Which are those integers $x, y$ for which the equality

$$
x^{6}-y^{2}=648
$$

holds?
# I. Solution.

$$
\begin{aligned}
648 & =x^{6}-y^{2}=\left(x^{3}\right)^{2}-y^{2}=\left(x^{3}+y\right)\left(x^{3}-y\right)= \\
& =2^{3} \cdot 3^{4}=2 \cdot 324=4 \cdot 162=6 \cdot 108=12 \cdot 54=18 \cdot 36
\end{aligned}
$$

Among the possible divisor pairs, we only listed those where both factors are even, since these factors are equal to the sum and difference of two numbers, and thus have the same parity. Moreover, since their product is even, they must both be even.

In the appropriate divisor pair, one factor is equal to $\left(x^{3}+y\right)$ and the other to $\left(x^{3}-y\right)$, so their difference is $2 y$. The following table provides the possible values of $y$, and from there, the possible values of $|x|$. Where we get an integer for $|x|$, the corresponding $x, y$ pairs satisfy the equation.

| $x^{3}+y$ | $x^{3}-y$ | $\left(x^{3}+y\right)-\left(x^{3}-y\right)=2 y$ | $y$ | $x^{6}=648+y^{2}$ | $\|x\|$ |
| ---: | ---: | ---: | ---: | ---: | ---: |
| 324 | 2 | 322 | 161 | 26569 | not an integer |
| 2 | 324 | -322 | -161 | 26569 | not an integer |
| 162 | 4 | 158 | 79 | 6889 | not an integer |
| 4 | 162 | -158 | -79 | 6889 | not an integer |
| 108 | 6 | 102 | 51 | 3249 | not an integer |
| 6 | 108 | -102 | -51 | 3249 | not an integer |
| 54 | 12 | 42 | 21 | 1089 | not an integer |
| 12 | 54 | -42 | -21 | 1089 | not an integer |
| 36 | 18 | 18 | 9 | 729 | 3 |
| 18 | 36 | -18 | -9 | 729 | 3 |

Thus, the solutions to the equation are: $x_{1}=3, y_{1}=9 ; x_{2}=-3, y_{2}=9 ; x_{3}=3, y_{3}=-9 ; x_{4}=-3, y_{4}=-9$.

II. Solution. Since $x$ and $y$ appear in even powers, if $(x ; y)$ is a solution pair, then $(-x ; y),(-x ;-y),(x ;-y)$ are also solutions.

Since 648 is even, $x^{6}$ and $y^{2}$ must have the same parity.

First, assume that both $x$ and $y$ are even. Then $x=2 n, y=2 m$.

$$
\begin{aligned}
(2 n)^{6}-(2 m)^{2} & =648 \\
2^{6} n^{6}-2^{2} m^{2} & =2^{3} \cdot 3^{4} \\
2^{4} n^{6}-m^{2} & =2 \cdot 3^{4}
\end{aligned}
$$

$m$ can only be even, since $2^{4} n^{6}$ and $2 \cdot 3^{4}$ are both even. Thus, $m^{2}$ is divisible by 4, so the left side is divisible by 4, but the right side is not. Therefore, there is no solution in this case.

Now consider the case where both $x$ and $y$ are odd. The squares of odd numbers can only end in 1, 5, or 9. Therefore, the last digit of $x^{6}$ must be 9, and the last digit of $y^{2}$ must be 1, because only the difference of numbers ending in these digits can result in the last 8 of 648. Thus, neither $x$ nor $y$ can end in 5.

Determine the minimum and maximum values of $|x|$. From the above, $|x| \geq 3$. Since $x^{6}$ is a square, it is worth examining the difference of two squares. The difference between the squares of consecutive integers continuously increases as the numbers increase. If we are looking for the maximum value of $|x|$, we can proceed as follows:

$$
a^{2}-(a-1)^{2} \leq 648
$$

from which $a \leq 324.5$. In this case, $a=x^{3}$, so $|x| \leq 6.87$.

According to the above, $|x|$ is odd, cannot be 5, so it can only be 3. In this case, from the equation

$$
3^{6}-y^{2}=648
$$

we get $|y|=9$.

Thus, the pairs

$$
(3 ; 9), \quad(-3 ; 9), \quad(-3 ;-9) \quad \text { and } \quad(3 ;-9)
$$

satisfy the equality.

Remark. After factoring the left side, most people arrived at the correct solution by examining all possible divisor pairs. Some provided appropriate lower and upper bounds for the possible values of $x$, then tried all possibilities within these bounds.