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
 A: 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.
A: By using the Euler's Criterion, we know that for odd prime $p$:
$$\left(\dfrac{a}{p}\right)\equiv a^\frac{p-1}{2}\pmod p$$
while $\left(\dfrac{a}{p}\right)$ is the Legendre's symbol.
Therefore, we just need to find the number of quadratic residue of $17$ between $1$ and $10$. We find that the quadratic residues are $1,4,9,16(rej),8,2,15(rej),13(rej)$, which has five are between $1$ and $10$. Therefore, the sum is congruent to $1\times 5+(-1)\times 5=0$, which means that it is divisible by $17$ 
A: If you know some theory of quadratic residues, then you know that $a^8\equiv\left( a\over17\right)$ mod $17$. You also know that there are $8$ quadratic residues and $8$ quadratic nonresidues mod $17$, and that $\left(17-a\over17\right)=\left(a\over17\right)$, which means that from $1$ to $8$ there must be $4$ quadratic residues and $4$ quadratic nonresidues, and thus $1+2^8+\cdots+8^8\equiv0$ mod $17$. Since $9$ is obviously a square (hence a quadratic residue), it remains to show that $10$ is a quadratic nonresidue. One easy way to do this uses quadratic reciprocity:
$$\left(10\over17\right)=\left(7\over17\right)=\left(17\over7\right)=\left(3\over7\right)=-\left(7\over3\right)=-\left(1\over3\right)=-1$$
