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

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Please update your equation. –  Patrick Li Jan 30 '13 at 6:32
What have you tried? –  mixedmath Jan 30 '13 at 6:40
Upps, is this my homework-assignment for today? –  Gottfried Helms Jun 24 '13 at 9:25

2 Answers 2

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

EDIT: I found these 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,
> G:=Basis(eqs,plex(s,t,u,v,x,y));
> factor(G[1]);

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

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);


So there are no real roots there either.

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Thank you but you can able to complete? help me –  Maria Petrova Jan 30 '13 at 10:16

I will suppose we are only looking for real solutions.

Without loss of generality, we can assume that $x,y\ge0$. (We can than 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 $\sqrt6$.)

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