## Task $5 / 86$

We are looking for all pairs of real numbers $(x ; y)$ that satisfy the equation:

$$
2 \sin \left(2 \pi x^{2}+y\right)=\sqrt{2\left(x^{2}+x^{-2}\right)}
$$
Due to $\sin \alpha \leq 1$, we have $2 \sin \left(2 \pi x^{2}+y\right) \leq 2$. Since $x^{2}+x^{-2} \geq 2$, it follows that $\sqrt{2\left(x^{2}+x^{-2}\right)} \geq \sqrt{2 \cdot 2}=2$. Therefore,

$$
2 \sin \left(2 \pi x^{2}+y\right)=\sqrt{2\left(x^{2}+x^{-2}\right)}=2
$$

which implies $\sin \left(2 \pi x^{2}+y\right)=1$ and $x^{2}+x^{-2}=2$. From the last equation, we get $x^{2}= \pm 1 ; x= \pm 1$ (non-real values are excluded!), and from the second-to-last equation, we get

$$
\sin \left(2 \pi x^{2}+y\right)=\sin (y \pm 2 \pi)=\sin y=1 ; y=(0.5 \pm 2 k) \pi
$$

with $k \in \mathbb{N}$. Therefore, all pairs $(x ; y)$ with $x= \pm 1 ; y=(0.5 \pm 2 k) \pi$ satisfy the equation.