Problem 1303

1. Let $8n + 1 = k^2$ for some integer $k$. This implies:
$$8n = k^2 - 1$$
$$8n = (k-1)(k+1)$$

2. Since $8n$ is even, $(k-1)(k+1)$ must also be even. Note that $k-1$ and $k+1$ are two consecutive even numbers, so one of them must be divisible by 2, and the other must be divisible by 4. Therefore, $(k-1)(k+1)$ is divisible by 8.

3. Let us check if $n$ must be odd. Suppose $n$ is even, say $n = 2m$. Then:
$$8(2m) + 1 = k^2$$
$$16m + 1 = k^2$$
This implies $k^2 \equiv 1 \pmod{16}$. The quadratic residues modulo 16 are $0, 1, 4, 9$, so $k^2 \equiv 1 \pmod{16}$ is possible. Therefore, $n$ can be even.

4. Now, let us check if $n$ can be a perfect square. Suppose $n = m^2$. Then:
$$8m^2 + 1 = k^2$$
$$k^2 - 8m^2 = 1$$
This is a Pell's equation, which has infinitely many solutions in integers $(k, m)$. Therefore, $n$ can be a perfect square.

5. Since $n$ can be even and $n$ can be a perfect square, options (A), (B), and (C) are incorrect. Therefore, the correct answer is:
$$\boxed{\text{None of the above}}$$