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\centerline
{
\bf  Solutions to Quiz \# 8 for Dr. Z.'s Number Theory
Course for Nov. 14, 2013
}
\bigskip

{\bf 1.} (i) (2 pts.) Define $\sigma_2(n)$ (ii) (2 pts.) State the formula for $\sigma_2(n)$
in terms of the expression of $n$ as a product of prime powers
(iii) (2 pts.) Verify it for $n=15$ by using the definition and the formula

{\bf Sol. to 1}:

(i)
$$
\sigma_2(n):= \sum_{d|n} d^2 \quad.
$$

(ii) If 
$$
n= \prod_{i=1}^{k} p_i^{\alpha_i} \quad
$$
is the factorization of $n$ into products of prime powers guaranteed by
the {\bf Fundamental Theorem of Arithmetic}, then
$$
\sigma_2(n) = \prod_{i=1}^{k} (1+ p_i^2 + p_i^4 + \dots + p_i^{2\alpha_i})
$$
Another way (summing the geometric series) is
$$
\sigma_2(n) = \prod_{i=1}^{k} \frac{p_i^{2\alpha_i+2}-1}{p_i^2-1} \quad.
$$

(iii) Since
$$
Div(15)= \{ 1,3,5,15 \} \quad,
$$
From the {\bf definition}
$$
\sigma_2(15)=1^2 + 3^2 + 5^2 + 15^2= 1+9+25+225=260 \quad .
$$
From the formula, since $15=3^1 \cdot 5^1$,
$$
\sigma_2(15)= (1+3^2) \cdot (1+5^2)= 10 \cdot 26= 260 \quad .
$$
Yea!

{\bf 2.} (4 pts.) Prove that if $p$ is a prime, and $2^p-1$ is also a prime, then
$$
2^{p-1} \cdot (2^p -1) \quad 
$$
is a perfect number.

{\bf Sol. of 2}: Since {\bf both} $2$ and $2^p-1$ are prime, it follows from
the formula for $\sigma(n)$, that
$$
\sigma(2^{p-1} \cdot (2^p -1))=
(1+2+ \dots + 2^{p-1}) \cdot (1 + (2^p-1)) \quad.
$$
By the famous geometric series formula
$$
1+q+ \dots + q^N = \frac{q^{N+1} -1}{q-1} \quad,
$$
with $q=2$ and $N=p-1$ we have
$$
1+2+ \dots + 2^{p-1} = \frac{2^{p-1+1} -1}{2-1} = 2^p-1 \quad.
$$
By the famous theorem
$$
1+(N-1)=N \quad ,
$$
applied to $N=2^p-1$, we have
$$
1+(2^p-1)=2^p \quad .
$$
Going back above we have
$$
\sigma(2^{p-1} \cdot (2^p -1))=
(2^p -1) \cdot (2^p)= 2 \cdot (2^{p-1} (2^p-1)) \quad.
$$
Hence $n=2^{p-1} (2^p-1)$ satisfies the condition
$$
\sigma(n)=2n \quad,
$$
that this means that $n$ is perfect. QED.


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