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QUESTION

An integer $n$ is called 3-perfect if $\sigma(n)=3n$. Show that
120 is 3-perfect. Find all 3-perfect numbers $n$ of the form
$2^k.3.p$ where $p$ is an odd prime.

[Hints:\begin{description}
\item[(i)]
Treat the case $p=3$ separately, so that you can multiplicity to
calculate $\sigma(n)$.

\item[(ii)]
Evaluate $\sigma(n)$ and use the equation $\sigma(n)=3n$ to get an
expression for $2^k$ in terms of $p$. Hence deduce $p\leq8$.

\item[(iii)]
Deal with the possible values of $p$ separately.]

\end{description}



ANSWER

$\sigma(120)=\sigma(2^3,3,5)=\frac{(2^4-1)}{(2-1)}.
\frac{(3^2-1)}{(3-1)}.\frac{(5^2-1)}{(5-1)}=15.\frac{8}{2}
.\frac{24}{4}=15.4.6=2^3.3^2.5=3.(2^3.3.5)=3.120.$

Thus 120 is 3-perfect.

Now suppose $n=2^k.3.p$ is 3-perfect.

CASE 1: $p=3$, so $n=2^k.3^2$ and
$\sigma(n)=\frac{2^{k+1}-1}{2-1}.\frac{3^3-1}{3-1}=(2^{k+1}-1).\frac{26}{2}=13.(2^{k+1}-1)$.
As $\sigma(n)=3n$, we have $13(2^{k+1}-1)=3.2^k.3^2=2^k.3^3$ which
is clearly impossible, as the prime 13 divides the left-hand side
but not the right.

CASE 2: $p\neq3$, so $n=2^k.3.p$, and
$\sigma(n)=\frac{2^{k+1}-1}{2-1}.\frac{3^2-1}{3-1}.\frac{p^2-1}{p-1}=(2^{k+1}-1).4.(p+1)$
( using $p^2-1=(p-1)(p+1)$.) AS $\sigma(n)=3n,
(2^{k+1}-1).4.(p+1)=2^k.3^2.p$. Thus $2^k(3^2p-2.4.(p+1))=-4(p+1)$
( using $2^{k+1}=2.2^k$), i.e. $2^k(p-8)=-4(p+1)$. As the
right-hand side of this equation is negative, so is the left, so
$p<8$. As we know $p$ is odd, and $p\neq3$, only $p=5$ and $p=7$
are possible.

If $p=5$. the equation $2^k(p-8)=-4(p+1)$ gives $2^k(-3)=-4.6$,
giving $k=3$ and $n=2^3.3.5=120$ (the case we've already dealt
with.)

If $p=7$, the equation $2^k(p-8)=-4(p+1)$ gives $2^k(-1)=-4.8$,
giving $k=5$. Thus $n=2^5.3.7=671$, so 120 and 672 are the only
3-perfect numbers of the form $2^k.3.p$ where $p$ is an odd prime.




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