Number of Residues in Modulo 2027

Let $p = 2027$ and observe that this is prime. Let $a_i$ for $0 \leq i \leq p-1$ be the number of the $2^{1013}$ numbers which are $i$ in modulo $p$ . Then, $N = \prod_{i = 1}^{\frac{p-1}{2}} (\varepsilon^k + \varepsilon^{-k}) = a_0 + ... + a_{p-1} \varepsilon^{p-1}$ where $\varepsilon = \exp \left ( \frac{2\pi}{p}i \right )$ . Now, observe that $\begin{aligned} \prod_{i = 1}^{p-1} (\varepsilon^k + \varepsilon^{-k} ) & = N \prod_{i = \frac{p+1}{2}}^{p-1} (\varepsilon^k + \varepsilon^{-k} ) \\ & = N \prod_{i = \frac{p+1}{2}}^{p-1} (\varepsilon^{k-p} + \varepsilon^{p-k} ) \\ & = N \prod_{i = 1}^{\frac{p-1}{2}} (\varepsilon^k + \varepsilon^{-k}) \\ & = N^2 \end{aligned}$ But, $\prod_{i = 1}^{p-1} (\varepsilon^k + \varepsilon^{-k} ) = \prod_{i = 1}^{p-1} (\varepsilon^{2k} + 1) = \prod_{i = 1}^{p-1} (1 + \varepsilon^{k} ) = 1$ where we use the fact that $k \rightarrow 2k$ is bijective in $\mathbb{Z}/p\mathbb{Z}$ and $1 + x + ... + x^{p-1} = \prod_{1 \leq k \leq p-1} (x - \varepsilon^k)$ . So $N^2 = 1 \implies N = \pm 1$ . So, $a_0 + ... + a_{p-1} \varepsilon^{p-1} = \pm 1 \implies a_0 \pm 1 = a_1 = ... = a_{p-1} = q$ for some integer $q$ . Then, $pq = a_0 + ... + a_{p-1} \pm 1 = 2^{\frac{p-1}{2}} \pm 1.$ Since $q$ must be an integer and $2^{\frac{p-1}{2}} \equiv \left ( \frac{2}{p} \right )$ , $q = \frac{2^{1013} - \left (\frac{2}{2027} \right )}{2027} = \frac{2^{1013} + 1}{2027} \equiv 859 \pmod{1000}.$ This is our answer.