4. Find all integers $x$ and $y$ that satisfy the equation $x^{4}-2 y^{2}=1$.
4. If $(x, y)$ is a solution to the equation $x^{4}-2 y^{2}=1$, then $(-x, y),(x,-y)$, and $(-x,-y)$ are also solutions. Therefore, we will only look for non-negative solutions. It is clear that $x$ is an odd number, $x=2 t+1$. Rewrite the equation as

$x^{4}-1=(x-1)(x+1)\left(x^{2}+1\right)=2 t \cdot(2 t+2) \cdot\left(4 t^{2}+4 t+2\right)=2 y^{2}$.

Now it is clear that $y$ is an even number, $y=2 u$. Dividing both sides of the equation by 8, we get the equation

$$
t(t+1)(2 t(t+1)+1)=u^{2}
$$

It is not difficult to verify that the numbers $t, t+1$, and $2 t(t+1)+1$ are pairwise coprime. Indeed, suppose, for example,

$$
d \mid t+1 \quad \text { and } \quad d \mid 2 t(t+1)+1
$$

then $d$ divides $2 t(t+1)$, and therefore the difference

$$
1=(2 t(t+1)+1)-(2 t(t+1))
$$

(see fact 5). The coprimality of the other two pairs is proved similarly.

The product of these coprime numbers is a perfect square. According to a known theorem (see comment), each of them is also a perfect square.

Thus, $t$ and $t+1$ are perfect squares. This is only possible when $t=0$. Indeed, if $t=\alpha^{2}, t+1=\beta^{2}$, where $\beta \geqslant 0, \alpha \geqslant 0$, then

$$
(\beta-\alpha)(\beta+\alpha)=1,
$$

so $\beta-\alpha=1, \beta+\alpha=1$, hence $\alpha=0$, therefore, $t=0$. Then $u=0$ as well. Thus, $x= \pm 1, y=0$.

Comment. We will provide a detailed proof of the used statement. First, consider the case of two factors.

Theorem. If $a b=d^{2}, a, b$ and $d$ are natural numbers, and the numbers $a$ and $b$ are coprime, then $a$ and $b$ are perfect squares.

Write the factorizations of the numbers $a, b$, and $d$ into prime factors:

$$
a=p_{1}^{n_{1}} p_{2}^{n_{2}} \ldots p_{k}^{n_{k}}, \quad b=q_{1}^{m_{1}} q_{2}^{m_{2}} \ldots q_{l}^{m_{l}}, \quad d=r_{1}^{t_{1}} r_{2}^{t_{2}} \ldots r_{u}^{t_{u}} .
$$

Here $p_{i}$ are pairwise distinct prime numbers, $q_{i}$ are pairwise distinct prime numbers, and $r_{i}$ are pairwise distinct prime numbers (see fact 10).

Since the numbers $a$ and $b$ are coprime, no $p_{i}$ equals any $q_{j}$. Therefore,

$$
a b=p_{1}^{n_{1}} p_{2}^{n_{2}} \ldots p_{k}^{n_{k}} q_{1}^{m_{1}} q_{2}^{m_{2}} \ldots q_{l}^{m_{l}}
$$

is the factorization of the number $a b$ into pairwise distinct prime factors. On the other hand,

$$
d^{2}=r_{1}^{2 t_{1}} r_{2}^{2 t_{2}} \ldots r_{u}^{2 t_{u}}
$$

is the factorization of the number $d^{2}$ into pairwise distinct prime factors. By the theorem on the uniqueness of factorization, these factorizations can only differ in the order of the factors. Therefore, the numbers $n_{i}$ are some of the numbers $2 t_{j}$. Hence, all $n_{i}$ are even, but then $a$ is a product of prime numbers in even powers, and therefore is a perfect square. Similarly, it is proved that $b$ is a perfect square. The theorem is proved.

Now suppose $a, b$, and $c$ are pairwise coprime numbers, and $a b c=d^{2}$. It is easy to see that the numbers $a$ and $b c$ are coprime (assume they have a common prime divisor $p$ and use fact 9). Applying our theorem to the numbers $a$ and $b c$, we see that each of them is a perfect square. Similarly, it is proved that the numbers $b$ and $c$ are perfect squares.