## Task A-4.1.

Determine all natural numbers $m$ for which $m! + 8$ is a power of some prime number.
## Solution.

For $m=1$ we have $m!+8=3^{2}$, so this is one solution. 1 point

For $m \geqslant 2$ the number $m!+8$ is an even number, so it can only be a power of 2. 2 points

We can write $m!+8=2^{k}$.

For $m=2$ and $m=3$, the left side is 10 and 14 respectively, and these are not powers of a prime number.

For $m=4$ the left side gives $32=2^{5}$, so we have found another solution.

1 point

For $m=5$ the left side gives $128=2^{7}$, so we have found another solution.

For $m \geqslant 6$ the number $m!$ is divisible by 16 (since $m!$ contains the factors 2, 4, and 6). Therefore, the number $m!+8$ is not divisible by 16.

On the other hand, we have $2^{k}=m!+8>8$, which means $k \geqslant 4$, so the number $2^{k}$ is divisible by 16.

1 point

Thus, for $m \geqslant 6$ the equation $m!+8=2^{k}$ has no solutions because the right side of the equation is divisible by 16, while the left side is not.

1 point

Therefore, all solutions are $m=1,4,5$.