Prove that $n$ is the median of all positive divisor of $n^2$ for any positive integer $n$.

Question

For all positive integer $n$, we have $\tau(n^2)$ is odd. Hence, median of all positive divisor of $n^2$ is located at $\frac{\tau(n^2)+1}{2}$th.

Now, for $n=4$, then $n^2=4^2=16$ and $\tau(16)=5$, i.e., $1,2,4,8,$ and $16$. Here, the median is $4$ which located at the $\frac{5+1}{2} = \frac{\tau(4^2)+1}{2} = 3$rd position.

How to generelize this? Any ideas? Many thanks in advanced.

Answer 1

Let $n$ be a positive integer. Also, let $A_n$ and $B_n$ be the set of all positive divisors of $n^2$ which less than $n$ and which greater than $n$, respectively. Since $\tau(n^2)$ is odd, it suffices to prove that $|A_n| = |B_n|$.

Define a mapping $f:A_n \to B_n$ by $x \mapsto \frac{n^2}{x}$ for all $x \in A_n$. Let $x,y \in A_n$ for which $f(x)=f(y)$. Then $\frac{n^2}{x} = \frac{n^2}{y}$ which implies $x=y$. Hence, $f$ is injective. Let $y \in B_n$. Then $y>n$ and so, $\frac{1}{y} < \frac{1}{n}$ which implies $\frac{n^2}{y} < \frac{n^2}{n} = n$. Since $y \in B_n$, then $y$ is a divisor of $n^2$ and so is $\frac{n^2}{y}$. Hence, for all $y \in B_n$, there is $x=\frac{n^2}{y} \in A_n$ such that $f(x)=y$. Hence, $f$ is surjective. Thus, $f$ is bijective and so, $|A_n| = |B_n|$.

Therefore, $n$ is the median of all positive divisors of $n^2$, for any positive integer $n$. Q.E.D.

EXAMPLE: Let $n=6$, then $n^2=36$ and $\tau(n^2)=9$, that is, $1,2,3,4,6,9,12,18,$ and $36$. Here, $A_6=\{1,2,3,4\}$ and $B_6=\{9,12,18,36\}$ and clearly, $|A_6|=|B_6|$. On the other view, through $f$, we can map one-one all elements of $A_6$ onto $B_6$ as follows: \begin{align*} 1 &\mapsto 36 \\ 2 &\mapsto 18 \\ 3 &\mapsto 12\\ 4 &\mapsto 9. \end{align*}

Hope it helps.

Answer 2

$xy = n^2$. For every $x$ there's a divisor pair $y$.

Distribution of divisors: $1$ to $n^2$

What happens when $x = y = n$? $xy = n \cdot n = n^2$

For every $x < n$ there's $y > n$ (divisor pairs) and notice they're on opposite sides of $n$. In other words, there are as many divisors $< n$ as there are divisors $> n$. We know that $n$ is its own divisor pair i.e. when you list down the divisors of $n^2$, you list it down only once, but even if you do the honorable thing and write down both instances of $n$, we see that $\frac{n + n}{2} = n$.

$xy = n^2$ implies that $x = y = n$ is some kinda threshold or, formally a line of symmetry, beyond which, for any divisor pair $a$ and $b$, $ab = n^2$ becomes $ba = n$

$xy = 4^2 = 16, x = y$ is the line of symmetry. $(4, 4)$ splits the curve into identical halves" />