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QUESTION
\begin{description}
\item[(i)]
Prove that $d(n)$ is odd if and only if $a$ is the square of some
integer $m$.
\item[(ii)]
Prove that if $n>1$, then $\sigma(n)\geq n+1$, with equality
holding if and only if $n$ is prime.
\item[(iii)]
Prove $\sum_{d|n}\frac{1}{d}=\frac{\sigma(n)}{n}$ for every $n\in
N$.
\item[(iv)]
Describe all positive integers $n$ which satisfy $d(n)=12$. (There
are infinitely many of them- you need to find a way to describe
them all.
\end{description}
ANSWER
\begin{description}
\item[(i)]
If $n=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_k^{\alpha_k}$ then
$d(n)=(\alpha_1+1)(\alpha_2+1)\ldots(\alpha_k+1)$. Thus $d(n)$ is
odd $\Leftrightarrow$ each $\alpha_i$ is even $\Leftrightarrow n$
is a square.
\item[(ii)]
$\sigma(n)$= sum of positive divisors of $n$.
Now 1 and $n$ are both positive divisors of $n$, and are distinct
as $n>1$. Thus $\sigma(n)\geq1+n$, and equality holds if and only
if there are no other positive divisors of $n$,
i.e.$\Leftrightarrow n$ is prime.
\item[(iii)]
Consider $n.\sum_{d|n}\frac{1}{d}=\sum_{d|n}\frac{n}{d}$.
As $d$ varies through the divisors of $n$, so does $\frac{n}{d}$.
Thus the sum $\sum_{d|n}\frac{n}{d}$ is just the sum of the
positive divisors of $n$, so it can be rewritten as
$\sum_{d|n}d=\sigma(n)$. Hence $n\sum_{d|n}\frac{1}{d}=\sigma(n)$,
so $\sum_{d|n}\frac{1}{d}=\frac{\sigma(n)}{n}$.
\item[(iv)]
Suppose $n=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_k^{\alpha_k}$,
with each $\alpha_i\geq1$. Then
$d(n)=(\alpha_1+1)(\alpha_2+1)\ldots(\alpha_k+1)=12$.
Now each $\alpha_i+1$ is $\geq2$, so we wish to find in how many
ways we can write 12 as a product of $k$ factors each $\geq2$. Now
$12=2^2.3$, so certainly there can't be more than 3 factors, so
$k=1,2$ or 3.
If $k=1$, then $(\alpha_1+1)=12$, so $\alpha_1=11$, and so all $n$
of the form $p^{11}$ satisfy $d(n)=12$.
If $k=2$, then $(\alpha_1+1)(\alpha_2+1)=12$. If we allow the
primes $p_I$ to take any values, we may assume that they are
arranged in order of increasing $\alpha_i$, so we may as well
assume $|alpha_1\leq\alpha_2$. Thus the possible solutions are
$\alpha_1=1,\ \alpha_2=5$ and $\alpha_1=2,\ \alpha_2=3$ and the
possible forms of integers $n$ given by these are $pq^5$ and
$p^2q^3 (p$ and $q$ distinct primes).
If $k=3$, then $(\alpha_1+1)(\alpha_2+1)(\alpha_3+1)=12$, and
taking the $p_i$ to be arranged so that the $\alpha_I$ increase,
as above, the only possibility is $\alpha_1=1,\ \alpha_2=1,\
\alpha_3=2$, so the relevant $n$'s are $pqr^2$, where $p$, $q$ and
$r$ are distinct primes.
Thus our final list, with $p,\ q$ and $r$ any distinct primes is
$p^{11},pq^5,p^2q^3$ and $pqr^2$.
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