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


{\bf 1.} Find $\sigma(200)$

{\bf Sol. to 1}: First we factorize 200 
$$
200=2^3 \cdot 5^2 \quad
$$

Now
$$
\sigma(200)=(1+2+2^2+2^3) \cdot (1+5+5^2)=15 \cdot 31=465 \quad.
$$

{\bf Answer to 1}: $\sigma(200)=465$

{\bf 2} (a) Define a perfect number

{\bf Sol to 2}: A positive integer  is perfect if it equals to the sum of its proper divisors, or equivalenly
the sum of the divisors is twice that number. In symbols: $\sigma(n)=2n$.


(b) Prove that if both $p$ and $2^p-1$ are primes, then
$$
2^{p-1} \cdot (2^p -1) \quad 
$$
is a perfect number.

{\bf Sol. of 2(b)}: 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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