# Proof

• By hypothesis, $p > 2$ is an odd prime number.
• By the Euler's criterion for quadratic residues we have the congruence $\left(\frac {-1}p\right)(p)\equiv (-1)^{\frac{p-1}{2}}(p).$
• Now, the difference $\left(\frac {-1}p\right)-(-1)^{\frac{p-1}{2}}$ equals one of the integers $0$, $-2$ or $2.$
• Because $p$ is odd, only $0$ is possible.
• It remains to be shown that the identity $\left(\frac {-1}p\right)=(-1)^{\frac{p-1}{2}}$ is equivalent to $$\left(\frac {-1}p\right)=\begin{cases}1&\text{if }p\equiv 1\mod 4,\\-1&\text{if }p\equiv -1\mod 4.\end{cases}$$
• If $p(4)\equiv 1(4)$ then $p=4k+1$ for some $k\in\mathbb Z.$ * Then $\frac{p-1}2=2k$ is even and $(-1)^{2k}=1.$ * Thus, in this case $\left(\frac {-1}p\right)=1,$ as required.
• If $p(4)\equiv -1(4)$ then $p=4k-1$ for some $k\in\mathbb Z.$ * Then $\frac{p-1}2=2k-1$ is odd and $(-1)^{2k-1}=-1.$ * Thus, in this case $\left(\frac {-1}p\right)=-1,$ as required.
• Since $p$ is odd, no other congruence classes modulo $4$ except $\pm 1$ exist.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

1. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927