3-3 z^2+51 z+63=0$$ where $z=1$ is solution again. Another division and arrive to $$-111 z^2-114 z-63=0$$ which does not show real solutions.

So, the real solutions are all possible combinations of $(\pm \sqrt{6},\pm \sqrt{6})$

Answer 3

From my answer to the deleted question:

The real solutions are $x=\pm \sqrt{6}$, $y = \pm \sqrt{6}$. There are also complex solutions.

I found this using Maple's "solve" command. A somewhat more "hands-on" approach:

> with(Groebner):
> eqs:= [s^4-x^4/3 - 4,t^2-3/2*y^2,u^4-y^4/3-4,
       v^2-3/2*x^2,2*s-1-t,2*u-1-v];
> G:=Basis(eqs,plex(s,t,u,v,x,y));
> factor(G[1]);

$$ \left( 1369\,{y}^{4}+660\,{y}^{2}+15876 \right) \left( 12065393290011975315218089\,{y}^{24}+31241483903922756527916216\,{y}^{ 22}+1916822411606153357786575368\,{y}^{20}+ 2366873160958375355737537632\,{y}^{18}+108410617754815247626287499632 \,{y}^{16}+77702138286145060148486883072\,{y}^{14}+ 3076294432635247223638434223872\,{y}^{12}+ 977270871111943893941425093632\,{y}^{10}+ 46998177944708847603452419729152\,{y}^{8}+ 1724702245845492185473755174912\,{y}^{6}+ 369903763273618971503529112700928\,{y}^{4}- 28583510894171781210603713470464\,{y}^{2}+ 1174425086245411511666388854083584 \right) \left( {y}^{2}-6 \right) ^ {2} $$

The first factor obviously has no real roots, the last has $y = \pm \sqrt{6}$. For the middle factor:

> sturm(sturmseq(op(2,%),y),y,-infinity,infinity);

$$0$$

So there are no real roots there either.