Problem 1585

To solve the problem, we need to find the number of integer triples $(x, y, z)$ that satisfy the following system of congruences modulo 13:
$$\begin{array}{rcl} x - yz^2 &\equiv & 1 \pmod{13} \\ xz + y &\equiv & 4 \pmod{13} \end{array}$$
where $0 \leq x < 13$, $0 \leq y < 13$, and $0 \leq z < 13$.

1. **Express $x$ in terms of $y$ and $z$:**
From the first congruence, we have:
$$x \equiv 1 + yz^2 \pmod{13}$$

2. **Substitute $x$ into the second congruence:**
Substitute $x$ from the first congruence into the second congruence:
$$(1 + yz^2)z + y \equiv 4 \pmod{13}$$
Simplify the equation:
$$z + yz^3 + y \equiv 4 \pmod{13}$$
Combine like terms:
$$y(z^3 + 1) + z \equiv 4 \pmod{13}$$
Rearrange to solve for $y$:
$$y(z^3 + 1) \equiv 4 - z \pmod{13}$$

3. **Analyze the equation $y(z^3 + 1) \equiv 4 - z \pmod{13}$:**
We need to determine when $z^3 + 1 \equiv 0 \pmod{13}$. This occurs when $z^3 \equiv -1 \pmod{13}$.

4. **Find values of $z$ such that $z^3 \equiv -1 \pmod{13}$:**
We need to check which values of $z$ satisfy this condition. We can test each value of $z$ from 0 to 12:
$$\begin{array}{c|c} z & z^3 \pmod{13} \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 8 \\ 3 & 27 \equiv 1 \\ 4 & 64 \equiv 12 \\ 5 & 125 \equiv 8 \\ 6 & 216 \equiv 8 \\ 7 & 343 \equiv 5 \\ 8 & 512 \equiv 5 \\ 9 & 729 \equiv 1 \\ 10 & 1000 \equiv 12 \\ 11 & 1331 \equiv 12 \\ 12 & 1728 \equiv 12 \\ \end{array}$$
We see that $z^3 \equiv -1 \pmod{13}$ when $z \equiv 4, 10, 12 \pmod{13}$.

5. **Case analysis for $z$:
- Case $z \equiv 4$:**
$$y(4^3 + 1) \equiv 4 - 4 \pmod{13}$$
$$y(64 + 1) \equiv 0 \pmod{13}$$
$$y \cdot 65 \equiv 0 \pmod{13}$$
Since $65 \equiv 0 \pmod{13}$, any value of $y$ works. There are 13 possible values for $y$ (0 to 12). For each $y$, $x$ is uniquely determined by $x \equiv 1 + yz^2 \pmod{13}$. Thus, there are 13 solutions in this case.

- **Case $z \equiv 10$ or $z \equiv 12$:**
$$y(z^3 + 1) \equiv 4 - z \pmod{13}$$
For $z = 10$:
$$y(10^3 + 1) \equiv 4 - 10 \pmod{13}$$
$$y(1000 + 1) \equiv -6 \pmod{13}$$
$$y \cdot 1001 \equiv -6 \pmod{13}$$
Since $1001 \equiv 0 \pmod{13}$, there is no solution because $-6 \not\equiv 0 \pmod{13}$.

For $z = 12$:
$$y(12^3 + 1) \equiv 4 - 12 \pmod{13}$$
$$y(1728 + 1) \equiv -8 \pmod{13}$$
$$y \cdot 1729 \equiv -8 \pmod{13}$$
Since $1729 \equiv 0 \pmod{13}$, there is no solution because $-8 \not\equiv 0 \pmod{13}$.

6. Count the total number of solutions:
- There are 13 solutions when $z \equiv 4$.
- There are no solutions when $z \equiv 10$ or $z \equiv 12$.

Therefore, the total number of solutions is:
$$13 + 0 + 0 = 13$$

The final answer is $\boxed{13}$