Problem 1449

To prove that the sum $1^{p+2} + 2^{p+2} + \cdots + (p-1)^{p+2}$ is divisible by $p^2$ for a prime number $p > 3$, we will use properties of modular arithmetic and Fermat's Little Theorem.

1. Using Fermat's Little Theorem:
Fermat's Little Theorem states that for any integer $a$ and a prime $p$, we have:
$$a^p \equiv a \pmod{p}$$
This implies:
$$a^{p-1} \equiv 1 \pmod{p} \quad \text{for } a \not\equiv 0 \pmod{p}$$

2. **Simplifying $a^{p+2} \mod p$:**
For any integer $a$ not divisible by $p$, we can write:
$$a^{p+2} = a^{p+1} \cdot a = (a^p \cdot a^2) \equiv a \cdot a^2 = a^3 \pmod{p}$$
Therefore:
$$a^{p+2} \equiv a^3 \pmod{p}$$

3. Pairing terms:
Consider the sum $1^{p+2} + 2^{p+2} + \cdots + (p-1)^{p+2}$. We can pair terms as follows:
$$a^{p+2} + (p-a)^{p+2}$$
Using the binomial expansion and properties of modular arithmetic, we have:
$$(p-a)^{p+2} \equiv (-a)^{p+2} = (-1)^{p+2} \cdot a^{p+2} \pmod{p^2}$$
Since $p+2$ is odd (because $p$ is an odd prime), we have:
$$(-1)^{p+2} = -1$$
Thus:
$$(p-a)^{p+2} \equiv -a^{p+2} \pmod{p^2}$$

4. Summing the pairs:
Therefore, each pair sums to:
$$a^{p+2} + (p-a)^{p+2} \equiv a^{p+2} - a^{p+2} = 0 \pmod{p^2}$$
Since this holds for each $a$ from $1$ to $(p-1)/2$, the entire sum:
$$1^{p+2} + 2^{p+2} + \cdots + (p-1)^{p+2} \equiv 0 \pmod{p^2}$$

Conclusion:
$$\boxed{0}$$