Number Theory

What is the remainder when the sum

$1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2$ 

is divided by 17?

$\begin{array}{|rcll|} \hline 1^2 \pmod {17} &=& 1 \\ 2^2 \pmod {17} &=& 4 \\ 3^2 \pmod {17} &=& 9 \\ 4^2 \pmod {17} &=& 16 \\ 5^2 \pmod {17} &=& 8 \\ 6^2 \pmod {17} &=& 2 \\ 7^2 \pmod {17} &=& 15 \\ 8^2 \pmod {17} &=& 13 \\ 9^2 \pmod {17} &=& 13 \\ 10^2 \pmod {17} &=& 15 \\ 11^2 \pmod {17} &=& 2 \\ 12^2 \pmod {17} &=& 8 \\ 13^2 \pmod {17} &=& 16 \\ 14^2 \pmod {17} &=& 9 \\ 15^2 \pmod {17} &=& 4 \\ 16^2 \pmod {17} &=& 1 \\ 17^2 \pmod {17} &=& 0 \\ \hline \end{array}$

cycle mod 17: $(1, 4, 9, 16, 8, 2, 15, 13, 13, 15, 2, 8, 16, 9, 4, 1, 0)$

circle mod 17

$\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9+ 4+ 1+ 0\pmod {17} \\ &=& 136 \pmod {17} \\ &=& \mathbf{0 \pmod {17}} \\ \hline \end{array}$

2020/17 = 118 cycles remainder 14

remainder 14, this is the sum in the cycle from 1 to 14  mod 17

$\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9\pmod {17} \\ &=& 131 \pmod {17} \\ &=&\mathbf{ 12 \pmod {17}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && 1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2\pmod {17} \\ &=& \underbrace{0*118 \pmod {17}}_{118 \text{ cycles}} + \underbrace{\mathbf{ 12 \pmod {17}}}_{\text{remainder 14}} \\ &=& \mathbf{ 12 \pmod {17}} \\ \hline \end{array}$

The remainder when the sum $1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2$ is divided by 17 is $\mathbf{12}$