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Let $p$ be a prime number of the type $1 \pmod 4$. Prove that


Proof: $$\sum_{j=1}^{\frac{p-1}{2}}(\frac{j}{p})=\sum_{j=1}^{\frac{p-1}{2}}(\frac{p-j}{p})$$

Where do I go from here?

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2 Answers 2

up vote 2 down vote accepted

$$\sum_{j=1}^{\frac{p-1}{2}}(\frac{j}{p})=\sum_{j=1}^{\frac{p-1}{2}}(\frac{p-j}{p}) = \frac{1}{2} \sum_{j=1}^{p}(\frac{j}{p}) = 0. $$

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Hint: Note that since $(-1/p)=1$, we have that $j$ is a QR of $p$ if and only if $p-j$ is a QR of $p$, and $j$ is an NR of $p$ if and only if $p-j$ is an NR of $p$.

Since there are equal numbers of QR and NR, it follows that if $p$ is of the form $4k+1$, there are equal numbers of QR and NR in the interval from $1$ to $\frac{p-1}{2}$. The result follows.

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