Can you prove that if $\mid D_n \mid = m$, where $m$ is odd, then $n$ is a perfect square.

Question

$n,m \in \mathbb{N}$ and $D_n$ is the set that contains all divisors of $n$, including $1$ and $n$.

My Claim:

if $\mid D_n \mid=$ an odd number $\iff n$ is a perfect square. and indeed Using Ordinal Numbers.the number $$\frac{\mid D_n \mid+1}{2}.$$ in the set $D_n$: is the $\sqrt{n}$.

I didn't found a counterexample so far. if it exists just post it and I will delete the question.

Thank You.

Answer 1

Note that if $d$ is a divisor of $n$, so is $\frac{n}{d}$. Also we have that $d=\frac{n}{d}$ iff $d^2=n$. Hence if $n$ is not a square then all divisors come in pairs of the form $(d,\frac{n}{d})$, so $|D_n|$ is even. On the other hand if $n$ is a square then we get another divisor that is not in such a pair, namely $d=\sqrt{n}$.
Another way to see it is to look at the prime factorization: If $n=p_1^{a_1}\dots p_k^{a_k}$, then $|D_n|=(a_1+1)\dots(a_k+1)$, so $D_n$ is odd iff all $a_i$ are even iff $n$ is a square.