Sum of Squares in 2017

Let $n = 2d + 1,$ where $d$ is a positive integer. A sequence of $n$ consecutive perfect squares can be written as

$\left \{(x - d)^2, \dots, (x - 1)^2, x^2, (x + 1)^2, \dots (x + d)^2 \right \},$

where $x$ is a positive integer greater than $d.$ The $i$ th term of this sequence is $(x - (d - i + 1))^2.$

The sum of the terms of the sequence is equal to

$\begin{aligned} \sum_{i = 1}^{2d + 1} (x - (d - i + 1))^2 &= \sum_{i = 1}^{2d + 1} x^2 - 2x \sum_{i = 1}^{2d + 1} (d - i + 1) + \sum_{i = 1}^{2d + 1} (d - i + 1)^2 \\ &= (2d + 1)x^2 - 2x \left [ d(2d + 1) - \dfrac{(2d + 1)(2d + 2)}{2} + (2d + 1) \right ] + 2 \sum_{i = 1}^{d} i^2 \\ &= (2d + 1)x^2 - 2x[(d + 1)(2d + 1) - (2d + 1)(d + 1)] + 2 \left [ \dfrac{d(d + 1)(2d + 1)}{6} \right ] \\ &= (2d + 1)x^2 + \dfrac{d(d + 1)(2d + 1)}{3}. \end{aligned}$

Assume that this sum is a multiple of 2017 i.e. it is equivalent to 0 modulo 2017. This means

$\begin{aligned} (2d + 1)x^2 + \dfrac{d(d + 1)(2d + 1)}{3} &\equiv 0 \! \! \! \! \pmod{2017} \\ x^2 + \dfrac{d(d + 1)}{3} &\equiv 0 \! \! \! \! \pmod{2017} \\ x^2 + \dfrac{672d(d + 1)}{672(3)} &\equiv 0 \! \! \! \! \pmod{2017} \\ x^2 + \dfrac{672d(d + 1)}{2016} &\equiv 0 \! \! \! \! \pmod{2017} \\ x^2 - 672d(d + 1) &\equiv 0 \! \! \! \! \pmod{2017} \\ x^2 &\equiv 672d(d + 1) \! \! \! \! \pmod{2017}. \end{aligned}$

Let $\left ( \dfrac{a}{p} \right )$ denote the Legendre symbol . Since $2017 \equiv 1 \! \pmod{8},$ we have $\left ( \dfrac{2}{2017} \right ) = 1.$ Since $2017 \equiv 1 \! \pmod{12},$ we have $\left ( \dfrac{3}{2017} \right ) = 1.$ Also, by the law of quadratic reciprocity,

$\begin{aligned} \left ( \dfrac{7}{2017} \right ) \left ( \dfrac{2017}{7} \right ) &= (-1)^{(2016)(6)/4} \\ \left ( \dfrac{7}{2017} \right ) \left ( \dfrac{1}{7} \right ) &= 1 \\ \left ( \dfrac{7}{2017} \right ) &= 1. \end{aligned}$

Using these values together with the properties of the Legendre symbol, we have

$\begin{aligned} \left ( \dfrac{672d(d+1)}{2017} \right ) &= \left ( \dfrac{672}{2017} \right ) \left ( \dfrac{d}{2017} \right ) \left ( \dfrac{d + 1}{2017} \right ) \\ &= \left ( \dfrac{2}{2017} \right )^5 \left ( \dfrac{3}{2017} \right ) \left ( \dfrac{7}{2017} \right ) \left ( \dfrac{d}{2017} \right ) \left ( \dfrac{d + 1}{2017} \right ) \\ &= \left ( \dfrac{d}{2017} \right ) \left ( \dfrac{d + 1}{2017} \right ). \end{aligned}$

Now, notice that

$\begin{aligned} \left ( \dfrac{5}{2017} \right ) \left ( \dfrac{2017}{5} \right ) &= (-1)^{(2016)(4)/4} \\ \left ( \dfrac{5}{2017} \right ) \left ( \dfrac{2}{7} \right ) &= 1 \\ -\left ( \dfrac{5}{2017} \right ) &= 1 \\ \left ( \dfrac{5}{2017} \right ) &= -1. \end{aligned}$

Thus, $a = 5$ is the smallest value of $a$ such that $\left ( \dfrac{a}{2017} \right ) = -1,$ since $\left ( \dfrac{4}{2017} \right ) = \left ( \dfrac{2}{2017} \right )^2 = 1.$ If we set $d = 4,$ we find that $\left ( \dfrac{d}{2017} \right ) \left ( \dfrac{d + 1}{2017} \right ) = \left ( \dfrac{4}{2017} \right ) \left ( \dfrac{5}{2017} \right ) = -1,$ meaning that there does not exist $x$ such that $x^2 \equiv 672d(d + 1) \! \pmod{2017}.$

Thus, the smallest value of $n$ such that the sum of $n$ consecutive perfect squares is not divisible by 2017 is $n = 2(4) + 1 = \boxed{9}.$