## Task A-3.4.

Determine all natural numbers $m$ for which $2^{m}-63$ is the cube of some integer.
## Solution.

We seek all $m \in \mathbb{N}$ for which $2^{m} - 63 = n^{3}$ for some $n \in \mathbb{Z}$.

Consider the remainders that both sides of this equation give when divided by 7. 1 point

By listing all possible remainders for $n$ when divided by 7, we see that the cube of a natural number $n^{3}$ gives remainders 0, 1, or 6.

By listing all remainders that $2^{m}$ gives when divided by 7, we see that they are 2, 4, and 1.

Since 63 is divisible by 7, and both sides of the equation $2^{m} - 63 = n^{3}$ must give the same remainder when divided by 7, we conclude that the only possibility is that both $2^{m}$ and $n^{3}$ give a remainder of 1 when divided by 7. This is only achieved when $m$ is divisible by 3.

Let $k \in \mathbb{N}$ be such that $m = 3k$. Using the difference of cubes, we get

$$
63 = 2^{3k} - n^{3} = (2^{k} - n)(2^{2k} + 2^{k} \cdot n + n^{2})
$$

We check the possible factorizations of the number 63 into factors $a := 2^{k} - n$ and $b := 2^{2k} + 2^{k} \cdot n + n^{2}$. Since $b$ is a natural number, $a$ must also be a natural number.

We can conclude that $n$ must be a natural number. If $n \leq 0$, then from $2^{m} - 63 = n^{3} \leq 0$ we see that $m$ must be 0, which is not a natural number. From $2^{m} - 63 = n^{3} \leq 0$ we see that $m$ must be 0, which is not a natural number. Since $a b = 63$, we get $a^{3} < 63$, or $a \leq 3$. Therefore, the only possibilities for $(a, b)$ are the pairs $(1, 63)$ and $(3, 21)$.

We discard the possibility $(1, 63)$ because then the number $b - a^{2} = 62$ is not divisible by 3, and thus is not of the form $3 \cdot 2^{k} \cdot n$.

We are left to check the case $a = 2^{k} - n = 3$, $b = 2^{2k} + 2^{k} \cdot n + n^{2} = 21$. We also have $b - a^{2} = 3 \cdot 2^{k} \cdot n = 12$, or $2^{k} \cdot n = 4$. From this, we have two possibilities: $2^{k} = 2$, $n = 2$, or $2^{k} = 4$, $n = 1$.

By substituting into the remaining equations, only the second possibility gives a solution to the equation for $k = 2$, or $m = 6$. Therefore, $m = 6$ is the only natural number that satisfies the condition of the problem.

Note: The solution consists of proving that $m$ is divisible by 3 (4 points), factoring the equation using the difference of cubes (1 point), reducing the factorizations of the number 63 to only one case $3 \cdot 21$ (4 points), and finally finding the only such number $m$ (1 point). The reduction of the factorizations of the number 63 can be performed in multiple ways, including checking all possible factorizations. Such solutions should be graded according to the official solution.

The part of the solution that confirms $m = 6$ satisfies the condition of the problem (without proving it is the only such natural number) earns 1 point, corresponding to the last point in the grading scheme.

The part of the solution that directly verifies that no natural number $m$ less than 6 is a solution (or equivalently, that $n$ must be a natural number) earns 1 point, corresponding to the sixth point in the grading scheme.

The proof that $m$ is divisible by 3 can also be derived by looking at the remainders that both sides of the equation $2^{m} - 63$ give when divided by 9. The left side of the equation $2^{m} - 63 = n^{3}$ gives remainders 2, 4, 1, 7, 5, and 8, while the right side gives remainders 0, 1, and 8. The number $2^{m}$ gives remainders 1 and 8 only when $m$ is divisible by 3.