Problem 1346

1. Form of the Matrix:
Let $S$ be the set of $2 \times 2$ matrices over $\mathbb{F}_p$ with trace 1 and determinant 0. A general $2 \times 2$ matrix is given by:
$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
The trace condition $\text{tr}(A) = 1$ implies:
$$a + d = 1$$
The determinant condition $\det(A) = 0$ implies:
$$ad - bc = 0$$
Substituting $d = 1 - a$ into the determinant condition, we get:
$$a(1 - a) - bc = 0 \implies a(1 - a) = bc$$

2. **Case $a = 0$:**
If $a = 0$, then the equation $a(1 - a) = bc$ simplifies to $0 = bc$. This means either $b = 0$ or $c = 0$. Therefore, the possible matrices are:
$$\begin{pmatrix} 0 & b \\ c & 1 \end{pmatrix} \quad \text{with} \quad b = 0 \text{ or } c = 0$$
For each $b$ (where $b$ can take $p$ values) and each $c$ (where $c$ can take $p$ values), we have $2p - 1$ possible matrices (since the pair $(b, c) = (0, 0)$ is counted twice).

3. **Case $a = 1$:**
If $a = 1$, then the equation $a(1 - a) = bc$ simplifies to $0 = bc$. This means either $b = 0$ or $c = 0$. Therefore, the possible matrices are:
$$\begin{pmatrix} 1 & b \\ c & 0 \end{pmatrix} \quad \text{with} \quad b = 0 \text{ or } c = 0$$
Similarly, for each $b$ and each $c$, we have $2p - 1$ possible matrices.

4. **Case $a \neq 0, 1$:**
If $a \neq 0$ and $a \neq 1$, then $a(1 - a) \neq 0$. Hence, $bc = a(1 - a) \neq 0$. This implies $b \neq 0$ and $c = \frac{a(1 - a)}{b}$. For each non-zero $b$ (which can take $p - 1$ values), $c$ is uniquely determined. Therefore, for each $a$ (where $a$ can take $p - 2$ values), we have $p - 1$ possible matrices.

5. Counting the Total Number of Matrices:
- For $a = 0$, there are $2p - 1$ matrices.
- For $a = 1$, there are $2p - 1$ matrices.
- For $a \neq 0, 1$, there are $(p - 2)(p - 1)$ matrices.

Summing these, we get:
$$|S| = (2p - 1) + (2p - 1) + (p - 2)(p - 1)$$
Simplifying this expression:
$$|S| = 2(2p - 1) + (p - 2)(p - 1) = 4p - 2 + p^2 - 3p + 2 = p^2 + p$$
Therefore:
$$|S| = p(p + 1)$$

The final answer is $\boxed{p(p + 1)}$.