Factor proofs problem The coolness of an integer is equal to the integer divided by the total number of factors that it has. For example, $48$ has $10$ factors therefore, coolness $(48) = \frac { 48 }{ 10 } =\quad 4.8$

1.    Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y$ are different prime numbers.

My attempt: Just from a couple of trials a found that the number of factors of $xy$ seems to always be $4$. Therefore, we can note that coolness$(xy)$ can only be a whole number if $xy$ is divisible by $4$. I am pretty sure that we can’t make a multiple of $4$ by multiplying any combination of 2 prime numbers. 
The above working is not valid without proofs and I am unsure of how to approach them. How can I prove that $xy$ has $4$ factors and that it is not possible to get a multiple of $4$ by multiplying two prime numbers?    

2.    $x$ and $y$ are different prime numbers. Identify the numbers of the form $x y^4$ which have a coolness that is equal to an
  integer.

My attempt: I am not sure how to solve this one but I think listing the factors of $xy^4$ might help. 

3.    Prove that the square of any prime number $x$ is equal to the coolness of some integer.

My attempt: No idea other than just listing a couple of prime numbers and then squaring them to check if the result is equal to the coolness of some integer.  
Please help me solve the above problems.  
 A: The divisors of $xy$ when $x$, $y$ are distinct primes are $1,\  x,\  y,\  xy$.  If $x$, $y$ are distinct prime numbers then either both are odd, in which case their product is odd and therefore not a multiple of $4$, or one of them is $2$ and the other is odd, and the product of $2$ and an odd number is never divisible by $4$.
For your second problem, the divisors of $xy^4$ are $\underbrace{1,\ y,\ y^2,\ y^3,\ y^4}_\text{No $x$ appears here.},\ \  \underbrace{x,\ xy,\ xy^2,\ xy^3,\ xy^4}_\text{One $x$ divides each of these.}$.
A: This answers Question 3.
If $p\ge5$, then 
$${24p^2\over\tau(24p^2)}={24p^2\over\tau(2^3)\tau(3^1)\tau(p^2)}={24p^2\over4\cdot2\cdot3}=p^2$$
The factorization in the denominator doesn't hold for $p=2$ or $3$, so this leaves the problem of finding (small?) integers $m$ and $n$ such that
$${m\over\tau(m)}=4\quad\text{and}\quad{n\over\tau(n)}=9$$
A quick look at the OEIS finds that these are solved by $m=36$ and $n=108$, respectively. (The general inequality $\tau(N)\le2\sqrt N$, obtained by pairing each divisor $d\gt\sqrt N$ with its companion divisor $N/d\lt\sqrt N$, reduces the search for solutions to $N/\tau(N)=k$ to the range $1\le N\le(2k)^2$.  Note that $1\le36\le8^2$ and $1\le108\le18^2$.)
A: 
1.    Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y$ are different prime numbers.

You are correct in your thinking for this part:
As $x$ and $y$ are distinct primes, $xy$ has $4$ factors: $1, x, y, xy$. This means that $xy$ must be divisible by $4$. However, as $2$ is the only even prime number and cannot be used twice, this means that at least one of $x$ or $y$ must be odd. The product of a integer with an odd number will never yield a multiple of $4$, therefore coolness$(xy)$ will never equal a integer where $x$ and $y$ are distinct primes.

2.    $x$ and $y$ are different prime numbers. Identify the numbers of the form $x y^4$ which have a coolness that is equal to an
  integer.

$xy^4$ has $10$ factors: $1, x, y, y^2, y^3, y^4, xy, xy^2, xy^3, xy^4$. This means that for coolness$(xy^4)$ to be an integer, $xy^4$ must be divisible by $10$. Then $x$ and $y$ must multiple to a factor of $10$. The only prime numbers which serve this are $5$ and $2$. Therefore $x=2, y=5$, or $x=5, y=2$.

3.    Prove that the square of any prime number $x$ is equal to the coolness of some integer.

Given the prime $x\neq 3$, the integer number which has its coolness as the square of any prime number $x$, can be written $n(x) = 3^2x^2$. It has $3^2 = 9$ factors ($1$, $3$, $x$, $3x$, $3^2x$, $3x^2$, $3^2$, $x^2$, $3^2x^2$) and 
$$
\mbox{Coolness}(n(x)) = \frac{3^2x^2}{3^2} = x^2
$$
If $x=3$ then the number $n(x)=2^2 3^3 = 108$. In this case
$$
\mbox{Coolness}(108) = \frac{2^23^3}{2^2 3} = 3^2
$$
Therefore the square of any prime number $x$ is equal to the coolness of some integer.
I hope this helps you on your problems!
