Problem 1318

1. We start with the polynomial $n^2 + 8n + 15$ and need to determine if it is divisible by $n + 4$ for any positive integer $n$.

2. To check for divisibility, we can use polynomial division or factorization. Let's rewrite the polynomial in a form that makes it easier to analyze:
$$n^2 + 8n + 15 = (n + 4)^2 - 1$$
This can be verified by expanding the right-hand side:
$$(n + 4)^2 - 1 = (n^2 + 8n + 16) - 1 = n^2 + 8n + 15$$

3. If $n^2 + 8n + 15$ is divisible by $n + 4$, then there must exist a polynomial $q(n)$ such that:
$$n^2 + 8n + 15 = (n + 4)q(n)$$

4. Given the factorization $n^2 + 8n + 15 = (n + 4)^2 - 1$, we can express it as:
$$n^2 + 8n + 15 = (n + 4)(n + 4) - 1$$

5. For $n^2 + 8n + 15$ to be divisible by $n + 4$, the remainder when $n^2 + 8n + 15$ is divided by $n + 4$ must be zero. Let's perform the division:
$$n^2 + 8n + 15 = (n + 4)(n + 4) - 1$$
Simplifying, we get:
$$n^2 + 8n + 15 = (n + 4)^2 - 1$$

6. If $n + 4$ divides $n^2 + 8n + 15$, then it must also divide the constant term $-1$ (since the remainder must be zero). Therefore, we have:
$$(n + 4) \mid -1$$

7. The only divisors of $-1$ are $\pm 1$. However, $n + 4$ is always greater than 4 for any positive integer $n$, and thus cannot be $\pm 1$.

8. Therefore, it is impossible for $n + 4$ to divide $-1$ for any positive integer $n$.

$\blacksquare$