477*. What is the greatest common divisor of all numbers $7^{n+2}+8^{2 n+1}(n \in N)$?
$\triangle$ Let's denote the greatest common divisor of all such numbers by $d$. Replacing $n$ with $n+1$ in the given sum, we get the sum $7^{n+3} + 8^{2n+3}$, which must also be divisible by $d$. Additionally, let's multiply the sum $7^{n+2} + 8^{2n+1}$ by 7:

$$
\left(7^{n+2} + 8^{2n+1}\right) \cdot 7 = 7^{n+3} + 7 \cdot 8^{2n+1}
$$

Now we get:

$$
\begin{aligned}
& \left(7^{n+3} + 8^{2n+3}\right) : d \\
& \frac{-\left(7^{n+3} + 7 \cdot 8^{2n+1}\right) : d}{8^{2n+1} \cdot (64 - 7) : d}
\end{aligned}
$$

Since $d$ is odd, $57 : d$, so $d$ can take the values $1, 3, 19$, and 57.

Do all the found values of $d$ work? Let's set $n = 1$ in the sum $7^{n+2} + 8^{2n+1}$:

$$
7^{3} + 8^{3} = 343 + 512 = 855
$$

and 855 is divisible by 57 (check this!). Moreover, from the solution process, it is clear that if the sum $7^{n+2} + 8^{2n+1}$ is divisible by 57 for some natural $n$, then it is also divisible by 57 when this $n$ is replaced by $n+1$. Therefore, the given sum is divisible by 57 for any natural $n$.

(Actually, the proof here was conducted using mathematical induction - see, for example, [18], § 14, point 14.7.)

Answer: 57