Why every second even number have same number of even and odd Divisors. Question: I was looking for natural numbers with same number of odd and even Divisors.Then I observed that, every second number in the list $2,4,6,8,10,12,14,...$ is such a number. Following list shows that, $2,6,10,14,...$ are even natural numbers that have same number of odd and even Divisors.
$2 \qquad 1 \qquad 1\quad\ast$
$4 \qquad 1 \qquad 2$
$6 \qquad 2 \qquad 2\quad\ast$
$8 \qquad 1 \qquad 3$
$10 \qquad 2 \qquad 2\quad\ast$
$12 \qquad 2 \qquad 4$
$14 \qquad 2 \qquad 2\quad\ast$
$16 \qquad 1 \qquad 4$
$18 \qquad 3 \qquad 3\quad\ast$
.........
How we can prove this in general? Is there is something special about this?
Note: I just copied the above list from  answer given here, https://math.stackexchange.com/a/48887/168676
 A: First of all, what property does every 2nd even number have? if we look at 2, 6, 10, 14, etc we see that they are all divisible by 2, but not by 4. The so the even numbers which dont have that property are all divisible by 4.
can we prove that all numbers with the property are going to be all numbers which are even and not divisible by 4? Yes, it isn't too difficult so I would recommend you give it a try, use the fundamental theorem of arithmetic. But if you dont want to try it, here's an outline of a proof:
Lets start by seeing why every number which is divisible by 2 but not by 4 will have this property.
Let $n\in\mathbb{N}$ such that $n$ is even but not divisible by 4. Using the fundamental theorem of arithmetic, we know that
$n=p_1^{a_1}\cdot p_2^{a_2}\cdot...\cdot p_k^{a_k}$
where each $p_i$ is a distinct prime number, and $a_i$ is some natural number. Since we know $n$ is even, we know one of those primes is 2, so WLOG let $p_1=2$ then we also know that $a_1=1$ as otherwise $n$ would be divisible by 4.
Now, how many numbers divide n? well if $m$ is a natural number that divides $n$, then $k=p_1^{b_1}\cdot p_2^{b_2}\cdot...\cdot p_k^{b_k}$
where $0\leq b_i\leq a_i$.
At this point it should be easy to see that the number of possible dividers will depend on the number of ways to assign all of the $b_i$s in particular, there are 2 options for $b_1$, and $a_i+1$ different options for each $b_i$. Meaning the total number of dividers will just be $(a_1+1)\cdot (a_2+1)\cdot...\cdot(a_k+1)$ However, for exactly half of these, $b_1=0$ which will be an odd divisor, and in the other half $b_1=1$ which will be an even divisor.
It should now be easy to see why odd numbers, and numbers divisible by 4 will not have the same number of even factors as odd factors.
