To find all integers \( n \) such that \( 0 \leq n < 163 \) and \( n \) is its own inverse modulo \( 163 \), we start with the condition:
\[ n \cdot n \equiv 1 \pmod{163} \]
This implies:
\[ n^2 \equiv 1 \pmod{163} \]
Rearranging gives us:
\[ n^2 - 1 \equiv 0 \pmod{163} \]
Factoring gives:
\[ (n - 1)(n + 1) \equiv 0 \pmod{163} \]
This means that \( n - 1 \) and \( n + 1 \) must be multiples of \( 163 \). Thus, we have two cases:
- \( n - 1 \equiv 0 \pmod{163} \)
- \( n + 1 \equiv 0 \pmod{163} \)
Case 1: \( n - 1 \equiv 0 \pmod{163} \)
This implies:
\[ n \equiv 1 \pmod{163} \]
Thus, the solution is:
\[ n = 1 \]
Case 2: \( n + 1 \equiv 0 \pmod{163} \)
This implies:
\[ n \equiv -1 \pmod{163} \]
Since \( -1 \) is equivalent to \( 162 \) in modulo \( 163 \), the solution is:
\[ n = 162 \]
The values we found are \( n = 1 \) and \( n = 162 \).
To confirm, we check if both values indeed satisfy \( n \cdot n \equiv 1 \mod{163} \):
- For \( n = 1 \):
\[ 1 \cdot 1 = 1 \equiv 1 \pmod{163} \]
- For \( n = 162 \):
\[ 162 \cdot 162 = 26244 \] We can compute \( 26244 \mod 163 \): \[ 26244 \div 163 \approx 161 \quad \text{(as integer part)} \] \[ 161 \cdot 163 = 26203 \] \[ 26244 - 26203 = 41 \] Thus, \( 162 \cdot 162 \equiv 41 \) is incorrect. Therefore, we directly review:
Correctly calculated: \[ n \cdot n = 162^2 = (163 - 1)^2 = 163^2 - 2\cdot163 + 1 \equiv 1 \pmod{163} \]
Finally, we've confirmed they satisfy the condition. Hence the integers \( n \) such that \( n \) is its own inverse modulo \( 163 \) are:
\[ \boxed{1} \quad \text{and} \quad \boxed{162} \]