Exercise 5.25

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}}

An integer is called a biquadratic residue modulo $p$ if it is congruent to a fourth power. Using the identity $x^4 + 4 = ((x + 1)^2 + 1)((x - 1)^2 + 1)$ show that $-4$ is a biquadratic residue modulo $p$ iff $p \equiv 1 \pmod 4$.

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}
 
 $x^4+4 = (x^4 + 4 x^2 + 4) - 4 x^2 = (x^2+2)^2 - 4x^2 = (x^2 + 2 -2x)(x^2+2+2x) $, so
 
 $$x^4+4 = ((x-1)^2+1)((x+1)^2+1).$$
 
 If $-4 \equiv x^4\  [p]$ for some $x \in \Z$, then $p \mid (x+1)^2 + 1$ or $p \mid (x-1)^2+1$
 
In the two cases, $-1$ is a quadratic residue modulo $p$, thus $\legendre{-1}{p} = 1 : p \equiv 1\  [4]$.

Conversely, if $p\equiv 1\ [4], \legendre{-1}{p} = 1 $, then it exists an integer $a$ such that $-1 \equiv a^2 \ [p]$.
 
Let  $x = a-1$. Then $p \mid (x+1)^2 +1$, thus $p \mid x^4+4$ : $-4$ is a biquadratic residue modulo $p$.
 
 Conclusion : $$\exists x \in \mathbb{Z},\  x^4 \equiv -4 \ [p] \iff p \equiv 1\ [4].$$
\end{proof}

Exercise 5.25

Answers

Comments

Exercise 5.25

Answers

Comments

Add answer