Let $a=43120$ How many positive divisors does a have? I am doing a review assignment and I'm stuck on this problem.
a) How many positive divisors does $a$ have? I got $60$
b) How many positive integers less than $a$ are relatively prime to $a$? I got $720$
c) What is the smallest positive integer $m$ such that $a^2m$ is a cube?
d) list all positive divisors $b$ of a for which a divides $b^2$ is also true.
Any help and advice would be greatly appreciated. Thank you for your time and help!
 A: First, we factor $$43120=2^45^17^211^1$$
Solutions to a,  c, d, are obtained from the exponents $u=(4,1,2,1)$.  
a) A positive divisor is a $4$-vector of nonnegative integers majorized by $u$.  That is $$\{(a,b,c,d):0\le a\le 4, 0\le b\le 1, 0\le c\le 2, 0\le d\le 1\}$$
There are $(4+1)\times(1+1)\times (2+1)\times (1+1)=60$ positive divisors.  
c) $a^2$ corresponds to $(8,2,4,2)$.  To make a cube, each must be a multiple of $3$, i.e. $(9,3,6,3)$.  Subtracting, we get $(1,1,2,1)$, i.e. $2^15^17^211^1=m$.
d) For $a$ to divide $b^2$, $u$ must be majorized by the $4$-vector of $b^2$.  For example, if $b$ corresponds to $(5,1,2,2)$, then $b^2$ corresponds to $(10,2,4,4)$, which majorizes $u$, since it is larger in each component.  I leave to you to count how many there are of these.
b) You seek $\phi(43120)$, the Euler totient.  This is multiplicative, so you want $\phi(2^4)\phi(5)\phi(7^2)\phi(11)$.
A: First, note that $43120 = 2^\color{red}{4}\cdot 5^\color{blue}{1}\cdot 7^\color{green}{2}\cdot 11^\color{purple}{1}$.
Part A
The number of positive integer divisors is the product of one plus each exponent in the prime factorization.  That is, 
$$d(43120) = \color{red}{(4+1)}\color{blue}{(1+1)}\color{green}{(2+1)}\color{purple}{(1+1)} = (5)(2)(3)(2) = 60$$
Part B
The number of positive integers coprime to $43120$ can be found using Euler's Totient function, $\phi(n)$.  Since $\phi$ is multiplicative for coprime integers:
$$\phi(43120) = \phi(2^4)\phi(5)\phi(7^2)\phi(11)$$
Also, $\phi(p^n) = p^{n-1}(p-1)$ for prime integers $p$.  Then:
$$\begin{align}\phi(43120) &= \left[2^3(2-1)\right]\left[5-1\right]\left[7^1(7-1)\right]\left[11-1\right]\\
&= 8\cdot4\cdot 42\cdot 10\\
&= 13440
\end{align}$$
Part C
We want all exponents of $a^2$ to be multiples of three, and we want the smallest such exponents.
$$\left(2^\color{red}{4}\cdot 5^\color{blue}{1}\cdot 7^\color{green}{2}\cdot 11^\color{purple}{1}\right)^2 = 2^\color{red}{8}\cdot 5^\color{blue}{2}\cdot 7^\color{green}{4}\cdot 11^\color{purple}{2}$$
Well, it is easy to see that we need to add $1$ to the first exponent, $1$ to the second, $2$ to the third and $1$ to the fourth.  Thus, our integer $m$ is:
$$m= 2^\color{red}{1}\cdot 5^\color{blue}{1}\cdot 7^\color{green}{2}\cdot 11^\color{purple}{1}$$
Part D
This is a similar exponent-related trick as in part c if I'm thinking it through correctly.  I'll leave this one as an exercise to the reader. :)
