A conjecture about the number of divisors of a natural number Conjecture:  

$\tau(n)\mid\tau(n^2)\iff$ $n$ is a perfect square and $\sqrt n=p^2s$,
  where $p$ is prime and $s$ is a non prime squarefree number  such that $\gcd(p,s)=1$.

$\tau(n)$ is the number of factors of $n$.
See this question on MSE:
Number theory problem on divisors!
It's tested for $n<10,000$.

If $n=p^4s^2$ with $p$ prime, $s$ is a product of $m\ge1$ distinct primes $q_1,\ldots,q_m$ and $p\nmid s$, then
$$\tau(n) =\tau(p^4)\tau(q_1^2)\cdots \tau(q_m^2)=5\cdot 3\cdots 3=5\cdot 3^m$$
and
$$\tau(n^2) =\tau(p^8)\tau(q_1^4)\cdots \tau(q_m^4)=9\cdot 5\cdots 5=3^2\cdot 5^m.$$
It follows that for such $n$, we have $\tau(n)\mid\tau(n^2)$ only if $m\le 2$.
This allows us to find an explicit counterexample:
Let $n=420^2=176400$. 
Then $$\begin{align}\tau(n)&=\tau(2^4\cdot 3^2\cdot 5^2\cdot 7^2)=135\\\tau(n^2)&=\tau(2^8\cdot 3^4\cdot 5^4\cdot 7^4)=1125=8\cdot 135+45.\end{align}$$

It is also not hard to find counterexamples to the other direction: 
Let $$n=29674142746122490321305600000000000000000000.$$
Then $$\begin{align}
\tau(n)=\tau(2^{40}3^{24}5^{20}7^211^213^2)&=41\cdot25\cdot 21\cdot 3^3\\
\tau(n^2)=\tau(2^{80}3^{48}5^{40}7^411^413^4)&=81\cdot49\cdot 41\cdot 5^3=35\cdot \tau(n).\\
\end{align}$$
