Solve the following equation if $x$ and $y$ are integers:

$$
x^{2}-2 x y+2 y^{2}-4 y^{3}=0
$$
I. solution. On the left side of the equation, we can recognize the square of $(x-y)$. Thus, we get

$$
x^{2}-2 x y+2 y^{2}-4 y^{3}=(x-y)^{2}+y^{2}(1-4 y)=0
$$

or after rearrangement

$$
(x-y)^{2}=y^{2}(4 y-1)
$$

If $y=0$, then the right side of (1) is zero, so $x=0$. We will show that the equation has no other solution.

If $y \neq 0$, then since $y$ is an integer, $4 y-1$ is also not zero. Thus, according to (1), the product on the right side, $y^{2}(4 y-1)$, is a non-zero perfect square. One of the factors, $y^{2}$, is itself a perfect square, so the product can only be a perfect square if the second factor, $4 y-1=4(y-1)+3$, is also a perfect square. This, however, is impossible, since perfect squares give a remainder of 0 or 1 when divided by 4.

Therefore, the only solution to the equation is $x=y=0$.

II. solution. From the equation, we get

$$
\begin{gathered}
x^{2}=2 x y-2 y^{2}+4 y^{3}, \text { and } \\
2 y^{2}=2 x y-x^{2}+4 y^{3}
\end{gathered}
$$

From (2a), it is clear that $x$ is even, so the right side of (2b) is divisible by 4. Thus, $y^{2}$ is also even, and therefore divisible by 4. From (2a) again, it then follows that $8 \mid x^{2}$, which can only be true if $4 \mid x$. The reasoning can be continued, and we get that both $x$ and $y$ are divisible by any positive integer power of 2, which can only be true if $x$ and $y$ are both zero.

The above reasoning is merely an outline. The "continuability" of what is said must be proven in a real proof, as this is the key to the solution. It can be proven by induction; we do not detail this here, instead, we take a seemingly different path.

If either $x$ or $y$ is not zero, let $d$ be the largest exponent for which $2^{d}$ divides both $x$ and $y$. Then, according to (2a), $x^{2}$ is divisible by $2^{2 d+1}$, and since in the factorization of a perfect square, every prime number appears with an even exponent, $x^{2}$ is also divisible by $2^{2 d+2}$, so $x$ is divisible by $2^{d+1}$.

Using this in (2b), we get that $2 y^{2}$ is divisible by $2^{2 d+2}$, i.e., $y^{2}$ is divisible by $2^{2 d+1}$. This, however, is a contradiction, since by our assumption, $2^{d}$ was the largest power of 2 by which both $x$ and $y$ are divisible.

The contradiction implies that neither $x$ nor $y$ can be non-zero. The pair $(0,0)$ is clearly a solution, so we have solved the problem.

Remarks. 1. Both solutions used the fundamental theorem of arithmetic, which states that positive integers can be uniquely expressed as a product of primes.

2. In the second solution, we showed that if $x$ and $y$ are divisible by $2^{d}$, then they are also divisible by $2^{d+1}$. Since every integer is divisible by $2^{0}=1$, this is essentially the induction proof mentioned.