Problem 1425

### Part (a)
We need to prove that $f(n)$ divides each of the numbers $f(3n), f(5n), f(7n)$.

1. Define $f(n) = g^n + 1$, where $g$ is an even positive integer.
2. Consider $f(3n) = g^{3n} + 1$.
3. We need to show that $f(n) \mid f(3n)$, i.e., $g^n + 1 \mid g^{3n} + 1$.

Using the fact that $g$ is even, we can use the polynomial identity:
$$g^{3n} + 1 = (g^n + 1)(g^{2n} - g^n + 1)$$

4. Since $g^{3n} + 1 = (g^n + 1)(g^{2n} - g^n + 1)$, it is clear that $g^n + 1$ divides $g^{3n} + 1$.

5. Similarly, for $f(5n) = g^{5n} + 1$, we use the identity:
$$g^{5n} + 1 = (g^n + 1)(g^{4n} - g^{3n} + g^{2n} - g^n + 1)$$

6. Thus, $g^n + 1$ divides $g^{5n} + 1$.

7. For $f(7n) = g^{7n} + 1$, we use the identity:
$$g^{7n} + 1 = (g^n + 1)(g^{6n} - g^{5n} + g^{4n} - g^{3n} + g^{2n} - g^n + 1)$$

8. Therefore, $g^n + 1$ divides $g^{7n} + 1$.

Hence, $f(n)$ divides each of $f(3n), f(5n), f(7n)$.

### Part (b)
We need to prove that $f(n)$ is relatively prime to each of the numbers $f(2n), f(4n), f(6n), \ldots$.

1. Let $a = g^n$. Then $f(n) = a + 1$ and $f(2n) = a^2 + 1$.

2. We need to show that $\gcd(a + 1, a^{2k} + 1) = 1$ for $k \geq 1$.

3. Let $d = \gcd(a + 1, a^{2k} + 1)$.

4. Since $d \mid (a + 1)$, we have $a \equiv -1 \pmod{d}$.

5. Substituting $a \equiv -1 \pmod{d}$ into $a^{2k} + 1$:
$$(-1)^{2k} + 1 \equiv 1 + 1 \equiv 2 \pmod{d}$$

6. Therefore, $d \mid 2$. Since $g$ is even, $a = g^n$ is also even, and $a + 1$ is odd.

7. Hence, $d$ cannot be 2, so $d = 1$.

Thus, $\gcd(a + 1, a^{2k} + 1) = 1$, meaning $f(n)$ is relatively prime to each of $f(2n), f(4n), f(6n), \ldots$.

$\blacksquare$