# Solution

(related to Problem: Calculating Quadratic Residues)

• The congruence $x^2(p)\equiv 3\equiv(p)$ has a solution, if the Legendre symbol $\left(\frac 3p\right)=1.$
• According to the quadratic reciprocity law, we have $$\left(\frac{3}{p}\right)\cdot \left(\frac{p}{3}\right)=(-1)^{\frac{p-1}{2}\frac{3-1}{2}}=(-1)^{\frac{p-1}{2}}.$$
• We can divide both sides by $\left(\frac{p}{3}\right)=\pm 1$ and get $$\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)\cdot (-1)^{\frac{p-1}{2}}.\label{eq:E18781a}\tag{a}$$
• First of all, we have $$(-1)^{\frac{p-1}{2}}=\begin{cases}1&\text{if }p\equiv 1\mod 4,\\-1&\text{if }p\equiv -1\mod 4.\end{cases}\label{eq:E18781b}\tag{b}$$
• On the other hand, we can apply the Legendre symbols of equal residues rule to deduce, that if $p(3)\equiv 1(3),$ then $\left(\frac{p}{3}\right)=\left(\frac{1}{3}\right).$
• But $\left(\frac{1}{3}\right)=1,$ which can be verified directly.
• Note that if $p(3)\not\equiv 1(p)$, then otherwise we have $p(3)\equiv 2(3)\equiv-1(3).$ In this case, we can apply the first supplementary law to the quadratic reciprocity law to deduce that $\left(\frac{-1}{3}\right)=-1.$
• Altogether, we have $$\left(\frac{p}{3}\right)=\begin{cases}1&\text{if }p\equiv 1\mod 3,\\-1&\text{if }p\equiv -1\mod 3,\; p > 2.\end{cases}\label{eq:E18781c}\tag{c}$$
• Combining the results $(\ref{eq:E18781a}),$ $(\ref{eq:E18781b}),$ and $(\ref{eq:E18781c})$ together, we get $$\left(\frac{3}{p}\right)=\begin{cases}1&\text{if }p\equiv\pm 1\mod 12,\\-1&\text{if }p\equiv \pm 5\mod 12.\\\end{cases}$$
• Therefore, the congruence $x^2(p)\equiv 3\equiv(p)$ has a solution, if and only if $p\equiv\pm 1\mod 12.$

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### References

#### Bibliography

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