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I'm working on a function $\lambda(\xi)$ that takes in input an intger $\xi$ and calculate for how many values of $x\in N$, with $0\leq x < \xi$, the ratio: $$\psi = \frac{\xi x}{\xi -x}$$ is an integer $(\psi \in N)$.

Here there is a graph of $\lambda(\xi)$ for $2\leq \xi \leq113$, created in Excel: enter image description here

I have tried in a lot of diferent ways (supposing for example $\xi$ is odd and $x$ is even), but I can't faind a solution. Any idea?

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    $\begingroup$ This is equal to the number of positive integer divisors of $\xi^2$. $\endgroup$
    – Peter Foreman
    Jul 20, 2019 at 13:02
  • $\begingroup$ Why it's equal to the number of divisors of $\xi^2$? $\endgroup$
    – Matteo
    Jul 20, 2019 at 13:04
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    $\begingroup$ Because $\psi = \frac{\xi^2}{\xi-x}-\xi\in\mathbb{N}\iff \frac{\xi^2}{\xi-x}\in\mathbb{N}$ and the latter is true only when $(\xi-x)|\xi^2$ i.e. once for each divisor of $\xi^2$. $\endgroup$
    – Peter Foreman
    Jul 20, 2019 at 13:08
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    $\begingroup$ I think it's not correct because if you put $\xi = 60$, $\xi^2$ have 45 not between 20 and 25 $\endgroup$
    – Matteo
    Jul 20, 2019 at 13:18
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    $\begingroup$ Sorry, we need to emit the divisors which are greater than $\xi$ so the value is equal to $\frac{d(\xi^2)+1}2$ where $d(n)$ is the number of divisors of $n$. For $\xi=60$ this gives $\frac{45+1}{2}=23$. $\endgroup$
    – Peter Foreman
    Jul 20, 2019 at 13:27

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To prove the assertion of Peter Foreman in its comment, that's $$\lambda(\xi)=\frac{d(\xi^2)+1}2$$ let consider the sets \begin{align*} D&=\{d>0:d|\xi^2\}\\ L&=\{d\in D:d\leq\xi\}\\ U&=\{d\in D:d>\xi\} \end{align*} and the function \begin{align*} &\varphi:D\to D&&d\mapsto\xi^2/d \end{align*} Then \begin{align} D&=L\cup U& L\cap U&=\varnothing\\ &\text{$\varphi$ is bijective}&\varphi[U]&=L-\{\xi\} \end{align} Consequently, $d(\xi^2)=|D|=|L|+|U|$ and $|U|=|L|-1$ from which $$\lambda(\xi)=|L|=\frac{|D|+1}2=\frac{d(\xi^2)+1}2$$ thus proving the assertion.

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