Number Theory | The Problem Solver's Paradise

The world is simply full of numbers of all kinds. All the clutter on numbers was made by none but man. Fibonacci and Lucas numbers; perfect numbers; Fermat numbers; Multiply-perfect numbers; abundant and super-abundant numbers, practical and impractical (!) numbers, etc. However, there is an interesting correlation between all these junk of numbers. In this entry, I try to let us get an insight into Perfect, Multiply-perfect, Abundant and Super-abundant and Practical numbers.
As usual, all of us have realized from the definition of a perfect number that what a multiply-perfect number is.
Definition and properties of the $\sigma$ function
Let us, for the sake of clarity, define our notations first. We denote $\sigma (n)$ as the sum of the divisors of $n$ , and $\tau (n)$ as the number of the divisors of $n$ . We try to express $\tau(n)$ in terms of the prime divisors of $n$ . Note that, by the fundamental theorem we have that any $n$ can be expressed in the form of (where $p_i$ ‘s are the primes arranged in order or not in order)
$n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k};$
A simple combinatorial argument leads to the obvious and well-known fact that
$\tau (n)=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1)$ (including $1$ as a factor).
Now, for a formula of $\sigma(n)$ we see that all the divisors $d_i$ are of the form
$d_i=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k};$ and $0\leq \beta_i\leq \alpha_i$ . Thence the sum, can be factored into (or rather, found out to be) terms in the expansion of $\prod_{i=1}^n(1+p_i+\cdots+p_i^{\alpha_i})=\prod_{i=1}^n\frac{p_i^{\alpha_i+1}-1}{p_i-1}.$
Now it is evident that when $\text{gcd}(m,n)=1$ we have the multiplicity property of this $\sigma$ function, ie $\sigma(mn)=\sigma(m)\sigma(n)$ . The (obvious) proof is left to the reader.
Multiply-Perfect Numbers
Before discussing the general multiply-perfect numbers, we dive into the perfect numbers for a little time. Let us try to list some of the perfect numbers. They are $6,28,496,8128,\cdots$ written in order. The next values can be generated with a computer, but what do we see common in these numbers?
(1) (Obviously) All are even and $\sigma(n)=2n$ ; which makes them perfect.
(2) $6=2\cdot (2^2-1);$ and $3=2^2-1$ is a prime.
$28 = 2^2\cdot (2^3-1);$ and $2^3-1=7$ is a prime.
$496 = 2^4\cdot(2^5-1);$ and $2^5-1=31$ is a prime.
$8128 = 2^6\cdot(2^7-1);$ and $2^7-1=127$ is a prime.
Proceeding in this manner, Euler proved (again, how many theorems did he prove?) that if $n$ is a given number with the property that $2^n-1$ is a prime, that $2^{n-1}\cdot(2^{n}-1)$ is going to be a perfect number.
With this (extremely) brief idea of perfect numbers we are able to give a generalization. We consider a number $n$ such that its $\sigma (n)=kn$ where $k$ is some natural number.
Before getting some information and problems on multiply-perfect numbers, let us try a problem.
$\boxed{E1.}$ If $n$ is a perfect number and $n+1, n-1$ are primes, find all possible values of $n$
Solution.
Obviously considering modulo $6$ we have that $6|n\implies n=6n_1$ for some $n_1\in\mathbb N$ . If so, we consider $n_1>3.$ In this case,
$\sigma(6n_1)>1+2+3+6+n_1+2n_1+3n_1+6n_1>12n_1+12$ , which is way too bizarre for a perfect number $6n_1$ . So $n_1\leq 3$ . So we get $n=6,12,18$ . Indeed, $5,7$ and $11,13$ and $17, 18$ are all primes. $\Box$
Let us denote a multiply perfect number of the order $k$ by $x_k$ whereas $\sigma(x_k)=kx_k$ . We play around with a few multiply-perfect numbers of some orders. Just as in the previous case, we try to list some properties of a $x_k$ number. At the first glance into a $k-$ tuply perfect number like $120$ or $672$ (Verify with a calculator or the prime factorization formulae for $\sigma(n)$ ), we might try to conjecture a few things. Like, all $x_k$ are even. We might try to construct a sequence of these numbers for different $k$ , and conjecture that there does or does not exist such a sequence. These conjectures are needed in mathematics, but we still need a proof of those, that is why they are called conjectures. :lol:
Very few has been known about large $k$ ‘s, so let us start by noting some examples.
$\boxed{E2.}$ Let $n$ be a number such that $3$ does not divide $n$ yet it satisfies $\sigma(n)=3n$ , show that O $\sigma(3n)=12n$ , ie $3n$ is a perfect number of order $4$ .
Solution.
It is obvious from the multiplicative property of $\sigma(n)$ that
$\sigma(3n)=3\sigma(n)=12n;$ so that $3n$ is an $x_4-$ type number.
$\boxed{E3.}$ Every number such that $\sigma(n)=5n$ has at least $6$ different prime factors.
Solution.
This, also is obvious on using the prime decomposition of $n$ , which is
$n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ gives us
$\prod_{i=1}^n\frac{p_i^{\alpha_i+1}-1}{p_i^{\alpha_i}(p_i-1)}=5.$
Now using $\frac{p_1^{\alpha_1+1}-1}{p_1^{\alpha_1}(p_1-1)}>\frac{p_1}{p_1-1};$ which is obvious on expanding, we have
$\frac{p_1}{p_1-1}\cdots\frac{p_n}{p_n-1}>5.$
Now with the same old technique, we consider the product upto $5$ terms, which is surely going to be $<5$ (because the statement is modified to contradict this event by origin); therefore we are surely done.
Superabundant Numbers
A number $n$ is known as superabundant if and only if $\dfrac{\sigma(n)}{n}$ exceeds $\dfrac{\sigma(k)}{k}$ for all $k<n$ .
$\boxed{E4}.$ The sequence of numbers $t_n=\dfrac{\sigma(n)}{n}$ does not have any upper bound..
Solution.
If $d_i$ are the divisors of $n$ then we have, $d_i=\dfrac n{d_{n-i}}$ (why?) and summing up on this sequence we get $\sigma(n)$ as,
$\sigma(n)=\sum_{d_i|n}\dfrac 1{d_i}\implies \dfrac{\sigma(n)}{n}=\sum_{d_i|n}\dfrac 1{d_i}.$
Since we have to prove that this has no upper bound, let $n=m!$ for our convenience. Then we have,
$\dfrac{\sigma(n)}{n}=\sum_{d_i|n}\dfrac 1{d_i}=\sum_{d_i|m!}\dfrac1{d_i}\geq 1+\frac 12+\frac 13+\cdots+\frac 1m.$
Then we have $t_n\geq H_m$ where $H$ is the harmonic series. As we know the harmonic series on natural numbers does not have any upper bound as $n$ increases, and therefore $t_n$ is unbounded from above.
Abundant Numbers
The numbers $n$ with the property that $\sigma(n)\geq 2n$ are abundant ones.
$12, 18,20,24\cdots$ are abundant at our first instance, and we can also find out that all abundant numbers $\leq 100$ are even.
We have a property regarding the abundant numbers.
$\boxed{E5}$ Every number $n$ is abundant implies that $mn$ is abundant, $m\ge 2$
Solution to be found by the reader.

We will discuss on some Practical, Highly Composite Numbers and Quasi-perfect, Semi-perfect numbers soon.
Hope you do not get insane with a lot of numbers, we will get some exercises on these too.

(To Be Updated Soon)