If $x^2+y^2 = \sin(xy)$ find all $(x,y)\in\mathbb R^2$

Question

If $$x^2+y^2 = \sin(xy)$$ find all $(x,y)\in\mathbb R^2$

Since , $\sin(xy)≤1$ and $x^2+y^2≥0$ , hence: $x,y\in[-1,1]$.

I noticed that $(0,0)$ is a trivial solution. Also, since it's implicit, I can't sketch it's graph to even see number of solutions.

How can I proceed further ? If there are no more solutions, how can I prove so ?

Answer 1

Note that $x^2+y^2\geq 2|xy|$ (by $x^2+y^2\pm 2xy=(x\pm y)^2\geq0$), hence $$\sin(xy)\geq 2|xy|.\tag{1}$$ On the other hand, we have (see this post, or the picture in this answer) $$|\sin(xy)|\leq|xy|.\tag{2}$$ Combining $(1)$ and $(2)$ we obtain $2|xy|\leq \sin(xy)=|\sin(xy)|\leq|xy|$, hence $xy=0$. Plugging this back to the original equation we get $x^2+y^2=0$, thus $x=y=0$.

Yes. As Martin R points out in the comment: A slightly quicker way is to note that $$x^2+y^2=\sin(xy)\leq xy\leq \frac{x^2+y^2}2,$$ which directly gives $x^2+y^2=0$ hence $x=y=0$.

Answer 2

To find all real solutions $(x, y)$ to the equation $x^2 + y^2 = \sin(xy)$, we can start by analyzing the properties of the equation.

First, note that $x^2 + y^2$ represents a sum of squares, which is always non-negative. On the other hand, the range of the sine function $-1 \leq \sin(xy) \leq 1$. This implies that for any real numbers $x$ and $y$, $\sin(xy)$ is bounded between $-1$ and $1$.

Now, let's consider the equation:

$x^2 + y^2 = \sin(xy)$

Since the left-hand side of the equation is always non-negative, and the right-hand side is bounded between $-1$ and $1$, the only way this equation can be satisfied for real numbers $x$ and $y$ is if $\sin(xy) = 0$ because $0 \leq 0 + 0 \leq 1$.

So, we have:

$sin(xy) = 0$

To find all solutions to this equation, we need to consider when $\sin(xy)$ is equal to zero. The sine function $\sin(u) = 0$ when $u$ is any multiple of $\pi$. Therefore, we have:

$xy = n\pi$

Where $n$ is an integer. Now, we can solve for $x$ and $y$ as follows:

If $n = 0$, then $xy = 0$, which implies that either $x = 0$ or $y = 0$ or both.

If $n$ is a non-zero integer, then we can solve for $y$ as $y = \frac{n\pi}{x}$.

So, the solutions to the equation $x^2 + y^2 = \sin(xy)$ for all real numbers $(x, y)$ are:

1. $x = 0$ or $y = 0$.
2. For non-zero integers $n$, $x$ can take any real value except $x = 0$, and $y$ is given by $y = \frac{n\pi}{x}$.

These are the solutions to the given equation for all $(x, y)$ in the real number domain.