Find all integers $a, b, c, d$ such that $$\begin{cases} ab - 2cd = 3 \\ ac + bd = 1\end{cases}$$
We are given the system of equations:
\[
\begin{cases}
ab - 2cd = 3 \\
ac + bd = 1
\end{cases}
\]

1. **Square the first equation:**
\[
(ab - 2cd)^2 = 9
\]
Expanding the left-hand side, we get:
\[
a^2b^2 - 4abcd + 4c^2d^2 = 9
\]

2. **Square the second equation and multiply by 2:**
\[
2(ac + bd)^2 = 2
\]
Expanding the left-hand side, we get:
\[
2(a^2c^2 + 2abcd + b^2d^2) = 2
\]
Simplifying, we get:
\[
2a^2c^2 + 4abcd + 2b^2d^2 = 2
\]

3. **Add the two equations:**
\[
a^2b^2 - 4abcd + 4c^2d^2 + 2a^2c^2 + 4abcd + 2b^2d^2 = 9 + 2
\]
Simplifying, we get:
\[
a^2b^2 + 4c^2d^2 + 2a^2c^2 + 2b^2d^2 = 11
\]

4. **Factor the left-hand side:**
\[
(a^2 + 2d^2)(b^2 + 2c^2) = 11
\]

Since 11 is a prime number, the factors of 11 are 1 and 11. Therefore, we have two cases to consider:

**Case 1:**
\[
a^2 + 2d^2 = 1 \quad \text{and} \quad b^2 + 2c^2 = 11
\]

- For \(a^2 + 2d^2 = 1\):
  \[
  a = \pm 1 \quad \text{and} \quad d = 0
  \]

- For \(b^2 + 2c^2 = 11\):
  \[
  b = \pm 3 \quad \text{and} \quad c = \pm 1
  \]

Thus, the solutions for this case are:
\[
(a, b, c, d) = (1, 3, 1, 0), (-1, -3, -1, 0)
\]

**Case 2:**
\[
a^2 + 2d^2 = 11 \quad \text{and} \quad b^2 + 2c^2 = 1
\]

- For \(a^2 + 2d^2 = 11\):
  \[
  a = \pm 3 \quad \text{and} \quad d = \pm 1
  \]

- For \(b^2 + 2c^2 = 1\):
  \[
  b = \pm 1 \quad \text{and} \quad c = 0
  \]

Thus, the solutions for this case are:
\[
(a, b, c, d) = (3, 1, 0, 1), (-3, -1, 0, -1)
\]

Combining all the solutions, we get:
\[
(a, b, c, d) = (1, 3, 1, 0), (-1, -3, -1, 0), (3, 1, 0, 1), (-3, -1, 0, -1)
\]

The final answer is \( \boxed{ (1, 3, 1, 0), (-1, -3, -1, 0), (3, 1, 0, 1), (-3, -1, 0, -1) } \)