Exercise 5.35

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Use the preceding exercise to show that $3$ is a square modulo $p$ iff $p$ is congruent to $1$ or $-1$ modulo $12$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
We know from Ex. 5.34 that $
u = \mathrm{Card}\, \{j \in [1,(p-1)/2] \ \vert \ p/2 \leq 3j < p\} = \mu$. 
Therefore $
u$ is the number of $j$ such that $p/6 \leq j < p/3$, so $
u =\lfloor p/3 floor - \lfloor p/6 floor $.

If $p = 12k +1$, $
u = \lfloor p/3 floor - \lfloor p/6 floor = 4k - 2k = 2k :  (3/p) = (-1)^
u = 1$.

If $p = 12k +5$, $
u = \lfloor p/3 floor - \lfloor p/6 floor = 4k + 1 - 2k = 2k + 1:  (3/p) = (-1)^
u = -1$.

If $p = 12k - 5$, $
u = \lfloor p/3 floor - \lfloor p/6 floor = 4k - 2 - (2k-1) = 2k -1:  (3/p) = (-1)^
u = -1$.

If $p = 12k  - 1$, $
u = \lfloor p/3 floor - \lfloor p/6 floor = 4k  - 1 - (2k-1) = 2k :  (3/p) = (-1)^
u = 1$.

Therefore $3$ is a square modulo $p$ (where $p
e 2, p 
e 3$)  iff $p$ is congruent to $1$ or $-1$ modulo $12$.
\end{proof}

Exercise 5.35

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Exercise 5.35

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