\documentstyle[12pt]{article} \begin{document} \title{Analysis of the ratio $\frac{\sigma(n)}{n}$} \author{Kurt Ludwick} \date{Spring, 1994} \maketitle {\bf Introduction} \\ I began this thesis by researching the topic of odd perfect numbers. Before long, I found that much thought has been given to this topic by mathematicians, both in the present and in the past. Hundreds of years ago, in fact, Euler showed that an odd perfect number would have to have a prime factorization of the form $$n=p^{\alpha}q_1^{2\beta_1}q_2^{2\beta_2}...q_g^{2\beta_g},$$ where $p, q_1...q_g$ are distinct primes, ${\alpha}$, ${\beta_1}$...${\beta_g}$ are positive integers, and $p\equiv {\alpha}\equiv 1 \pmod{4}$. \\ More recent results have shown that if an odd perfect number exists, its largest prime factor is at least 100129, and its second-largest prime factor is at least 1000. Also, an odd perfect number must have at least eight distinct prime factors; if it is not divisible by 3, then it must have at least eleven distinct prime factors. \\ After convincing myself that I wouldn't be able to either find an odd perfect number or to prove that none exists anytime soon, I re-examined Euler's result. I realized it was saying that $$\frac{{\sigma}(n)}{n}=\frac{\sigma(p^{\alpha})}{p^{\alpha}}\frac{\sigma(q_1^{2 \beta_1})} {q_1^{2\beta_1}}...\frac{\sigma(q_g^{2\beta_g})}{q_g^{2\beta_g}} = 2.$$ The problem is finding the primes and exponents which result in a product of exactly 2. I decided to modify this problem, to look for other such products. For example, one could search for positive integers n such that: $$\frac{\sigma(n)}{n}=\frac{7}{2}.$$ This is just an example; we could search for positive integers satisfying the above equation for any rational number. This paper deals with the problem: $$\frac{\sigma(n)}{n}=\frac{a}{b},$$ where $(a,b) = 1$ and $a>b$ \\ This problem is interesting on its own, and it also has some potential usefulness in the search for an odd perfect number (or the proof that none exists.) \\ In the next section, several interesting and useful results are listed and proven. These results are useful in finding solutions to the above problem. \newpage {\bf \Large \noindent Results} \\ In this section, we will assume $a,b,c,g,h,i,j,k,l,m,n$ to be positive integers, and $p,q$ to be primes, unless otherwise stated. \\ {\bf \large \noindent Lemma 1:} If $a\frac{{\sigma}(a)}{a}$. \\ Proof: If $p\not|a$, then $(a,p)=1$, so $$\frac{{\sigma}(b)}{b}=\frac{{\sigma}(a)}{a}\frac{{\sigma}(p^h)}{p^h}>\frac{{\sigma}(a)}{a}.$$ If $p|a$, then there exist $j,k$ such that $a=jp^k$, and $(j,p)=1$ (where $k$ is the largest power of $p$ which divides $a$). Then, $b=jp^{k+h}$, so $$\frac{{\sigma}(b)}{b}=\frac{{\sigma}(j)}{j}\frac{{\sigma}(p^{k+h})}{p^{k+h}}. $$ Therefore, $$\frac{{\sigma}(b)}{b}-\frac{{\sigma}(a)}{a}=\frac{{\sigma}(j)}{j} \left[\frac{{\sigma}(p^{k+h})}{p^{k+h}}-\frac{{\sigma}(p^k)}{p^k}\right].$$ We know $k>0$, so $h+k>h$; thus, this difference is positive (by Lemma 1), and so $\frac{{\sigma}(b)}{b}>\frac{{\sigma}(a)}{a}$. \rule{2mm}{2mm} \\ {\bf \large \noindent Corollary 2a:} If $a|b$, then $\frac{{\sigma}(b)}{b}\geq\frac{{\sigma}(a)}{a}$ (with equality if and only if $a=b$). \\ Proof: $a|b$, so $b=ak$ for some $k\in {\bf Z^+}$. If $k=1$ (meaning $a=b$), then clearly $\frac{{\sigma}(b)}{b}=\frac{{\sigma}(a)}{a}$. It remains to be shown that when $a|b$ and $a\not=b$ (that is, when $k>1$), $\frac{{\sigma}(b)}{b}>\frac{{\sigma}(a)}{a}$. If $k>1$, then we can write the prime factorization of $k$ as $$k=p_1^{l_1}p_2^{l_2}...p_g^{l_g}$$ (so that $g$ is the number of distinct prime factors of $k$). By Lemma 2, $$\frac{{\sigma}(a)}{a}<\frac{{\sigma}(ap_1^{l_1})}{ap_1^{l_1}}<\frac{{\sigma}(a p_1^{l_1}p_2^{l_2})}{ap_1^{l_1}p_2^{l_2}}<...$$ By repeating this process, we end up with $$\frac{{\sigma}(a)}{a}< \frac{{\sigma}(ap_1^{l_1}p_2^{l_2}...p_g^{l_g})}{ap_1^{l_1}...p_g^{l_g}}.$$ But the right side of this inequality is just $\frac{{\sigma}(ak)}{ak}$, which is $\frac{{\sigma}(b)}{b}$. Thus, we have $\frac{{\sigma}(b)}{b}>\frac{{\sigma}(a)}{a}$. \rule{2mm}{2mm} \\ {\bf \large \noindent Corollary 2b:} If $a|b$, and $a\not=b$ (that is, $a\frac{a}{b}$, then there exists no positive integer n such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$. \\ Proof: Contrapositive of Lemma 5. \rule{2mm}{2mm} \\ {\bf \large \noindent Lemma 6:} Let $p$ be any prime integer, and $k$ any positive integer. Then $(p,{\sigma}(p^k))=1$. \\ Proof: We'll prove this lemma by finding integers u,v, such that $up+v{\sigma}(p^k)=1$. In particular, let $u=-{\sigma}(p^{k-1})$, and let $v=1$. Then, \begin{eqnarray*} up &=&-(1+p+p^2+...+p^{k-1})p\\ &=&-(p^k+p^{k-1}...+p)\\ &=&-({\sigma}(p^k)-1)\\ &=&1-{\sigma}(p^k). \end{eqnarray*} Thus, \begin{eqnarray*} up+v{\sigma}(p^k) &=&(1-{\sigma}(p^k))+{\sigma}(p^k)\\ &=&1. \end{eqnarray*} \rule{2mm}{2mm} \\ {\bf \large \noindent Lemma 7:} Let $a,b,c\in {\bf Z}^+$, and let $p$ be a prime integer, such that $(a,p)=1$, $b=p^c$ (so that $b$ is a prime power, and $(a,b)=1$), and $a\geq{\sigma}(b)$. Suppose we want to find a positive integer, $n$, such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$. This problem is equivalent to the problem of finding positive integers $m,k$ such that: \\ \begin{enumerate} \item $n=mp^k$, $k\geq c$, and (m,p)=1 (or equivalently, (m,b)=1). \item $\frac{{\sigma}(m)}{m}=\frac{ap^{k-c}}{{\sigma}(p^k)}$. \end{enumerate} Proof: First, we will show that if we can find $n\in {\bf Z}^+$ satistying $\frac{{\sigma}(n)}{n}=\frac{a}{b}$, then we can find $m,k\in {\bf Z}^+$ satisfying (1) and (2) above. Suppose we have $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$, By Lemma 3, $b|n$; that is, $p^c|n$. Since b is a prime power, there is some $k\in{\bf Z}^+$ such that $n=mp^k$, with $(m,b)=1$. Here, $k$ is the largest power of $p$ that divides $n$, so clearly $k\geq c$. This satisfies (1). Now, we have $$\frac{{\sigma}(n)}{n}=\frac{a}{b}=\frac{a}{p^c},$$ and we also have $$\frac{{\sigma}(n)}{n}=\frac{{\sigma}(p^k)}{p^k}\frac{{\sigma}(m)}{m}.$$ Thus, $$\frac{a}{p^c}=\frac{{\sigma}(p^k)}{p^k}\frac{{\sigma}(m)}{m},$$ and so $$\frac{{\sigma}(m)}{m}=\frac{ap^{k-c}}{{\sigma}(p^k)}.$$ This satisfies (2). Therefore, solving $\frac{{\sigma}(n)}{n}=\frac{a}{b}$ for n gives us $m,k\in {\bf Z}^+$ satisfying (1) and (2). \\ Conversely, we will show that if we can find $m,k\in {\bf Z}^+$ satisfying (1) and (2), then we can find $n\in {\bf Z}^+$ such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$. Suppose we have $m,k\in {\bf Z^+}$ safisfying (1) and (2). Let $n=mp^k$. Then, \begin{eqnarray*} \frac{{\sigma}(n)}{n} &=&\frac{{\sigma}(m)}{m}\frac{{\sigma}(p^k)}{p^k}\\ &=&\frac{ap^{k-c}}{{\sigma}(p^k)}\frac{{\sigma}(p^k)}{p^k}\\ &=&\frac{ap^{k-c}}{p^k}=\frac{a}{p^c}=\frac{a}{b}. \end{eqnarray*} \rule{2mm}{2mm} \\ (Note: In Lemma 7, we require that $a\geq{\sigma}(b)$ so that a solution to the given problem can exist. If we were to allow $a<{\sigma}(b)$, then Lemma 4 shows that no solution could exist. Since we are interested in finding solutions to the problem, we only consider the case of $a\geq{\sigma}(b)$ in Lemma 7.) \newpage {\bf \large \noindent Corollary 7a:} Let $a,b\in {\bf Z}^+$, such that $b$ is prime, $(a,b)=1$, and $a\geq b+1$ (or equivalently, $a\geq{\sigma}(b)$). Then the problem of finding $n\in {\bf Z}^+$ such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$ is equivalent to the problem of finding $m,k\in {\bf Z}^+$ such that: \\ \begin{enumerate} \item $n=mb^k$, and $(m,b)=1$. \item $\frac{{\sigma}(m)}{m}=\frac{ab^{k-1}}{{\sigma}(b^k)}$. \end{enumerate} Proof: This is simply the special case of Lemma 7 with $c=1$ (and $p=b$). \rule{2mm}{2mm} \\ (Note: Corollary 7a is of more use for the purposes of this paper than the more generalized Lemma 7. Usually, $b$ will be a prime, and so it will be more practical to simply refer to this corollary rather than the lemma from which it is derived.) \\ {\bf \large \noindent Lemma 8:} For $n\in {\bf Z^+}$, ${\sigma}(n)$ is odd if and only if $n$ is a perfect square times a power of 2 (including $2^0=1$). That is, ${\sigma}(n)$ is odd if and only if $n=2^hm$, where $h$ is a non-negative integer and $m$ is a perfect square. \\ Proof: We can write the prime factorization of $n$ as $$n=2^hp_1^{\alpha_1}p_2^{\alpha_2}...p_g^{\alpha_g}.$$ Thus, $${\sigma}(n)={\sigma}(2^h){\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})... {\sigma}(p_g^{\alpha_g}).$$ For any $h\in {\bf Z^+}$, \begin{eqnarray*} {\sigma}(2^h) &=&1+2+2^2+...+2^k\\ &=&1+2(1+2+...+2^{h-1}), \end{eqnarray*} which is clearly odd. Also, ${\sigma}(2^0)=1$, so ${\sigma}(2^h)$ is odd for all non-negative integers $h$. Therefore, ${\sigma}(2^h)$ times any odd number is odd (and of course, ${\sigma}(2^h)$ times any even number is even). This means that $${\sigma}(2^h){\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})...{\sigma}(p_g^{ \alpha_g})$$ is odd if and only if $${\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})...{\sigma}(p_g^{\alpha_g})$$ is odd; that is, ${\sigma}(n)$ is odd if and only if $${\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})...{\sigma}(p_g^{\alpha_g})$$ is odd. Now, to prove this lemma, we only need to show that $${\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})...{\sigma}(p_g^{\alpha_g})$$ is odd if and only if $$p_1^{\alpha_1}p_2^{\alpha_2}...p_g^{\alpha_g}$$ is a perfect square. \\ For any $i\in {\bf Z^+}$, $1\leq i\leq g$, \begin{eqnarray*} {\sigma}(p_i^{\alpha_i}) &\equiv& 1+p_i+p_1^2+...+p_i^{\alpha_i}\\ &\equiv& 1+1+...+1\\ &\equiv& ({\alpha_i}+1)\pmod{2} \end{eqnarray*} (since $p_i$ is odd, and thus $p_i\equiv 1\pmod{2}$). Thus, ${\sigma}(p_i^{\alpha_i})$ is odd if and only if ${\alpha_i}\equiv 0\pmod{2}$; that is, if and only if ${\alpha_i}$ is even. Now, for ${\sigma}(n)$ to be odd, we need every ${\sigma}(p_i^{\alpha_i})$ to be odd; if just one were even, then the entire product $${\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})...{\sigma}(p_g^{\alpha_g})$$ would be even. That is, $${\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})...{\sigma}(p_g^{\alpha_g})$$ is odd if and only if every ${\alpha_i}$ is even, $1\leq i\leq g$. But every ${\alpha_i}$ is even if and only if $$p_1^{\alpha_1}p_2^{\alpha_2}...p_g^{\alpha_g}$$ is a perfect square; thus, $${\sigma}(p_1^{\alpha_1}){\sigma}(p_2^{\alpha_2})...{\sigma}(p_g^{\alpha_g})$$ is odd if and only if $$p_1^{\alpha_1}p_2^{\alpha_2}...p_g^{\alpha_g}$$ is a perfect square. Therefore, ${\sigma}(n)$ is odd if and only if $n$ is of the form $n=2^hm$, where $h$ is a non-negative integer and $m$ is a perfect square. \rule{2mm}{2mm} \newpage {\bf \large \noindent Examples} \\ Given $a,b\in {\bf Z}^+$ such that $(a,b)=1$, $b=p^c$ for some prime integer $p$, and $a\geq{\sigma}(b)$, we can now reduce the problem of finding $n\in {\bf Z}^+$ such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$ to the problem of finding $m,k\in {\bf Z}^+$ satisfying the conditions given in Lemma 7. \\ {\bf \large \noindent Example 1:} Find $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{7}{2}$. \\ This problem, is equivalent to the problem of finding $m,k\in {\bf Z}^+$ such that: \\ \begin{enumerate} \item $n=2^km$, where $(2,m)=1$ (that is, $k$ is the largest power of 2 to divide $n$). \item $\frac{{\sigma}(m)}{m}=\frac{7\cdot2^{k-1}}{{\sigma}(2^k)}$. \end{enumerate} We will now check different values of $k\in {\bf Z^+}$, attempting each time to find $m\in {\bf Z}^+$ satisfying these conditions subject to the choice of $k$. For each $k$, we will proceed until one of the following happens: \\ \begin{enumerate} \item We find $m\in {\bf Z}^+$ satisfying (1) and (2). In this case, $n=2^km$ is a solution to our problem. \item We prove that there is no $m\in {\bf Z}^+$ satisfying (1) and (2). In this case, there is no solution to our problem of the form $n=2^km$, where $(2,m)=1$. \item The problem becomes impractical to pursue. Often a given value of $k$ will leave us with an problem which either {\it can't} be solved with this method, or which is too complicated to be solved in a reasonable amount of time. \end{enumerate} \newpage We'll start with $k=1$: Then our conditions are: \\ \begin{enumerate} \item $n=2m$, and (m,2)=1. \item $\frac{{\sigma}(m)}{m}=\frac{7\cdot2^{1-1}}{{\sigma}(2^1)}=\frac{7}{3}$ \end{enumerate} Thus, our problem is now to find $m\in {\bf Z}^+$ satisfying (1) and (2). Let's carry the process one step further for the case of $k=1$. To do this, we will treat m in the same manner in which we initially treated n. Our goal is to find $m_1,k_1\in {\bf Z^+}$ such that: \\ \begin{enumerate} \item $m=3^{k_1}m_1$, and $(m_1,3)=1$. (Note that since $m_1|m$, and $(m,2)=1$, we actually need $(m_1,6)=1$.) \item $\frac{{\sigma}(m_1)}{m_1}=\frac{7\cdot3^{k_1-1}}{{\sigma}(3^k_1)}$. \end{enumerate} Let's now check the case of $k=1, k_1=1$. Our goal in this case is to find $m_1\in {\bf Z}^+$ such that: \\ \begin{enumerate} \item $m=3m_1$, and $(m_1,6)=1$. \item $\frac{{\sigma}(m_1)}{m_1}=\frac{7}{4}$. \end{enumerate} Thus, we want to find some $m_1\in {\bf Z}^+$ such that $(m_1,6)=1$ and $\frac{{\sigma}(m_1)}{m_1}=\frac{7}{4}$. However, if $\frac{{\sigma}(m_1)}{m_1}=\frac{7}{4}$, then $4|m_1$ (by Lemma 3). If $4|m_1$, then $(m_1,6)\not=1$. Therefore, there is no such $m_1\in {\bf Z}^+$. Our method has shown that, in the case of $k=1, k_1=1$, there is no $n\in {\bf Z}^+$ which solves our problem. In particular, $n$ is not of the form \begin{eqnarray*} n &=&2^1m\\ &=&2^1(3^1(m_1))\\ &=&6m_1 \end{eqnarray*} where $(m_1,6)=1$. (Another way to say this is that 6 is not a {\it unitary divisor} of any solution to our problem.) (Note that this does not prove the non-existence of a solution to our problem; it only disproves the existence of a solution of the form given in the last paragraph. In order to disprove the existence of a solution of any given problem, it would have to be shown that no solution existed for {\it any} value of $k$. Here we have not even eliminated the case of $k=1$, but only the special case of this where $k_1=1$.) \\ Let us move on now to $k=2$: Our goal is to find $m\in {\bf Z}^+$ such that: \\ \begin{enumerate} \item $n=2^2m=4m$, with $(m,2)=1$ \item $\frac{{\sigma}(m)}{m}=\frac{7\cdot2^{2-1}}{{\sigma}(2^2)}=\frac{14}{7}=2$. \end{enumerate} Here we find that, for the case of $k=2$, we must find $m\in {\bf Z^+}$ such that $m$ is odd and $\frac{{\sigma}(m)}{m}=2$; that is, $m$ must be an odd perfect number. If $m$ is an odd perfect number, then $n=4m$ is a solution to our problem. This is not especially helpful in our search for a solution, so we'll move on to another case. \\ Here we will skip the cases $k=3$ and $k=4$, because they are not especially interesting compared to the next case we will deal with. \\ Consider $k=5$: Our goal now is to find $m\in {\bf Z}^+$ such that: \\ \begin{enumerate} \item $n=2^5m=32m$, and $(m,2)=1$. \item $\frac{{\sigma}(m)}{m}=\frac{7\cdot2^4}{{\sigma}(2^5)}=\frac{112}{63}=\frac{16} {9}$ \end{enumerate} We can now apply our method to m. Keep in mind that the process will be slightly different this time, since the denominator of $\frac{16}{9}$ is a prime power, not just a prime, so we will turn to the more general Lemma 7 rather than Corollary 7a (which we have used exclusively until now). Here, $b=9=3^2$, so we will use $p=3$ and $c=2$ (as they are used in Lemma 7). Our goal is to find $m_1,k_1\in {\bf Z}^+$ such that: \\ \begin{enumerate} \item $m=m_1\cdot3^{k_1}, k_1\geq 2$, and $(m_1,3)=1$. (Note that $m_1|m$, and $(m,2)=1$, so $(m_1,6)=1$.) \item $\frac{{\sigma}(m_1)}{m_1}=\frac{16\cdot3^{k_1-2}}{{\sigma}(3^{k_1})}$ \end{enumerate} We will consider 2 of the possible cases here: $k_1=2$ and $k_1=3$. \\ First, let $k_1=2:$ \\ $$\frac{{\sigma}(m_1)}{m_1}=\frac{16}{{\sigma}(3^2)}=\frac{16}{13}.$$ Carrying the process one step further, we will search such an $m_1\in {\bf Z}^+$. We must find $m_2,k_2\in {\bf Z}^+$ such that: \\ \begin{enumerate} \item $m_1=m_2\cdot13^{k_2}$, and $(m_2,13)=1$. (Note that $m_2|m_1$, so $(m_2,6)=(m_2,13)=1$.) \item $\frac{{\sigma}(m_2)}{m_2}=\frac{16\cdot13^{k_2-1}}{{\sigma}(13^{k_2})}$. \end{enumerate} Let $k_2=1$. Then $m_1=13m_2$, and $$\frac{{\sigma}(m_2)}{m_2}=\frac{16}{14}=\frac{8}{7}.$$ Let $m_2=7$. Then $m_2$ satisfies both (1) and (2) of the case $k=5, k_1=2, k_2=1$. This gives us a solution; all we have to do now is work backwards until we get $n$. First, $m_1=13m_2=13\cdot7$. Next, $m=3^{k_1}m_1=13\cdot7\cdot3^2$. Finally, \begin{eqnarray*} n &=&2^5m\\ &=&2^5\cdot3^2\cdot7\cdot13\\ &=&26208. \end{eqnarray*} Thus, we have found a solution to the problem $\frac{{\sigma}(n)}{n}=\frac{7}{2}$. \\ Now we will consider the case $k_1=3$, which will give us one more solution: \\ $$\frac{{\sigma}(m_1)}{m_1}=\frac{16\cdot3}{{\sigma}(3^3)}=\frac{48}{40}=\frac{6 }{5}.$$ That is, we need to find $m_1\in {\bf Z}^+$ such that $\frac{{\sigma}(m_1)}{m_1}=\frac{6}{5}$, and $(m_1,6)=1$. If we let $m_1=5$, then we have solved this problem, and have thus discovered another solution of the problem $\frac{{\sigma}(n)}{n}$, this time for the case $k=5, k_1=3$. Again, we will now work backwards until we get $n$. First, $m=3^{k_1}m_1=3^3\cdot5$. Next, \begin{eqnarray*} n &=&2^5m\\ &=&2^5\cdot3^3\cdot5\\ &=&4320. \end{eqnarray*} This is a second solution to the problem $\frac{{\sigma}(n)}{n}=\frac{7}{2}$. \\ We have used this method to find two solutions to the problem $\frac{{\sigma}(n)}{n}=\frac{7}{2}$. Another usage is to prove that, for certain values of $k$, there is no $m\in {\bf Z^+}$ such that $n=mb^k$ (or $n=mp^k$, in the case that b is a prime power, and $b=p^c$) is a solution to the problem. \\ {\bf \large \noindent Example 2:} Find values of $k\in {\bf Z^+}$ such that the problem $\frac{{\sigma}(n)}{n}=\frac{7}{2}$ has no solution of the form $n=2^km$, where $(m,2)=1$. \\ In this example, we work with the same problem as we worked with in Example 1, but with a different purpose in mind. Here we want to show that, for certain values of $k$, there is no solution. The goal here is to find a value of k such that no $m\in {\bf Z^+}$, with $(m,2)=1$, can satisfy the equation $\frac{{\sigma}(m)}{m}=\frac{7\cdot2^{k-1}}{{\sigma}(2^k)}$. After some trial-and-error calculations, it was found that $k=11$ worked nicely. To illustrate, let us consider that case. \\ {\bf \large \noindent Claim:} There is no solution to the problem when $k=11$. \\ Proof: If there is a solution to the problem for $k=11$, then there must exist $m\in {\bf Z^+}$ such that \begin{eqnarray*} \frac{{\sigma}(m)}{m} &=&\frac{7\cdot2^{10}}{{\sigma}(2^{11})}=\frac{7\cdot2^{10}}{4095}\\ &=&\frac{1024}{585}. \end{eqnarray*} However, ${\sigma}(585)=1092$, so ${\sigma}(585)>1024$. Thus, by Corollary 4a, there is no $m\in {\bf Z^+}$ such that $\frac{{\sigma}(m)}{m}=\frac{1024}{585}$. Therefore, there is no solution to the problem when $k=11$. \rule{2mm}{2mm} \\ Now notice that, for $k\equiv -1\pmod{12}$, ${\sigma}(2^k)={\sigma}(2^{11})\cdot(1+2^{12}+...+2^{12(h-1)})$, where $k=12h-1$. In particular, whenever $k\equiv -1\pmod{12}$, ${\sigma}(2^{11})|{\sigma}(2^k)$. We can use this fact to prove the following: \\ {\bf \large \noindent Claim:} There is no solution to the problem when $k\equiv -1\pmod{12}$. \\ Proof: If we write $k=12j-1$, then we have the equation \begin{eqnarray*} \frac{{\sigma}(m)}{m} &=&\frac{7\cdot2^{k-1}}{{\sigma}(2^k)}\\ &=&\frac{7\cdot2^{12j-2}}{{\sigma}(2^{11})\cdot(1+2^{12}+...+2^{12(j-1)})}\\ &=&\frac{7\cdot2^{12j-2}}{4095\cdot(1+2^{12}+...+2^{12(j-1)})}\\ &=&\frac{2^{12j-2}}{585\cdot(1+2^{12}+...+2^{12(j-1)})}. \end{eqnarray*} Now let us apply Corollary 5a: let $a=2^{12j-2}$, let $b=585(1+2^{12}+...+2^{12(j-1)})$, and let $h=585$. Thus, we have $a,b,h\in {\bf Z^+}$ such that $(a,b)=1$ (since $b$ is odd and $a$ is a power of 2) and $h|b$, so we can use Corollary 5a, which states that, if $\frac{{\sigma}(h)}{h}>\frac{a}{b}$, then there is no $m\in {\bf Z^+}$ such that $\frac{{\sigma}(m)}{m}=\frac{a}{b}$. Notice that \begin{eqnarray*} \frac{a}{b} &=&\frac{2^{12j-2}}{585(1+2^{12}+...+2^{12(j-1)})}\\ &\leq&\frac{2^{12j-2}}{585\cdot2^{12(j-1)}}\\ &=&\frac{2^{10}}{585}. \end{eqnarray*} Thus, $\frac{a}{b}\leq\frac{1024}{585}$. But \begin{eqnarray*} \frac{{\sigma}(h)}{h} &=&\frac{{\sigma}(585)}{585}\\ &=&\frac{1092}{585}\\ &>&\frac{1024}{585}\geq\frac{a}{b}. \end{eqnarray*} That is, $\frac{{\sigma}(h)}{h}>\frac{a}{b}$, so there is no $m\in {\bf Z^+}$ such that $$\frac{{\sigma}(m)}{m}=\frac{2^{12j-2}}{585\cdot(1+2^{12}+...+2^{12(j-1)})}.$$ Thus, we cannot solve the original equation for n when $k\equiv -1 \pmod{12}$. \rule{2mm}{2mm} \\ {\bf \large \noindent Example 3:} Consider the problem: $\frac{{\sigma}(n)}{n}=\frac{5}{3}$. \\ I found no solution to this problem; however, it is still of interest. In fact, we will show that any solution to this problem is one-fifth of an odd perfect number. \\ The problem is equivalent to the problem of finding $m,k\in {\bf Z^+}$ such that: \\ \begin{enumerate} \item $n=3^km$, where $(m,3)=1$ \item $\frac{{\sigma}(m)}{m}=\frac{5\cdot3^{k-1}}{{\sigma}(3^k)}$ \end{enumerate} As I mentioned, we will not find a solution to this problem. We will, however, prove the following: \\ {\bf \large \noindent Claim:} If there is a solution of the form $n=3^km$, with $(3,m)=1$, then $k$ is an even integer. \\ Proof: By Lemma 7, $\frac{{\sigma}(m)}{m}=\frac{5\cdot3^{k-1}}{{\sigma}(3^k)}$. We will show that $k$ cannot be an odd integer. \\ First, suppose $k\equiv -1\pmod{4}$ (where $k\in {\bf Z^+}$, and so $k\geq 3$). Since $4|(k+1)$, ${\sigma}(3^k)={\sigma}(3^3)(1+3^4+...+3^{k-3})$. Thus, \begin{eqnarray*} {\sigma}(3^k) &=&40(1+81+...+81^{(k-3)/4})\\ &=&40\cdot\frac{81^{1+(k-3)/4}}{81-1}\\ &=&\frac{1}{2}\cdot(81^{(k+1)/4}-1). \end{eqnarray*} Therefore, \begin{eqnarray*} \frac{{\sigma}(m)}{m} =\frac{5\cdot3^{k-1}}{{\sigma}(3^k)} =\frac{5\cdot3^{k-1}}{\left[\frac{81^{(k+1)/4}-1}{2}\right]} =\frac{3^{k-1}}{\frac{1}{10}\cdot(81^{(k+1)/4}-1)}. \end{eqnarray*} Let $a=3^{k-1}$, $b=\frac{1}{10}\cdot(81^{(k+1)/4}-1)$, and $h=8$. Note that $3\not|(81^{(k+1/4)}-1)$, so clearly $3\not|b$; so, since $a$ is a power of 3, $(a,b)=1$. $81\equiv 1\pmod{80}$, so $80|(81^{\frac{k+1}{4}}-1)$, and $8|(\frac{1}{10}\cdot(81^{\frac{k+1}{4}}-1))$; that is, $8|b$, meaning $h|b$. Thus, we have $a,b,h\in {\bf Z^+}$ such that $(a,b)=1$ and $h|b$. Now, if we can show that $\frac{{\sigma}(h)}{h}>\frac{a}{b}$, then we will have shown, by Corollary 5a, that there is no $m\in {\bf Z^+}$ such that $\frac{{\sigma}(m)}{m}=\frac{a}{b}$, and thus no solution to the problem $\frac{{\sigma}(n)}{n}=\frac{5}{3}$ of the form $n=3^km$, when $k\equiv -1\pmod{4}$. We have $$\frac{a}{b}=\frac{3^{k-1}}{\frac{1}{10}\cdot(81^{(k+1)/4}-1)}=\frac{10\cdot3^ {k-1}}{3^{k+1}-1}.$$ Notice that \begin{eqnarray*} \frac{3}{5}\cdot\frac{a}{b} &=&\frac{3}{5}\cdot\frac{10\cdot3^{k-1}}{3^{k+1}-1}\\ &=&\frac{2\cdot3^k}{3^{k+1}-1}\\ &=&\frac{2}{3-3^{-k}}<\frac{2}{3-1}\\ &=&1. \end{eqnarray*} Thus, $\frac{3}{5}\cdot\frac{a}{b}<1$, so $\frac{a}{b}<\frac{5}{3}$. But $\frac{{\sigma}(h)}{h}=\frac{{\sigma}(8)}{8}=\frac{15}{8}>\frac{5}{3}$, so $\frac{a}{b}<\frac{{\sigma}(h)}{h}$. Therefore, there is no solution when $k\equiv -1\pmod{4}$ (by Corollary 5a). \\ Now, suppose $k\equiv 1\pmod{4}$ (where $k\in {\bf Z^+}$, and so $k\geq 1$). Since $2|(k+1)$, \begin{eqnarray*} {\sigma}(3^k) &=&(1+3)(1+3^2+3^4+...+3^{k-1})\\ &=&4(1+9+9^2+...+9^{(k-1)/2})\\ &=&4\cdot\frac{9^{(k+1)/2}-1}{9-1}\\ &=&\frac{1}{2}\cdot(9^{(k+1)/2}-1). \end{eqnarray*} Therefore, $$\frac{{\sigma}(m)}{m}=\frac{5\cdot3^{k-1}}{\frac{1}{2}\cdot(9^{\frac{k+1}{2}} -1)}.$$ Let $a=5\cdot3^{k-1}$, $b=\frac{1}{2}\cdot(9^{(k+1)/2}-1)$, and $h=4$. Now, $(9^{(k+1)/2}-1)\equiv 2\pmod{3}$, so $b\equiv\frac{1}{2}\cdot(9^{(k+1)/2}-1)\equiv 1\pmod{3}$. Also, $9\equiv -1\pmod{5}$, and since $\frac{k+1}{2}$ is odd, $(9^{\frac{k+1}{2}}-1)\equiv -2\pmod{5}$, so $b\equiv -1\pmod{5}$. Thus, neither 3 nor 5 divides $b$. Clearly, the only prime factors of $a$ are 3 and 5. Since $a$ and $b$ share no common prime factors, $(a,b)=1$. Thus, we have $a,b,h\in {\bf Z^+}$ such that $(a,b)=1$ and $h|b$. As in the first part of this proof (where $k\equiv -1\pmod{4}$), we want to show that $\frac{{\sigma}(h)}{h}>\frac{a}{b}$. This will prove (by Corollary 5a) that there is no $m\in {\bf Z^+}$ such that $\frac{{\sigma}(m)}{m}=\frac{a}{b}$, and thus no solution to the problem $\frac{{\sigma}(n)}{n}=\frac{5}{3}$ of the form $n=3^km$, when $k\equiv 1\pmod{4}$. We have \begin{eqnarray*} \frac{a}{b} &=&\frac{5\cdot3^{k-1}}{\frac{1}{2}\cdot(9^{(k+1)/2}-1)}\\ &=&\frac{10\cdot3^{k-1}}{9^{(k+1)/2}-1}. \end{eqnarray*} Notice that \begin{eqnarray*} \frac{3}{5}\cdot\frac{a}{b} &=&\frac{3}{5}\cdot\frac{10\cdot3^{k-1}}{9^{(k+1)/2}-1}\\ &=&\frac{2\cdot3^k}{3^{k+1}-1}\\ &=&\frac{2}{3-3^{-k}}<\frac{2}{3-1}\\ &=&1. \end{eqnarray*} Thus, $\frac{3}{5}\cdot\frac{a}{b}<1$, so $\frac{a}{b}<\frac{5}{3}$. But $\frac{{\sigma}(h)}{h}=\frac{{\sigma}(4)}{4}=\frac{7}{4}>\frac{5}{3}$, so $\frac{a}{b}<\frac{{\sigma}(h)}{h}$. Therefore, there is no solution when $k\equiv 1\pmod{4}$ (by Corollary 5a). \\ We have proven that $k\not\equiv 1\pmod{4}$ and $k\not\equiv -1\pmod{4}$. Therefore, if there is some $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{5}{3}$ and $n=3^km$, with $(3,m)=1$, then $k$ must be an even integer. \rule{2mm}{2mm} \\ \\ Some equivalent problems to finding $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{5}{3}$ are: \\ $k=2$: $\frac{{\sigma}(m)}{m}=\frac{5\cdot3^{2-1}}{{\sigma}(3^2)}=\frac{15}{13}$ \\ $k=4$: $\frac{{\sigma}(m)}{m}=\frac{5\cdot3^{4-1}}{{\sigma}(3^4)}=\frac{135}{121}$ \\ $k=6$: $\frac{{\sigma}(m)}{m}=\frac{5\cdot3^{6-1}}{{\sigma}(3^6)}=\frac{1215}{1093}$ \\ (In general, \begin{eqnarray*} \frac{{\sigma}(m)}{m} &=&\frac{5\cdot3^{k-1}}{{\sigma}(3^k)}\\ &=&\frac{5\cdot3^{k-1}}{\frac{1}{2}\cdot(3^{k+1}-1)}\\ &=&\frac{10\cdot3^{k-1}}{3^{k+1}-1}\\ &=&\frac{10}{9-3^{1-k}}.) \end{eqnarray*} There is one more interesting note to make about this problem before moving on to the next example: \\ {\bf \large \noindent Claim:} If $\frac{{\sigma}(n)}{n}=\frac{5}{3}$ (for some $n\in {\bf Z^+}$), then $(n,2)=(n,5)=1$. Furthermore, if such an integer $n$ exists, then $5n$ is an odd perfect number. \\ Proof: We will first show that, if there exists $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{5}{3}$, then $(n,2)=(n,5)=1$. Notice that $\frac{{\sigma}(18)}{18}=\frac{39}{18}>\frac{5}{3}$; that is, $\frac{{\sigma}(18)}{18}=\frac{39}{18}>\frac{{\sigma}(n)}{n}$. But Corollary 2a implies that if $18|n$, then $\frac{{\sigma}(18)}{18}\leq\frac{39}{18}<\frac{{\sigma}(n)}{n}$; thus, $18\not|n$. This means that either $2\not|n$ or $9\not|n$; however, we already know that $9|n$, so $2\not|n$. Therefore, $(n,2)=1$. Now notice that $\frac{{\sigma}(45)}{45}=\frac{78}{45}>\frac{5}{3}$; that is, $\frac{{\sigma}(45)}{45}=\frac{78}{45}>\frac{{\sigma}(n)}{n}$. Thus, by Corollary 2a, $45\not|n$. This means that either $9\not|n$ or $5\not|n$; however we already know that $9|n$, so $5\not|n$. Therefore, $(n,5)=1$. \\ Since $(n,2)=1$, $n$ must be odd. Since $(n,5)=1$, $5\not|n$, so ${\sigma}(5n)={\sigma}(5){\sigma}(n)$, and \begin{eqnarray*} \frac{{\sigma}(5n)}{5n} &=&\frac{{\sigma}(5)}{5}\cdot\frac{{\sigma}(n)}{n}\\ &=&\frac{6}{5}\cdot\frac{5}{3}\\ &=&2. \end{eqnarray*} Thus, $5n$ is an odd perfect number. \rule{2mm}{2mm} \\ {\bf \large \noindent Example 4:} Consider the problem: $\frac{{\sigma}(n)}{n}=\frac{2p-1}{p}$, where $p$ and ${2p-1}$ are odd primes. \\ Clearly, $(p,2p-1)=1$ (since these are distinct odd primes), so $\frac{{\sigma}(n)}{n}=\frac{2p-1}{p}$ is in lowest terms. Corollary 7a states that this problem is equivalent to the problem of finding $m,k\in {\bf Z^+}$ such that $$\frac{{\sigma}(m)}{m}=\frac{(2p-1)p^{k-1}}{{\sigma}(p^k)}.$$ \\ {\bf \large \noindent Claim:} If there exists some $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{2p-1}{p}$ and \\ $(n,2p-1)=(n,2)=1$, then $(2p-1)n$ is an odd perfect number. \\ Proof: It's clear that, since $(n,2p-1)=1$, \begin{eqnarray*} \frac{{\sigma}(n\cdot(2p-1))}{n\cdot(2p-1)} &=&\frac{{\sigma}(n)}{n}\cdot\frac{{\sigma}(2p-1)}{2p-1}\\ &=&\frac{2p-1}{p}\cdot\frac{2p}{2p-1}\\ &=&\frac{2p}{p}\\ &=&2. \end{eqnarray*} Since $n$ is odd, $(2p-1)n$ is odd, so $(2p-1)n$ is an odd perfect number. \rule{2mm}{2mm} \\ We will use this discovery to prove that, if there exists some odd positive integer $n$ such that $\frac{{\sigma}(n)}{n}=\frac{2p-1}{p}$, then $n$ is a perfect square. First, though, we have another claim to prove. \\ Back in the Introduction, we saw a result proven by Euler: if $n\in {\bf Z^+}$ is an odd perfect number, then $$n=p^{\alpha}q_1^{2\beta_1}...q_g^{2\beta_g},$$ where $p,q_1...q_g$ are distinct odd primes, ${\alpha},{\beta_1}...{\beta_g}$ are positive integers, and $p\equiv{\alpha}\equiv 1\pmod{4}$. (This is the prime factorization of $n$.) We will give a proof of this here, then use this result to show that \\ {\bf \large \noindent Claim: (Euler)} If there exists $n\in {\bf Z^+}$ such that $n$ is an odd perfect number, then $$n=p^{\alpha}q_1^{2\beta_1}...q_g^{2\beta_g},$$ where $p,q_1...q_g$ are distinct odd primes, ${\alpha},{\beta_1}...{\beta_g}$ are positive integers, and $p\equiv{\alpha}\equiv 1\pmod{4}$. \\ Proof: Suppose an odd perfect number $n$ exists. Then, ${\sigma}(n)=2n$, so $2|{\sigma}(n)$. However, it is clear that for $h>1$, $2^h\not|2n$, and so $2^h\not|{\sigma}(n)$. Now, we can write the prime factorization of $n$ as $$n=p_1^{k_1}p_2^{k_2}...p_{g+1}^{k_{g+1}};$$ thus, $${\sigma}(n)={\sigma}(p_1^{k_1})...{\sigma}(p_{g+1}^{k_{g+1}}).$$ We know that 2 divides ${\sigma}(n)$ exactly one time. This means that there is exactly one $i\in {\bf Z^+}, 1\leq i\leq (g+1)$ such that $2|{\sigma}(p_i^{k_i})$; furthermore, it means that $4\not|{\sigma}(p_i^{k_i})$. When $j\in {\bf Z^+}, j\not= i, {\sigma}(p_j^{k_j})$ must be odd. Lemma 8 tells us that this is true if and only if $p_j^{k_j}$ is a perfect square times a power of 2; or, in this case, a perfect square, since each $p_j$ is odd. Thus, when $j\not=i$, $k_j$ must be even. \newpage We now must determine what conditions $p_i$ and $k_i$ must fulfill so that ${\sigma}(p_i^{k_i})$ is divisible by 2, but not by 4. That is, we need $p_i,k_i\in {\bf Z^+}$ such that ${\sigma}(p_i^{k_i})\equiv 2\pmod{4}$. We know that $p_i\equiv 1$ or $-1 \pmod{4}$, and that $k_i\equiv 1$ or $-1 \pmod{4}$. We will cover all possibilities using 2 separate cases: \\ Case 1: Suppose $k_i\equiv -1\pmod{4}$. \\ $${\sigma}(p_i^{k_i})=1+p_i+p_i^2+...+p_i^{k_i}.$$ Since $k_i\equiv -1\pmod{4}$, $4|(k_i-3)$, so we can write $${\sigma}(p_i^{k_i})={\sigma}(p_i^3)(1+p_i^4+...+p_i^{k_i-3}).$$ Thus, ${\sigma}(p_i^3)|{\sigma}(p_i^{k_i})$. Notice that \begin{eqnarray*} {\sigma}(p_i^3) &=&1+p_i+p_i^2+p_i^3\\ &=&(1+p_i)(1+p_i^2), \end{eqnarray*} which is a product of two even numbers. But this means that 2 divides ${\sigma}(p_i^3)$ at least twice; that is, $2^2|{\sigma}(p_i^3)$. So, since $4|{\sigma}(p_i^3)$ and ${\sigma}(p_i^3)|{\sigma}(p_i^{k_i})$, $4|{\sigma}(p_i^{k_i})$. Thus, when $k_i\equiv -1\pmod{4}$, ${\sigma}(p_i^{k_i})\not\equiv 2\pmod{4}$. Therefore, $k_i\not\equiv -1\pmod{4}$. \\ Case 2: Suppose $k_i\equiv 1\pmod{4}$. \\ $${\sigma}(p_i^{k_i})=1+p_i+p_i^2+...+p_i^{k_i}$$. Since $k_i$ is odd, we can write $${\sigma}(p_i^{k_i})=(1+p_i)(1+p_i^2+...+p_i^{k_i-1}).$$ Thus, $(1+p_i)|{\sigma}(p_i^{k_i})$. \\ Case 2a: Suppose $p_i\equiv -1\pmod{4}$. Then $(1+p_i)\equiv 0\pmod{4}$; that is, $4|(1+p_i)$. Since $(1+p_i)|{\sigma}(p_i^{k_i})$, this means that $4|{\sigma}(p_i^{k_i})$, so ${\sigma}(p_i^{k_i})\not\equiv 2\pmod{4}$. Therefore, $p_i\not\equiv -1\pmod{4}$ when $k_i\equiv 1\pmod{4}$. \\ Case 2b: Suppose $p_i\equiv 1\pmod{4}$. Then $(1+p_i)\equiv 2\pmod{4}$. Also, \begin{eqnarray*} (1+p_i^2+...+p_i^{k_i-1}) &\equiv&(1+1^2+...+1^{k_i-1})\\ &\equiv&\frac{k_i+1}{2}\pmod{4}. \end{eqnarray*} But $k_i+1\equiv 2\pmod{4}$, so $\frac{k_i+1}{2}\equiv 1$ or $-1\pmod{4}$. This means that $(1+p_i^2+...+p_i^{k_i-1})\equiv 1$ or $-1\pmod{4}$. Thus, $${\sigma}(p_i^{k_i})\equiv(1+p_i)(1+p_i^k+p_i^{k_i-1})\equiv 2\pmod{4}$$ (since $2\cdot1\equiv 2\cdot-1\equiv 2\pmod{4}$). Therefore, ${\sigma}(p_i^{k_i})\equiv 2\pmod{4}$ when $k_i\equiv p_i\equiv 1\pmod{4}$. \\ The only way we can get ${\sigma}(p_i^{k_i})\equiv 2\pmod{4}$ is if $k_i\equiv p_i\equiv 1\pmod{4}$. Thus, we've proven that if an odd perfect number, $n$, exists, then it has prime factorization $$n=p_1^{k_1}p_2^{k_2}...p_{g+1}^{k_{g+1}},$$ with the following conditions: \\ 1) For some $i\in {\bf Z^+}$, $1\leq i\leq (g+1)$, $p_i\equiv k_i\equiv 1\pmod{4}$. 2) For any $j\in {\bf Z^+}$ such that $1\leq j\leq (g+1)$ and $j\not=i$, $k_j$ is even. \\ Let us revise our notation as follows: \\ \begin{enumerate} \item Rewrite $p_i$ as $p$, and $k_i$ as ${\alpha}$. \item Relabel the other $g$ distinct prime factors as (in any order) $q_1,q_2...q_g$; also, relabel the other $g$ exponents (corresponding to the labeling of the prime factors) as $2{\beta_1},2{\beta_2}...2{\beta_g}$. (We may write them in this way because we have shown that these exponents must all be even.) \end{enumerate} After this revision, we can write the prime factorization of odd perfect number $n$ in the form $$n=p^{\alpha}q_1^{2\beta_1}q_2^{2\beta_2}...q_g^{2\beta_g},$$ where $p\equiv{\alpha}\equiv 1\pmod{4}$. \rule{2mm}{2mm} \\ We can now prove the following: \\ {\bf \large \noindent Claim:} If there exists $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{2p-1}{p}$ and $(2p-1,n)=(2,n)=1$, then $n$ is a perfect square. (Here $p,2p-1$ are odd primes, just as before.) \\ Proof: We proved earlier that if such a positive integer $n$ exists, then $(2p-1)n$ is an odd perfect number. Therefore, $(2p-1)n$ must have prime factorization of the form $$n=P^{\alpha}q_1^{2\beta_1}...q_g^{2\beta_g}=P^{\alpha}m^2.$$ (Here, $m=q_1^{\beta_1}...q_g^{\beta_g}$). (The capital $P$ is to avoid ambiguity; the lower-case $p$ had already been defined separately.) Now, $p$ is an odd prime, so $2p\equiv 2\pmod{4}$, so $(2p-1)\equiv 1\pmod{4}.$ Also, $(2p-1)$ is raised to the first power, and $1\equiv 1\pmod{4}$. Clearly, $(2p-1)^1$ does appear in the prime factorization of $n$, and $(2p-1)\equiv 1\equiv 1\pmod{4}$, so we must have $P=(2p-1)$ and ${\alpha}=1$. Thus, \begin{eqnarray*} (2p-1)n &=&P^{\alpha}n\\ &=&P^{\alpha}m^2, \end{eqnarray*} and so $n=m^2$. Therefore, $n$ is a perfect square. \rule{2mm}{2mm} \\ Note: This gives us an alternate proof of an earlier result -- if there is a solution of the form $n=3^km$, $(m,3)=1$, to the problem $\frac{{\sigma}(n)}{n}=\frac{5}{3}$, then $k$ is even. \\ {\bf \large \noindent Claim:} If there is a solution of the form $n=3^km$, with $(3,m)=1$, then $k$ is an even integer. \\ Proof: We showed earlier that $(5,m)=(2,m)=1$. Also, if we let $p=3$, then this problem is of the form $\frac{{\sigma}(n)}{n}=\frac{2p-1}{p}$, with $(n,2p-1)=(n,2)=1$. Therefore, by the previous result, we know that $n$ must be a perfect square. Since $n=3^km$ is a perfect square (and $(3,m)=1$), $k$ must be even. \rule{2mm}{2mm} \newpage {\bf Summary} \\ Given $a,b\in {\bf Z^+}$, where $a\geq {\sigma}(b)$, (a,b)=1 and $b$ is a prime power ($b=p^c$, $p$ prime), the problem of finding $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$ is equivalent to the problem of finding $m,k\in {\bf Z^+}$ such that $mp^k=n$, $k\geq c$, $(m,p)=1$, and $\frac{{\sigma}(m)}{m}=\frac{ap^{k-c}}{{\sigma}(p^k)}$. In practice, this is done by selecting first selecting a value of $k$, then trying to find $m$ based on this value of $k$. For each value of $k$, we continue to search for $m$ until (and if) one of the following occurs: \begin{enumerate} \item We find some $m\in {\bf Z^+}$ such that $n=mp^k$ is a solution to $\frac{{\sigma}(n)}{n}=\frac{a}{b}$. \item We manage to prove that, given $k$, there exists no $m\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$ when $n=mp^k$. \end{enumerate} We also proved the following useful facts about the ratio $\frac{{\sigma}(n)}{n}$: \\ (For the following, assume $a,b,g,h$ to be positive integers, and $p,q$ to be primes.) \begin{enumerate} \item If $a\frac{{\sigma}(n)}{n}$, then there exists no $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{a}{b}$. \end{enumerate} \newpage These interesting results were also proven: \begin{enumerate} \item ${\sigma}(n)$ is odd if and only if $n=2^hm$, where $h$ is a non-negative integer and $m$ is a perfect square. \item If there exists some $n\in {\bf Z^+}$ such that $\frac{{\sigma}(n)}{n}=\frac{2p-1}{p}$, where $p$ and $2p-1$ are both prime, then $n$ is a perfect square, and $(2p-1)n$ is an odd perfect number. This was proven using Euler's discovery that and odd perfect number, $n$, is of the form $$n=p^{\alpha}q_1^{2\beta_1}...q_g^{2\beta_g},$$ where $p,q_1...q_g$ are distinct primes, ${\alpha},{\beta_1}...{\beta_g}$ are positive integers, and $p\equiv{\alpha}\equiv 1\pmod{4}$. \end{enumerate} \end{document}