Problem 1545

1. Lemma Proof:
We need to prove that for an odd prime $p$, and for all $0 \le n \le \frac{p-1}{2}$, the following congruence holds:
$${2n \choose n} \equiv (-4)^n { \frac{p-1}{2} \choose n} \pmod{p}$$
First, recall the binomial coefficient:
$${2n \choose n} = \frac{(2n)!}{(n!)^2}$$
We can express the product in the numerator as:
$$(2n-1)(2n-3)\cdots 3 \cdot 1$$
This product can be rewritten modulo $p$ as:
$$(2n-1)(2n-3)\cdots 3 \cdot 1 \equiv (-1)^n (p-2n+1)(p-2n+3)\cdots (p-3)(p-1) \pmod{p}$$
Simplifying further, we get:
$$(p-2n+1)(p-2n+3)\cdots (p-3)(p-1) = (-2)^n \left( \frac{p-1}{2} - n + 1 \right) \left( \frac{p-1}{2} - n + 2 \right) \cdots \left( \frac{p-1}{2} \right)$$
Therefore:
$${2n \choose n} \equiv (-1)^n \cdot 4^n \cdot \frac{\left( \frac{p-1}{2} - n + 1 \right) \left( \frac{p-1}{2} - n + 2 \right) \cdots \left( \frac{p-1}{2} \right)}{n!} \equiv (-4)^n { \frac{p-1}{2} \choose n} \pmod{p}$$

2. Back to the Original Problem:
Given that $2c - 1$ is divisible by $p$, we have $2c \equiv 1 \pmod{p}$, which implies $c \equiv \frac{1}{2} \pmod{p}$.

3. Using the Lemma:
We need to evaluate:
$$(-1)^{\frac{p+1}{2}} + \sum_{n=0}^{\frac{p-1}{2}} {2n \choose n} c^n$$
Using the lemma, we substitute ${2n \choose n} \equiv (-4)^n { \frac{p-1}{2} \choose n} \pmod{p}$:
$$\sum_{n=0}^{\frac{p-1}{2}} {2n \choose n} c^n \equiv \sum_{n=0}^{\frac{p-1}{2}} (-4)^n { \frac{p-1}{2} \choose n} \left( \frac{1}{2} \right)^n \pmod{p}$$
Simplifying the expression inside the sum:
$$(-4)^n \left( \frac{1}{2} \right)^n = (-2)^n$$
Thus, we have:
$$\sum_{n=0}^{\frac{p-1}{2}} (-2)^n { \frac{p-1}{2} \choose n}$$
Recognizing this as the binomial expansion of $(1-2)^{\frac{p-1}{2}}$:
$$\sum_{n=0}^{\frac{p-1}{2}} (-2)^n { \frac{p-1}{2} \choose n} = (1-2)^{\frac{p-1}{2}} = (-1)^{\frac{p-1}{2}}$$

4. Combining Results:
Therefore, the original expression becomes:
$$(-1)^{\frac{p+1}{2}} + (-1)^{\frac{p-1}{2}}$$
Since $\frac{p+1}{2}$ and $\frac{p-1}{2}$ differ by 1, one of them is even and the other is odd. Thus:
$$(-1)^{\frac{p+1}{2}} + (-1)^{\frac{p-1}{2}} = 0 \pmod{p}$$

The final answer is $\boxed{ 0 \pmod{p} }$