Find  all nonzero  integers  $a, b, c, d$   with   $a>b>c>d$  that  satisfy  $ab+cd=34$ and $ac-bd=19.$
1. We start with the given equations:
   \[
   ab + cd = 34
   \]
   \[
   ac - bd = 19
   \]

2. We use the identity for the sum of squares:
   \[
   (a^2 + d^2)(b^2 + c^2) = (ac - bd)^2 + (ab + cd)^2
   \]
   Substituting the given values:
   \[
   (a^2 + d^2)(b^2 + c^2) = 19^2 + 34^2 = 361 + 1156 = 1517
   \]

3. Factorize 1517:
   \[
   1517 = 37 \times 41
   \]
   Since 37 and 41 are prime numbers, the possible values for \(a^2 + d^2\) and \(b^2 + c^2\) are 37 and 41.

4. Determine the pairs of squares that sum to 37 and 41:
   \[
   37 = 6^2 + 1^2
   \]
   \[
   41 = 5^2 + 4^2
   \]
   Thus, the possible values for \(|a|\), \(|b|\), \(|c|\), and \(|d|\) are 1, 4, 5, and 6.

5. Test the combinations to satisfy \(ab + cd = 34\):
   - If \(|a| = 1\) and \(|b| = 4\), then \(|c| = 5\) and \(|d| = 6\):
     \[
     ab + cd = 1 \cdot 4 + 5 \cdot 6 = 4 + 30 = 34
     \]
     This combination works.

6. Verify \(ac - bd = 19\) with the above values:
   - If \(a = 1\), \(b = 4\), \(c = 5\), and \(d = 6\):
     \[
     ac - bd = 1 \cdot 5 - 4 \cdot 6 = 5 - 24 = -19
     \]
     This does not satisfy the equation.
   - If \(a = 1\), \(b = 4\), \(c = -5\), and \(d = -6\):
     \[
     ac - bd = 1 \cdot (-5) - 4 \cdot (-6) = -5 + 24 = 19
     \]
     This satisfies the equation.

7. Check the other possible signs:
   - If \(a = -1\), \(b = -4\), \(c = 5\), and \(d = 6\):
     \[
     ac - bd = (-1) \cdot 5 - (-4) \cdot 6 = -5 + 24 = 19
     \]
     This also satisfies the equation.

Thus, the valid solutions are:
\[
(a, b, c, d) = (1, 4, -5, -6) \quad \text{or} \quad (a, b, c, d) = (-1, -4, 5, 6)
\]

The final answer is \( \boxed{ (1, 4, -5, -6) } \) or \((-1, -4, 5, 6)\)