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Lecture 11 - 2025-02-24
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That is, at most \( \deg(f) \) elements \( a \in \Z/p\Z \) satisfy \( f(a) \equiv 0 \pmod{p} \).
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\end{remark}
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% \begin{proof}
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% Let \( m_p(d) \) be the number of elements of degree \( d \) in \( \big( U( \Z/p\Z ), \cdot \big) \).
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% WTS \( m_p(p - 1) \neq 0 \), because \( U(\Z/p\Z) \) has order \( p - 1 \).
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% We observe that for any \( d \mid p - 1 \), \( m_p(d) \leq \varphi{d} \).
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% Indeed, we have two options,
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% \begin{enumerate}
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% \item There are no elements of order \( d \), and so \( m_p(d) = 0 \).
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% \item There is at least one element of order \( d \).
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% That is, \( x^d \equiv \pmod{p} \) has at least one solution.
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% However, \( x^d \equiv 1 \pmod{p} \) has at most \( d \) solutions by the remark above.
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% This means that there is at most one copy of cyclic group of order \( d \) in \( U(\Z/p\Z) \). Otherwise, \( \exists g_1, g_2 \in U(\Z/p\Z) \) of order \( d \) and \( g_1, g_2 \) cannot generate each other, which means element of order \( d \) is at least \( d + 1 \), contradicting with the remark above.
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% We could further constraint \[
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% m_p(d) \in \{ 0, \varphi(d) \}
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% \]
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% Now, we have \begin{align*}
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% \left| U(\Z/p\Z) \right|
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% &
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% = p - 1
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% \\
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% & = \sum_{d \mid p - 1} m_p(d) \leq \sum_{d \mid p - 1} \varphi(d) = p - 1
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% \end{align*}
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% This could only happen if \( m_p(d) = \varphi(d) \) for all \( d \mid p - 1 \).
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% In particular, \( m_p(p - 1) = \varphi(p - 1) \geq 1 \neq 0 \).
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% \end{enumerate}
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% \end{proof}
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\begin{proof}[Proof (start).]
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Size of \( U(\Z / p\Z) \) is \( \phi(p^n) = (p - 1)p^{n-1} \).
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WTS \( \exists g \in U(\Z/p\Z) \) of order \( (p - 1)p^{n-1} \iff \begin{cases}
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g^{q p^{n-1}} \not\equiv 1 \pmod{p^n} & \text{for every} q \mid p - 1 \\
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g^{(p-1) p^{n-2}} \not\equiv 1 \pmod{p^n}
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\end{cases} \)
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\end{proof}
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\begin{remark}
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Suppose \( h \in \Z \) is a generator of \( U(\Z / p\Z) \).
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Now consider \( h \) in \( U( \Z / p^2 \Z) \) and suppose it is not a generator.
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Then,
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\begin{itemize}
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\item \( h^{(p-1) p} \equiv 1 \pmod{p^2} \)
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\item Maybe \( h^{p-1} \equiv 1 \pmod{p^2} \)
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\end{itemize}
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Can \( h^{qp} \equiv 1 \pmod{p^2} \) for some \( q \mid p - 1 \)?
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No. If \( h^{qp} \equiv 1 \pmod{p^2} \), then \( h^{qp} \equiv 1 \pmod{p} \) and \( h^{q} \equiv 1 \pmod{p} \) which means \( h \) is not a generator of \( U(\Z / p\Z) \).
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Conclusion: \( h \) fails to be a generator modulo \( p^2 \) if and only if \( h^{p-1} \equiv 1 \pmod{p^2} \).
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\end{remark}
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\begin{proof}[Proof (for \(n = 2\)).]
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Suppose the above happens. Consider \( h + p \) as a different lift of \( h \) in \( U(\Z / p^2 \Z) \).
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\begin{align*}
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(h + p)^{p-1} & = h^{p-1} + h^{p-2} p(p - 1) + h^{p-3}p^2 (p - 1) + \cdots
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\\
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& = h^{p-1} + h^{p-2} p(p - 1) \pmod{p^2}
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\\
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& = 1 + p \left[ h^{p-2}(p-1) \right] \pmod{p^2}
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\\
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& \not\equiv 1 \pmod{p^2}
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& h^{p-2}(p-1) \text{ coprime to } p
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\end{align*}
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So \( h + p \) generates \( U(\Z / p^2 \Z) \).
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\end{proof}
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\begin{lemma}
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Suppose \( h \in \Z \) generates \( U(\Z / p\Z) \).
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Then, \( h \) can fail to generate \( U(\Z / p^n \Z) \) if and only if \( h^{p-1} p^{n-2} \equiv 1 \pmod{p^n} \).
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\end{lemma}
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\begin{proof}
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Let \( m_p(d) \) be the number of elements of degree \( d \) in \( \big( U( \Z/p\Z ), \cdot \big) \).
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Proof of lemma.
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WTS \( m_p(p - 1) \neq 0 \), because \( U(\Z/p\Z) \) has order \( p - 1 \).
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\begin{itemize}
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\item One direction is clear.
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We observe that for any \( d \mid p - 1 \), \( m_p(d) \leq \varphi{d} \).
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\item In the other direction, suppose \( h^{qp^{n-1}} \pmod{p^n} \) for \( q \) a proper divisor of \( p - 1 \).
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\( (h^1)^{p^{n-1}} \equiv 1 \pmod{p^n} \) so \( h^q \equiv 1 \pmod{p^n} \) which means \( h \) is not a generator of \( U(\Z / p \Z) \), a contradiction.
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\end{itemize}
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\end{proof}
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\begin{proof}[Proof (theorem, contd).]
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Pick \( h \) to be a lift of a generator of \( \Z / p\Z \).
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Now, there are two cases:
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Indeed, we have two options,
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\begin{enumerate}
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\item There are no elements of order \( d \), and so \( m_p(d) = 0 \).
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\item \( h^{(p-1)p^{n-2}} \not\equiv 1 \pmod{p^n} \implies g = h \) by lemma will generate \( U(\Z / p^n \Z) \).
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\item \( h^{(p-1)p^{n-2}} \equiv 1 \pmod{p^n} \)
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\item There is at least one element of order \( d \).
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Take \( g = h + p \).
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That is, \( x^d \equiv \pmod{p} \) has at least one solution.
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However, \( x^d \equiv 1 \pmod{p} \) has at most \( d \) solutions by the remark above.
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This means that there is at most one copy of cyclic group of order \( d \) in \( U(\Z/p\Z) \). Otherwise, \( \exists g_1, g_2 \in U(\Z/p\Z) \) of order \( d \) and \( g_1, g_2 \) cannot generate each other, which means element of order \( d \) is at least \( d + 1 \), contradicting with the remark above.
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We could further constraint \[
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m_p(d) \in \{ 0, \varphi(d) \}
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\]
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Now, we have \begin{align*}
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\left| U(\Z/p\Z) \right|
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&
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= p - 1
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\\
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& = \sum_{d \mid p - 1} m_p(d) \leq \sum_{d \mid p - 1} \varphi(d) = p - 1
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Check: \begin{align*}
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(h+p)^{(p-1)p^{n-2}}
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& = h^{(p-1)p^{n-2}} + p h^u (p-1) p^{n-2} + p^2 h^{u-1} {{(p-1) p^{n-1}}\choose{2}} + \cdots
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\end{align*}
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This could only happen if \( m_p(d) = \varphi(d) \) for all \( d \mid p - 1 \).
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In particular, \( m_p(p - 1) = \varphi(p - 1) \geq 1 \neq 0 \).
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which is in the form of \( 1 + p^{n-1} ( h^u (p-1) ) \) which is not congruent to \( 1 \pmod{p^n} \).
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\end{enumerate}
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\end{proof}
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\end{proof}
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\begin{lemma}
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Suppose \( p \geq 3 \), \( x = 1 + p^{n-2} y \) for \( y \) coprime to \( p \).
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Then, \( x^p = 1 + p^{n-1} z \) for \( z \) coprime to \( p \).
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\end{lemma}
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\begin{proof}
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\( x = (h + p)^{(p-1)p^{n-3}} \) and work by induction.
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\begin{itemize}
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\item \( h^{p-1} \equiv 1 \pmod{p} \) so \( (h + p)^{p-1} = 1 + py \pmod{p^2} \).
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\item \( (h + p)^{(p-1)p} = 1 + pz \pmod{p^2} \) for \( z \) coprime to \( p \).
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\end{itemize}
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\( x^p = ( 1 + p^{n-2} y )^p = 1 + p \cdot p^{n-2} y + {p \choose 2} (p^{n-2} y)^2 + \cdot = 1 + p^{n-1}y + p^n w = 1 + p^{n-1}(y+pw) \).
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\end{proof}
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\begin{proof}[Proof (theorem, for \(p = 2\)).]
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If \( h \) generates \( \Z / 4\Z \) or \( Z / p\Z \), \( p \) odd, then \( h \) (or maybe \(h + p \)) generate \( \Z / p^n \Z \).
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We prove this by computing the order of \( h \) (or \(h + p\)).
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Look at \( p = 2 \) in more detail.
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\( U(\Z/8\Z) =\{1, 3, 5, 7\} \), \( x_2 = 1 \pmod 8\), so \( U(\Z/8\Z) \cong \Z/2\Z \times \Z/2\Z \).
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This is also true for \( U( \Z / 16\Z ) \cong \Z/2\Z \times \Z/4\Z = \langle \pm 1 \rangle \times \langle 3 \rangle \).
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\end{proof}
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Correlation of proof: If \( <x> = U(\Z p^n \Z) \), then \( <x> = U(\Z p^{n-1} \Z) \), \( p \) odd, \(n \geq 3\)
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\section{Quadratic Residues}
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Suppose \( p \) odd. then \( U(\Z / p\Z) \) is cyclic of order \( p - 1 \), which is even.
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Them, an element in \( U(\Z/p\Z) \) is a square makes sense.
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Any \( x \) in \( U(\Z/p\Z) \) is of the form \( g^k \), \( k \pmod{p-1} \), then \( k \pmod{2} \) makes sense.
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Caution: \( (\Z/3\Z, +) \) asking if an element is a multiple of \( 2 \) is meaningless.
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\begin{example}
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\( x^2 + x + 1 \equiv 0 \pmod{11} \iff (x + 1/2)^2 - \frac{1}{4} + 1 \equiv 0 \pmod{11} \)
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This means \( ( 1 + 1/2)^2 = -\frac{3}{4} \pmod 11 \).
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\end{example}
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\begin{definition}[Quadratic Residue]\index{Quadratic Residue}
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A \term{quadratic residue} modulo \( p \) is just another term for a square in \( U(\Z/p\Z) \). That is, \[
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QR(p) = \{ x \in U(\Z/p\Z) | \exists y \in U(\Z/p\Z) \text{ such that } y^2 \equiv x \pmod{p} \}
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\]
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\end{definition}
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\begin{lemma}
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For \( p \) odd, \( | QR(p) | = \frac{p-1}{2} \).
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\end{lemma}
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\begin{proof}
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\( x^2 \equiv y^2 \pmod{p} \implies (x-y)(x+y) \equiv 0 \pmod{p} \implies x \equiv \pm y \pmod{p} \).
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The map \( U(\Z/p\Z)^{x \to x^2} \to QR \) is 2 to 1.
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\end{proof}
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\begin{definition}[Legendre Symbol]\index{Legendre Symbol}
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The \term{Legendre symbol} is defined as \[
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\left( \frac{a}{p} \right) = \begin{cases}
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-1 & \text{if } a \not\in QR(p) \\
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0 & \text{if } a \equiv 0 \pmod{p} \\
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1 & \text{if } a \in QR(p)
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\end{cases}
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\]
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\end{definition}
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\begin{lemma}
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\( \left( \frac{a}{p} \right) \equiv a^{\frac{p-1}{2}} \pmod{p} \).
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\end{lemma}
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\begin{proof}
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We consider three cases.
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\begin{itemize}
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\item \( a \equiv 0 \pmod{p} \) is trivial.
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\item \( a \in QR(p) \), so \( a = g^{2k} \), \( kj \in \N \).
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Thus, \( a^{\frac{p-1}{2}} \equiv g^{k(p-1)} \equiv 1 \pmod{p} = \left( \frac{a}{p} \right) \).
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\item \( a \notin QR(p) \), so \( a = g^{2k+1} \), \( k \in \N \).
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Thus, \( a^{\frac{p-1}{2}} \equiv g^{(2k+1)(\frac{p-1}{2})} \)
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\( \frac{(2k+1)(p-1)}{2} \) is not a multiple of \( p - 1 \), so \( a^{\frac{p-1}{2}} \not\equiv 1 \pmod{p} \).
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But it's square equiv 1 modulo \( p \).
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Thus, \( a^{\frac{p+1}{2}} \equiv -1 \pmod{p} = \left( \frac{a}{p} \right) \).
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\end{itemize}
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\end{proof}
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\begin{lemma}
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\( \left( \frac{ab}{p} \right) = \left( \frac{a}{p} \right) \left( \frac{b}{p} \right) \).
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\end{lemma}
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\begin{proof}
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By the previous lemma, \[
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\left( \frac{ab}{p} \right) = (ab)^{\frac{p-1}{2}} \pmod{p} \equiv a^{\frac{p-1}{2}} b^{\frac{p-1}{2}} \pmod{p} = \left( \frac{a}{p} \right) \left( \frac{b}{p} \right)
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\]
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\end{proof}
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\begin{table}[H]
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\centering
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\begin{tabular}{c|c|c}
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& QR & QNR \\
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\hline
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QR & QR & QNR \\
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\hline
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QNR & QNR & QR
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\end{tabular}
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\end{table}
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For the last case (QNR, QNR), we have \( x = g^{2k+1}, y=g^{2j+1} \), so \( xy = g^{2(k+j)+2} \) is a square.
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\begin{theorem}[Quadratic Reciprocity]
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For \( p, q \) odd primes, \[
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\left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}
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\]
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\end{theorem}
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\begin{example}
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Suppose for simplicity \( p, q \equiv 1 \pmod{4} \). Then \( \frac{(p-1)(q-1)}{4} \) is even.
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By the theorem, \( \left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = 1 \).
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That is, \( \left( \frac{p}{q} \right) = \left( \frac{q}{p} \right) \).
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That is, \( p \in QR(q) \iff q \in QR(p) \).
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\end{example}
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\begin{remark}
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A supplement to the theorem states that \[
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-1 \in QR(p) \iff p \equiv 1 \pmod{4}
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\]
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\end{remark}
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