Legendre symbol sum

Since $\left( \dfrac{p}{p} \right) = 0$ , the desired sum is equal to $\displaystyle \sum_{a=1}^{p-1} \left( \frac{a}{p} \right)$ .

Observe that every term $\left( \dfrac{a}{p} \right)$ is paired up with a term $\left( \dfrac{p-a}{p} \right) = \left( \dfrac{-a}{p} \right) = \left( \dfrac{-1}{p} \right) \left( \dfrac{a}{p} \right)$ . Because $a^{(p-1)/2} \equiv \left( \dfrac{a}{p} \right) \pmod{p}$ , and here p = 163 is in the form 4k+3, $\left( \dfrac{-1}{p} \right) \left( \dfrac{a}{p} \right) = (-1)^{(p-1)/2} \left( \dfrac{a}{p} \right) = -\left( \dfrac{a}{p} \right)$ .

Hence, the total sum is 0.

Follows by a simple counting argument. First of all, note that barring $p$ , there are an even number of residues (or equivalence classes) $\{1,2, \ldots, p-1\}$ , modulo any odd prime $p$ . Now a residue $q$ can be paired up with $p-q$ , both of which produces the same quadratic residue $q^2 \mod (p)$ . Since $p$ is prime, it ensures that the quadratic residues produced in this way are distinct. Hence, there are exactly $\frac{p-1}{2}$ quadratic residues and an equal number of quadratic non-residues. Hence, $\sum_{a=1}^{p} \left( \frac{a}{p} \right) = \frac{p-1}{2}- \frac{p-1}{2}= 0.$

Since $\left( \dfrac{163}{163} \right) = 0$ , we need to find sum of even amount of odd numbers. Therefore the answer is 0, because it is the only even number available :P (I know how to solve it properly)

• A computation in SAGE shows:

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>>> sum(kronecker(a, 163) for a in range(1, 163 + 1))
0

• More generally, a computation in SAGE for primes 2, 3, 5, ..., 163 shows:

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>>> f = lambda p: sum(kronecker(a, p) for a in range(1, p + 1))
>>> [f(p) for p in primes(164)]
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

• That is, for $p > 2$ , $\sum_{a=1}^{p} \left( \dfrac{a}{p} \right) = 0$