If n is a perfect number, prove that $\sum_{d \mid n} l / d=2$.

Gauss (1796) discovered that a regular polygon with p sides, where p is a prime, can be constructed with ruler and compass if and only if $p-1$ is a power of 2 . Show that this condition is equivalent to requiring that $p$ be a Fermat prime.

Verify each of the statements below:
(a) No power of a prime can be a perfect number. (b) A perfect square cannot be a perfect number. (c) The product of two odd primes is never a perfect number. [Hint: Expand the inequality $(p-1)(q-1)>2$ to get $p q>p+q+1$.]

If $n_1, n_2, \cdots, n_r$ are distinct even perfect numbers, establish that

$$\phi\left(n_1 n_2 \cdots n_r\right)=2^{r-1} \phi\left(n_1\right) \phi\left(n_2\right) \cdots \phi\left(n_r\right)$$

Assuming that n is an even perfect number, say $n=2^{k-1}\left(2^k-1\right)$, prove that the product of the positive divisors of $n$ is equal to $n^k$; in symbols,

$$\prod_{d \mid n} d=n^k$$

Prove that if $2^k-1$ is prime, then the sum

$$2^{k-1}+2^k+2^{k+1}+\cdots+2^{2 k-2}$$

will yield a perfect number. For instance, $2^3-1$ is prime and $2^2+2^3+2^4=28$, which is perfect.

The twin primes 5 and 7 are such that one half their sum is a perfect number. Are there any other twin primes with this property? $\left[\right.$ Hint: Given the twin primes $p$ and $p+2$, with $p>5, \frac{1}{2}(p+p+2)=6 k$ for some $k>1$.

If the integer n>1 is a product of distinct Mersenne primes, show that $\sigma(n)=2^k$ for some $k$.

The harmonic mean H(n) of the divisors of a positive integer n is defined by the formula

$$\frac{1}{H(n)}=\frac{1}{\tau(n)} \sum_{d \mid n} \frac{1}{d}$$

Show that if $n$ is a perfect number, then $H(n)$ must be an integer. [Hint: Observe that $H(n)=n \tau(n) / \sigma(n)$.

If n is an odd perfect number, prove that $n$ has at least three distinct prime factors. [Hint: Assume that $n=p^k q^{2 j}$, where $p=k=1(\bmod 4)$. Use the inequality $2=$ $o(n) / n \leq[p /(p-1)][q /(q-1)]$ to reach a contradiction.]

Prove that
(a) Any odd perfect number $n$ can be represented in the form $n=p a^2$, where $p$ is a prime. (b) If $n=p a^2$ is an odd perfect number, then $n \equiv p(\bmod 8)$

Numbers n such that $\sigma(\sigma(n))=2 n$ are called superperfect numbers.
(a) If $n=2^k$ with $2^{k+1}-1$ a prime, prove that $n$ is superperfect; hence, 16 and 64 are superperfect. (b) Find all even perfect numbers $n=2^{k-1}\left(2^k-1\right)$ which are also superperfect. [Hint: First establish the equality $\sigma(\sigma(n))=2^k\left(2^{k+1}-1\right)$.]

A finite sequence of positive integers is said to be a sociable chain if each is the sum of the positive divisors of the preceding integer, excluding the number itself (the last integer is considered as preceding the first integer in the chain). Show that the following integers form a sociable chain:

$$14288,15472,14536,14264,12496$$

Only two sociable chains were known until 1970, when nine chains of four integers each were found.

For any even perfect number $n=2^{k-1}\left(2^k-1\right)$, show that $2^k \mid \sigma\left(n^2\right)+1$.