If $ c$ is a positive integer and $ p$ is an odd prime, what is the smallest residue (in absolute value) of \[  \sum_{n=0}^{\frac{p-1}{2}} \binom{2n}{n}c^n \;(\textrm{mod}\;p\ ) \ ?\] 

J. Suranyi
To solve the problem, we need to find the smallest residue (in absolute value) of the sum 

\[
\sum_{n=0}^{\frac{p-1}{2}} \binom{2n}{n}c^n \pmod{p}
\]

where \( c \) is a positive integer and \( p \) is an odd prime. Let's break down the steps:

1. **Understanding the Binomial Coefficient Modulo \( p \)**:
   The binomial coefficient \(\binom{2n}{n}\) modulo \( p \) can be analyzed using properties of binomial coefficients and Lucas' theorem. However, for simplicity, we will use the fact that \(\binom{2n}{n}\) is related to the central binomial coefficient.

2. **Generating Function Approach**:
   Consider the generating function for the central binomial coefficients:
   \[
   \sum_{n=0}^{\infty} \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}
   \]
   We are interested in the sum up to \(\frac{p-1}{2}\) terms, evaluated at \( x = c \).

3. **Modulo \( p \) Analysis**:
   We need to evaluate the sum modulo \( p \). The key insight is to consider the behavior of the generating function modulo \( p \). Specifically, we need to determine whether \( 1 - 4c \) is a quadratic residue modulo \( p \).

4. **Quadratic Residue Consideration**:
   - If \( 1 - 4c \) is a quadratic residue modulo \( p \), then there exists an integer \( k \) such that \( k^2 \equiv 1 - 4c \pmod{p} \). In this case, the sum will be congruent to \( 1 \) modulo \( p \).
   - If \( 1 - 4c \) is not a quadratic residue modulo \( p \), then the sum will be congruent to \( -1 \) modulo \( p \), which is equivalent to \( p - 1 \) modulo \( p \).

5. **Special Case \( c \equiv 4^{-1} \pmod{p} \)**:
   If \( c \equiv 4^{-1} \pmod{p} \), then \( 1 - 4c \equiv 0 \pmod{p} \). This implies that the sum is congruent to \( 0 \) modulo \( p \).

Therefore, the smallest residue (in absolute value) of the given sum modulo \( p \) is:
- \( 0 \) when \( c \equiv 4^{-1} \pmod{p} \)
- \( 1 \) if \( 1 - 4c \) is a quadratic residue modulo \( p \)
- \( p - 1 \) if \( 1 - 4c \) is not a quadratic residue modulo \( p \)

\(\blacksquare\)

The final answer is \( \boxed{ 0 } \) when \( c \equiv 4^{-1} \pmod{p} \), \( 1 \) if \( 1 - 4c \) is a quadratic residue modulo \( p \), and \( p - 1 \) if it is not.