Prove that the following system of equations has a unique solution

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I'm having a difficulty figuring a way to proving that the following system has a unique solution :

$$\begin{cases} -y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =0 \\ \space \space\space x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=0\end{cases}$$

I can obviously see that $O(0,0)$ is a solution, but I need to figure out if it's the only one.

One thing that look catchy to me, is that if we interchange variables on the first equation, $x$ and $y$ (which means $x:=y$ and $y:=x$) we get exactly the same equation but with a minus sign in front of the "free" variable that is not part of the product expression of the equation.

Also, since everything involves variables, bringing the system to a form to calculate determinants to decide if the solution is unique, would also be complicated if not impossible.

I would really appreciate any help or tip towards this, as I can't find a way to prove that the solution is unique.

Here's a link to Wolfram Alpha for this particular system.

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1
On BEST ANSWER

Assume $x,y\ne0$. Multiply the first equation by $-y$ and the second by $x$ and add. You get $y^2+x^2=0$, which is not compatible with the hypothesis.

Now assume $x\ne0,y=0$. You have

$$\begin{cases} x^3\cdot \sin\sqrt{x^2}=0\\x=0\end{cases}$$

which is also incompatible. Same with $x=0,y\ne0$.

Finally, $x=y=0$ is a solution.

1
On

You are right that swapping x and y in one equation gives the other. Another way of saying that is that if we multiply the first equation by y, we get $-y^2+ xy(x^2+ y^2)sin\sqrt{x^2+ y^2}= 0$ and if we multiply the second equation by x we get $x^2+ xy(x^2+ y^2)sin\sqrt{x^2+ y^2}= 0$. Subtracting the first of those from the second, $x^2+ y^2= 0$. The only (real) x and y that satisfy that are x= y= 0.

5
On

let $a = (x^2 + y^2) \sin \sqrt {x^2 + y^2}$

$ax - y = 0\\ x + ay = 0$

has a non-trival solution if $a^2 + 1 = 0$

Which requires $a$ be complex.

there are no non-trivial real solutions.

$(x,y) = 0$ is the unique solution