Question

Find all integers n, 0 <= n < 163, such that n is its own inverse modulo 163.

Answer 1

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:

1. \( n - 1 \equiv 0 \pmod{163} \)
2. \( n + 1 \equiv 0 \pmod{163} \)

**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 \( n \cdot n \equiv 1 \mod{163} \):

1. For \( n = 1 \):

\[
1 \cdot 1 = 1 \equiv 1 \pmod{163}
\]

2. 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:

\[
\boxed{1} \quad \text{and} \quad \boxed{162}
\]