Tau Tau?

Let's consider the statements one by one.

$[1]$ is not true. The number of integer solutions $(x, y)$ to $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$ is actually $2\tau(n^2)-1$ .

Notice that the equation can be re-written as $(x-n)(y-n)=n^2 \cdots (*)$ . That means for every divisor [positive and negative] of $n^2$ , we'll get a unique ordered pair $(x,y)$ that satisfies our equation. The number of divisors [positive and negative] $n^2$ has is $2\tau(n^2)$ . However, notice that $(0, 0)$ satisfies equation $(*)$ , but it doesn't satisfy our original equation. So, our final count is $2\tau (n^2)-1$ .

$[2]$ is not true because $\tau(n)$ is an odd number when $n$ is a perfect square. If $k$ divides $n$ , then $\frac{n}{k}$ also divides $n$ . This sentence is actually equivalent to saying that the positive divisors of $n$ come in pairs. However, if for some $k$ , $k=\frac{k}{n}$ , $n$ will have an odd number of positive divisors. That happens when $n$ is a perfect square and then $\tau (n)$ yields an odd number.

$[3]$ isn't true as well. Here's a quick counter-example. The positive divisors of $2$ are $1$ and $2$ . So, $\tau (2)=2$ . So, $\tau(n)$ is not necessarily less than $n$ .

So, none of these statements are correct.

All of them fail for $n=1$ .

Clearly, $x+y=xy$ in integers means $x|y$ and $y|x$ , so $x=y(=2)$ .

$\tau(1)=1$ is odd and equal to $n=1$ .