User:Cmerkel/Sandbox

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\title[]{Math 110, Homework 4}% \author{Celesse Merkel}% \address{Dept. of Mathematics, University of California, Santa Cruz, CA 95064}% \email{cmerkel AT ucsc DOT edu}%

%\date{}% %\dedicatory{}% %\commby{}% % \maketitle % \section{Bounding Solutions}

In this section, we prove the following: \begin{thm} Suppose that $a$, $b$, and $c$ are positive integers. Then the number of solutions to the Diophantine equation $ax^2 + by^2 = c$ is less than $(2\sqrt{\frac{c}{a}} + 1) (2 \sqrt{\frac{c}{b}} + 1)$. \end{thm} % Replace $N$ by a a formula involving $a$, $b$, and $c$. You have to find a correct formula! There are many possible answers. \proof By the equation $ax^2 + by^2 = c$ we find that $x^2\leq \frac{c}{a}$, hence, $-\sqrt{\frac{c}{a}}\leq x \leq \sqrt{\frac{c}{a}}$. Also, $y^2\leq \frac{c}{b}$, hence, $-\sqrt{\frac{c}{b}}\leq y \leq \sqrt{\frac{c}{b}}$. After finding these bounds we can calculate the total number of possible $x's$ and $y's$. So we get $2\sqrt{\frac{c}{a}} + 1$ possible $x's$ and $2 \sqrt{\frac{c}{b}} + 1$ possible $y's$. By multiplying the these expressions together we find that the number of solutions so the given equation is less than $(2\sqrt{\frac{c}{a}} + 1) (2 \sqrt{\frac{c}{b}} + 1)$.

% Prove the theorem, using the formula you gave. % Just guessing a correct formula is worth one point. Proving it is worth two more points. % A very good bound (as small as possible) is worth one more point. \qed

\section{A dry quadratic form} In this section, we prove the following: \begin{thm} If $x,y \in \ZZ$, and $11x^2 - 5y^2 = 0$, then $x = 0$ and $y = 0$. \end{thm} \proof Suppose $x,y\in\ZZ$ and $11x^2-5y^2=0$ then $\frac{y^2}{x^2} = \frac{11}{5}$ or $x=0$ which implies $y=0$. Let $z^2=\frac{y^2}{x^2}=\frac{11}{5}$. Hence $z=\frac{y}{x}$. Now, let $e_{-1}, e_{2}, e_{3}, e_{5},\ldots$ be the exponents in the canonical decomposition of $z$. where $e_{-1}\in\left\lbrace 0,1\right\rbrace $ and $e_{p}\in\ZZ$ for every prime p. So $z=(-1)^{e_{-1}} 2^{e_{2}} 3^{e_{3}} 5^{e_{5}} 7^{e_{7}} 11^{e_{11}}\cdots$ and $z^2=(-1)^{2e_{-1}}2^{2 e_{2}}3^{2e_{3}}5^{2e_{5}}7^{2e_{7}}11^{2e_{11}}\cdots=\dfrac{11}{5}=(-1)^{0}2^{0}3^{0}5^{-1}7^{0}11^{1}$. Since the exponents in the canonical decomposition of $z$ are even, by the definition of canonical decomposition, the exponents in the canonical decomposition of $\dfrac{11}{5}$ must be even as well. Hence, $-1=2s$ for some $s\in\ZZ$ and $1=2t$ for some $t\in\ZZ$. But this is a contradiction. Thus $x=0$ and $y=0$. % Prove the theorem. % Use what you know about rationality and irrationality, e.g., the ``n-q-z'' theorem. % A clear and concise proof is worth three points. \qed

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