Problem 1785

1. Lemma 1: For any integer $n$ and any non-integer real number $a$, we have $\lfloor a \rfloor + \lfloor n - a \rfloor = n - 1$.

2. Since $p$ is a prime number of the form $4k + 1$, it is known that $-1$ is a quadratic residue modulo $p$. Thus, there exists an integer $\lambda$ such that $\lambda^2 \equiv -1 \pmod{p}$.

3. Since $p \nmid \lambda$ (because otherwise, we would have $\lambda^2 \equiv 0 \pmod{p}$, contradicting $\lambda^2 \equiv -1 \pmod{p}$), multiplication with $\lambda$ is a permutation of the nonzero residues modulo $p$. Define a function $L$ from $\{1, 2, \ldots, p-1\}$ to $\{1, 2, \ldots, p-1\}$ by letting $L(i)$ be the remainder of $\lambda i$ upon division by $p$. This function $L$ is a permutation of $\{1, 2, \ldots, p-1\}$.

4. We have:
$$\sum_{i=1}^{p-1} \left( \left\lfloor \frac{2i^2}{p} \right\rfloor - 2 \left\lfloor \frac{i^2}{p} \right\rfloor \right) = \sum_{i=1}^{p-1} \left( \left\lfloor \frac{2(L(i))^2}{p} \right\rfloor - 2 \left\lfloor \frac{(L(i))^2}{p} \right\rfloor \right)$$

5. For every $i \in \{1, 2, \ldots, p-1\}$, both integers $i^2$ and $2i^2$ are coprime with $p$. Since $L(i) \equiv \lambda i \pmod{p}$, we have:
$$i^2 + (L(i))^2 \equiv i^2 + (\lambda i)^2 = (1 + \lambda^2)i^2 \equiv (1 + (-1))i^2 = 0 \pmod{p}$$
Hence, $\frac{i^2 + (L(i))^2}{p}$ is an integer. Applying Lemma 1 to $n = \frac{i^2 + (L(i))^2}{p}$ and $a = \frac{i^2}{p}$, we obtain:
$$\left\lfloor \frac{i^2}{p} \right\rfloor + \left\lfloor \frac{i^2 + (L(i))^2}{p} - \frac{i^2}{p} \right\rfloor = \frac{i^2 + (L(i))^2}{p} - 1$$
Equivalently,
$$\left\lfloor \frac{i^2}{p} \right\rfloor + \left\lfloor \frac{(L(i))^2}{p} \right\rfloor = \frac{i^2}{p} + \frac{(L(i))^2}{p} - 1$$

6. Since $2 \cdot \frac{i^2 + (L(i))^2}{p}$ is an integer, while $\frac{2i^2}{p}$ is a non-integer, applying Lemma 1 to $n = 2 \cdot \frac{i^2 + (L(i))^2}{p}$ and $a = \frac{2i^2}{p}$, we obtain:
$$\left\lfloor \frac{2i^2}{p} \right\rfloor + \left\lfloor 2 \cdot \frac{i^2 + (L(i))^2}{p} - \frac{2i^2}{p} \right\rfloor = 2 \cdot \frac{i^2 + (L(i))^2}{p} - 1$$
Equivalently,
$$\left\lfloor \frac{2i^2}{p} \right\rfloor + \left\lfloor \frac{2(L(i))^2}{p} \right\rfloor = \frac{2i^2}{p} + \frac{2(L(i))^2}{p} - 1$$

7. Now, we compute:
$$\sum_{i=1}^{p-1} \left( \left\lfloor \frac{2i^2}{p} \right\rfloor - 2 \left\lfloor \frac{i^2}{p} \right\rfloor \right) = \frac{1}{2} \left( \sum_{i=1}^{p-1} \left( \left\lfloor \frac{2i^2}{p} \right\rfloor - 2 \left\lfloor \frac{i^2}{p} \right\rfloor \right) + \sum_{i=1}^{p-1} \left( \left\lfloor \frac{2(L(i))^2}{p} \right\rfloor - 2 \left\lfloor \frac{(L(i))^2}{p} \right\rfloor \right) \right)$$
Using the fact that $L$ is a permutation:
$$= \frac{1}{2} \sum_{i=1}^{p-1} \left( \left( \left\lfloor \frac{2i^2}{p} \right\rfloor + \left\lfloor \frac{2(L(i))^2}{p} \right\rfloor \right) - 2 \left( \left\lfloor \frac{i^2}{p} \right\rfloor + \left\lfloor \frac{(L(i))^2}{p} \right\rfloor \right) \right)$$
Using the results from steps 5 and 6:
$$= \frac{1}{2} \sum_{i=1}^{p-1} \left( \left( \frac{2i^2}{p} + \frac{2(L(i))^2}{p} - 1 \right) - 2 \left( \frac{i^2}{p} + \frac{(L(i))^2}{p} - 1 \right) \right)$$
Simplifying:
$$= \frac{1}{2} \sum_{i=1}^{p-1} 1 = \frac{1}{2} (p - 1) = \frac{p - 1}{2}$$

The final answer is $\boxed{ \frac{p-1}{2} }$