New Year's Countdown Day 10: Square $\pm$ Ten

For this solution, we use the Legendre symbol and its properties.

The problem is equivalent to finding how many of the expressions $\left ( \dfrac{10}{2017} \right ), \left ( \dfrac{-10}{2017} \right ), \left ( \dfrac{10}{2027} \right ), \left ( \dfrac{-10}{2027} \right )$ are equal to 1. First, we calculate

$\left ( \dfrac{-1}{2017} \right ) = (-1)^{\frac{2017 - 1}{2}} = (-1)^{1008} = 1 \\ \left ( \dfrac{-1}{2027} \right ) = (-1)^{\frac{2027 - 1}{2}} = (-1)^{1013} = -1. \\$

From this, we determine that if 10 is a quadratic residue mod 2017, then -10 is too (and vice versa); and if 10 is a quadratic residue mod 2027, then -10 is not (and vice versa).

Now, we calculate $\left ( \dfrac{10}{2017} \right )$ and $\left ( \dfrac{10}{2027} \right )$ using the law of quadratic reciprocity and other properties of the Legendre symbol:

$\begin{aligned} \left ( \dfrac{10}{2017} \right ) &= \left ( \dfrac{2}{2017} \right ) \left ( \dfrac{5}{2017} \right ) \\ &= \left ( \dfrac{2}{2017} \right ) \left ( \dfrac{2017}{5} \right ) \\ &= \left ( \dfrac{2}{2017} \right ) \left ( \dfrac{2}{5} \right ) \\ &= (-1)^{\frac{2017^2 - 1}{8}} \cdot (-1)^{\frac{5^2 - 1}{8}} \\ &= 1 \cdot (-1) \\ &= -1. \\ \\ \left ( \dfrac{10}{2027} \right ) &= \left ( \dfrac{2}{2027} \right ) \left ( \dfrac{5}{2027} \right ) \\ &= \left ( \dfrac{2}{2027} \right ) \left ( \dfrac{2027}{5} \right ) \\ &= \left ( \dfrac{2}{2027} \right ) \left ( \dfrac{2}{5} \right ) \\ &= (-1)^{\frac{2027^2 - 1}{8}} \cdot (-1)^{\frac{5^2 - 1}{8}} \\ &= -1 \cdot (-1) \\ &= 1. \end{aligned}$

Thus, only $\left ( \dfrac{10}{2027} \right ) = 1$ , and only $\boxed{1}$ of the original statements is true.

A simple computer program reveals that $487^2 - 10 = 237159 = 117 \times 2027,$ among infinitely many other possible values of $a$ .