It looks like there might be a typographical error in your equation. You have provided two equations that appear to refer to the same variables \(x\) and \(y\):

1. \(x^2 - y^2 = 5\)
2. \(x^2 - y^2 = 5\) (it appears to be a repetition)

Since both equations are essentially the same, we can proceed with solving the first equation:

\[
x^2 - y^2 = 5
\]

This can be factored using the difference of squares:

\[
(x - y)(x + y) = 5
\]

To find specific integer or rational solutions, you can consider the pairs of factors of 5:

1. \(1 \cdot 5\)
2. \(-1 \cdot -5\)
3. \(5 \cdot 1\)
4. \(-5 \cdot -1\)

Setting up the equations for each factorization:

1. If \(x - y = 1\) and \(x + y = 5\):
- Adding these two equations:
\[
2x = 6 \Rightarrow x = 3
\]
- Substituting \(x = 3\) into \(x - y = 1\):
\[
3 - y = 1 \Rightarrow y = 2
\]

2. If \(x - y = -1\) and \(x + y = -5\):
- Adding these two equations:
\[
2x = -6 \Rightarrow x = -3
\]
- Substituting \(x = -3\) into \(x - y = -1\):
\[
-3 - y = -1 \Rightarrow y = -2
\]

3. If \(x - y = 5\) and \(x + y = 1\):
- Adding these two equations:
\[
2x = 6 \Rightarrow x = 3
\]
- Substituting \(x = 3\) into \(x - y = 5\):
\[
3 - y = 5 \Rightarrow y = -2
\]

4. If \(x - y = -5\) and \(x + y = -1\):
- Adding these two equations:
\[
2x = -6 \Rightarrow x = -3
\]
- Substituting \(x = -3\) into \(x - y = -5\):
\[
-3 - y = -5 \Rightarrow y = 2
\]

In conclusion, the integer solutions to the equation \(x^2 - y^2 = 5\) are:
- \((3, 2)\)
- \((-3, -2)\)
- \((3, -2)\)
- \((-3, 2)\)

If you were looking for a different form of the equation or additional constraints, please provide further details!