Solve the following equation in the set of integers:

$$
x^{3}+y^{3}=8^{30}
$$
We will solve the following equation, which is a more general form of the original, over the set of integers:

$$
x^{3}+y^{3}=8^{n}
$$

where $n$ can be any non-negative integer.

We will prove by complete induction on $n$ that the solutions are the pairs $\left(2^{n}, 0\right)$ and $\left(0,2^{n}\right)$. First, consider the case $n=0$:

$$
x^{3}+y^{3}=1
$$

Since the sum or difference of two positive cubes cannot be 1, and the same applies to two negative cubes, one of $x$ and $y$ must be zero, and the other must be $1=2^{0}$.

Now let's consider the inductive step: assume that the equation $x^{3}+y^{3}=8^{n}$ has only the two solutions mentioned above, and examine the equation $x^{3}+y^{3}=8^{n+1}$. It is clear that $x$ and $y$ must have the same parity, since the sum of their cubes is even. If both are even, then $x=2 x^{\prime}, y=2 y^{\prime}$, and thus $8^{n+1}=8 x^{\prime^{3}}+8{y^{\prime 3}}^{\prime 3}$, which simplifies to $8^{n}=x^{\prime^{3}}+y^{\prime 3}$. By the induction hypothesis, $\left(x^{\prime}, y^{\prime}\right)=\left(0,2^{n}\right)$ or $\left(2^{n}, 0\right)$, and thus $(x, y)=\left(0,2^{n+1}\right)$ or $\left(2^{n+1}, 0\right)$.

We will show that the other case is not possible, i.e., both cannot be odd. We will use an indirect proof: assume that $x$ and $y$ are indeed odd. Then

$$
x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=8^{n+1}=2^{3 n+3}
$$

Here, $x^{2}-x y+y^{2}$ is odd and a divisor of $2^{3 n+3}$, so it can only be $\pm 1$.

First, consider the system of equations $x+y=8^{n+1}, x^{2}-x y+y^{2}=1$. Since $x$ and $y$ are odd, they cannot be equal. Therefore, we can assume that $x>y$. This implies that $x>0$. From this, we get $x^{2}>x y$, and then $x^{2}-x y+y^{2}>y^{2} \geq 1$, which is a contradiction.

The only remaining case is the system of equations $x+y=-8^{n+1}, x^{2}-x y+y^{2}=-1$. However, $x^{2}-x y+y^{2}=\left(x-\frac{y}{2}\right)^{2}+\frac{3}{4} y^{2} \geq 0$, so this system also has no solution.

This completes the proof of our statement; in the special case $n=30$, the solutions are $\left(2^{30}, 0\right)$ and $\left(0,2^{30}\right)$.

Based on the work of Gergely Bérczi (Szeged, Ságvári E. Gymnasium, 2nd year)

Note. Our general problem is a special case of the equation appearing in Fermat's theorem:

$$
x^{3}+y^{3}=\left(2^{n}\right)^{3} \text {. }
$$

Fermat's theorem (or its special case for third powers, which was already proven in Euler's time) states that this equation has no solution in positive integers. Therefore, the case $x, y<0$ is also impossible, as is the case $x<0<y$: rewriting the equation as

$$
y^{3}=\left(2^{n}\right)^{3}+(-x)^{3}
$$

we see that all numbers are positive, and thus this has no solution. Therefore, one of $x$ and $y$ must be zero, leading to the two solutions we have already seen.