Is there a way to guss the sign of 1 in the expressions (2^8-1)/17=15 or (3^8+1)/17 =386 I found some expressions like above that are divisible by 17  as bellow:
(4^8-1)/17 ,(5^8+1)/17 ,(6^8+1)/17 ,(7^8+1)/17 ,(8^8-1)/17 ,(9^8-1)/17,(10^8+1)/17
then I sum and get (1+2^8+3^8+...+10^8) that are divisible by 17
My question is that how can I find the sign of (1) based on 17 without testing

We know that $a^{16}=1\mod17$ (for $17\nmid a$) by Fermat's Little Theorem. So $a^8=\pm1\bmod 17$.
It is not hard to check that $2^8=1,3^8=-1\bmod17$. Hence $4^8=8^8=1,6^8=(2^8)(3^8)=-1\bmod17$ and $9^8=(3^8)^2=1\bmod17$. Similarly $12^8=(2^8)^23^8=-1\bmod17$, so $5^8=(17-5)^8=-1\bmod17$ and $10^8=(2^8)(5^8)=-1\bmod17$. Hence $7^8=(17-10)^8=-1\bmod17$.
So we end up with $a^8=1\bmod17$ for $a=1,2,4,8,9$ and $-1$ for $a=3,5,6,7,10$. So the sum is 0.
