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Lecture 17 - 2025-03-17
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@@ -1296,4 +1296,165 @@ We get unique factorization, i.e., prime iff irreducible. \[
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\]
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Suppose we want to compute \( \left( \frac{7}{101} \right) \). We have \[ \left( \frac{7}{101} \right) = \left( \frac{101}{7} \right) = \left( \frac{3}{7} \right) = - \left( \frac{7}{3} \right) = - \left( \frac{1}{3} \right) = -1 \]
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\end{remark}
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\end{remark}
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\section{FLT for \( n = 3, 4 \)}
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\begin{theorem}
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\( x^4 + y^4 = z^4 \) has no solutions in \( \Z \) except \( x = y = z = 0 \).
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\end{theorem}
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\begin{note}
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If \( p \mid x, y, z \), then we can factor it out. Without loss of generality, we can assume \( \gcd(x, y, z) = 1 \).
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\end{note}
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\subsubsection{Divisibility at \( 2 \)}
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We have \( x^4 \equiv 0, 1 \pmod{4} \). We look at modulo \( p \), and we show that \( A + B = C \) has no solutions in \( A, B, C \in \{ x^4, x \in \Z / p \Z \} = Q_4(p) \).
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{~~~}
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In solving for \( x^2 + y^2 = p \), we used \( \Z[i] \) and factorization \( x^2 + y^2 = (x + iy)(x - iy) \). We try something similar, \[
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x^4 + y^4 = (x^2 + iy^2)(x^2 - iy^2) = x^4 \qquad \text{ in } \Z[i]
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\] take \( \eta = x^2 + iy^2 \), \( \bar{\eta} = x^2 - iy^2 \), and we want to solve \[
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\eta \bar{\eta} = z^4 \qquad \text{ in } \Z[i]
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\]
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\begin{remark}
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Primes in \( \Z[i] \)
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\begin{enumerate}[label=(\roman*)]
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\item \( 2i = (1 + i)^2 \), \( 1 + i \) is prime in \( \Z[i] \)
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\item \( p \in \Z \) is \( 1 \pmod{4} \), \( p = \pi \bar{\pi} \), \( \pi, \bar{\pi} \in \Z[i] \) are primes
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\item \( p \in Z \) is \( 3 \pmod{4} \), \( p \) itself is prime in \( \Z[i] \)
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\end{enumerate}
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\end{remark}
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Note that since \( x \) and \( y \) are coprime, \( \gcd(\eta, \bar{\eta}) = 1 \) in \( \Z[i] \).
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Indeed, suppose \( \pi | \eta, \bar{\eta} \), then \( \pi \mid \eta + \bar{eta} = 2x^2 \) and \( \pi \mid \eta - \bar{\eta} = 2iy^2 \) so \( \pi \mid \gcd(2x^2, 2y^2) \) so \( \pi \mid 2 \). Since \( \gcd(\eta, \bar{eta}) \) is bot dividible by \( 1 + i \), we can only have \( \pi = 1 \).
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{~~~}
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We conclude that
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\begin{enumerate}
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\item Away from \( 2 \), \( \eta \) and \( \bar{\eta} \) are both 4th powers in \( \Z[i] \).
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\item If \( p \equiv 3 \pmod{4} \), then \( p \mid \eta \implies p = \bar{p} \mid \bar{\eta} \) so \( p \mid \gcd(\eta, \bar{\eta}) \mid 2 \) which cannot happen.
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\end{enumerate}
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\begin{remark}
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\( z \) cannot be even.
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If \( z \) is even, then we want to solve \( z^4 \equiv 0 \pmod{4} \) so \( ? + ? \equiv 0 \pmod{4} \) where \{ ? = \{ 0, 1 \} \}.
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\end{remark}
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Take \( \pi = a + bi \), and suppose \( \pi \mid \eta \implies \pi^4 \mid \eta = x^2 + y^2 \).
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\( \pi^4 = a^4 + b^4 + 4a^3bi - 4ab^3i - 6a^2b^2 = (a^4 + b^4 - 6a^2b^2) + 4iab(a^2 - b^2) \).
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Let's ignore \( 2 \) and suppose that \( \pi^4 = x^2 + iy^2 \).
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Then, \begin{align*}
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x^4 + b^4 - 6a^2b^2 & = x^2 & \text{ up to some power of } 2 \\
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4ab(x^2 + b^2) & = y^2 \\
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ab(x^2 + b^2) & = y^2 & \text{ ignore } 2
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\end{align*}
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We can assume that \( \gcd(a, b) = 1 \) since \( \gcd(x, y) = 1 \).
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\begin{remark}
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Since \( 4ab(x^2 + b^2) = y^2 \), we must have \( y \) even. Since \( a^2 - b^2 = w^2 \) we must have \( a \) odd and \( b \) even.
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\end{remark}
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We know that \( a, b, a^2 - b^2 \) are all mutually coprime so \( a, b, a^2 - b^2 \) are all squares. \[
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a = u^2 \qquad b = v^2 \qquad a^2 - b^2 = w^2
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\] and we have \[
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u^4 - v^4 = w^2
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\]
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{~~~}
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We now have \begin{align*}
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a^4 + b^4 - 6a^2b^2 & = x^2 \\
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(a^2 - b^2)^2 - 4a^2b^2 & = x^2 \\
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(a^2 - 2ab - b^2)(a^2 + 2ab - b^2) & = x^2
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\end{align*}
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If \( \ell \mid a^2 - 2ab - b^2 \) and \( \ell \mid a^2 + 2ab - b^2 \), then \( \ell \mid 2(a^2 - b^2), 4ab \) (the sum and difference of the two terms)
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\begin{itemize}
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\item If \( \ell = 2 \), \[
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2 \mid a^2 - b^2
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\] nad since \( a \) odd \( b \) even, this cannot happen.
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\item If \( \ell \neq 2 \),\[
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\ell \mid ab \implies \ell \mid a \text{ or } b
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\] and \[
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\ell \mid a^2 - b^2 \implies \ell \mid a \text{ and } b
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\] so \[
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\ell = 1
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\]
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\end{itemize}
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% Now \( a^2 - 2ab - b^2 \) is coprime to \( a^2 + 2ab - b^2 \). Thus, \[
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% a^2 - 2ab - b^2 = c^2 \qquad \text{ for some } c \in \Z
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% \] so \[
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% b^2 + 2ab + (c^2 - a^2) = 0
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% \] We solve for \( b \) and obtain \[
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% = \frac{-2a \pm \sqrt{4a^2 - 4(C^2 - a^2)}}{2} = -a \pm \sqrt{2C^2}
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% \]
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TODO: See Picture
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\subsection{Review: M\"obius Inversion}
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Suppose \[
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f(n) = \sum_{d \mid n} g(d) \qquad \forall n \in \N_{\geq 1}
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\] then \[
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g(n) = \sum_{d \mid n} \mu\left( \frac{n}{d} \right) f(d)
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\]
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Think of the \( g(d) \) as variables, \( f(n) \) as knowns, and \( f(n) = \sum_{d \mid n} g(d) \) as a system of linear equations.
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The simpliest equations are uppar-triangular. We conclude that this system is indeed uppar triangular.
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\begin{itemize}
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\item For \( n = 1 \), \( g(1) = f(1) \)
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\item For \( n = p \), \( f(p) = g(1) + g(p) \) so \[
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g(p) = f(p) - f(1) \qquad \text{ since } f(1) = g(1)
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\]
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\item For \( n = p^2 \), \( g(p^2) + g(p) + g(1) = f(p^2) \) so \[
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g(p^2) = f(p^2) - f(p) \qquad \text{ since } f(p) = g(1) + g(p)
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\]
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For \( n = pq \) distinct primes, \( g(pq) + g(p) + g(q) + g(1) = f(pq) \) so \[
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g(pq) = f(pq) - f(p) - f(q) + f(1)
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\]
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\end{itemize}
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\begin{proof}
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Because the equations are triangular, there is at most one solution. Sufficies to verify that the solution works.
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\begin{align*}
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\sum_{d \mid n} g(d)
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& = \sum_{d \mid d} \sum_{e \mid n = d} f(e) \mu(m)
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\\
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& = \sum_{e \mid n} f(e) \sum_{m \mid n/e} \mu(m)
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\end{align*}
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Take \( N = n/e \), WTS \[
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\sum_{m \mid N} \mu(m) = \begin{cases}
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1 & \text{if } N = 1 \\
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0 & \text{otherwise}
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\end{cases}
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\]
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It's easy to prove since \( \mu \ast 1 \left( = \sum_{m \mid N} \mu(m) \right) \) is multiplicative, so check on prime powers \( N = p^k \), \[
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\sum_{m \mid N} \mu(m) = \begin{cases}
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\mu(1) - \mu(p) = 0 & \text{ if } k \geq 1 \\
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\mu(1) = 1 & \text{ if } k = 0
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\end{cases}
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\]
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\end{proof}
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