Prove that N is a perfect square

Question

The integer $N$ is positive. There are exactly $2005$ ordered pairs $(x, y)$ of positive integers satisfying:

$\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{1}{N}$

Prove that $N$ is a perfect square.

Please explain as if I don't know anything about the topic or please tell me the math topics i need to know to solve this.

Thank you.

Answer 1

Rewriting the equation by expressing $x$ in terms of $y$ and $N$, $$ x=N+\frac{N^2}{y-N} $$ Since $x$ and $ N$ are integers, therefore $y-N|N^2$. There are $2005$ such ordered pairs $(x,y)$ implies that $N^2$ has $2005$ divisors. Let the prime factorisation of $N$ be $$ N=p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_n^{\alpha_n}, \textrm{ then }N^2=p_1^{2\alpha_1} p_2^{2\alpha_2} \cdots p_n^{2\alpha_n}, $$ and the number of divisor of $N^2$ is $2005=\left(2 \alpha_1+1\right)\left(2 \alpha_2+1\right) \cdots\left(2 \alpha_n+1\right) =5 \times 401.$ Consequently, $$ \begin{aligned} & 2 \alpha_1+1=5 \text { and } 2 \alpha_2+1=401 \\ \Rightarrow \quad & \alpha_1=2 \text { and } \alpha_2=200 \end{aligned} $$ Hence $ N=p_1^2 p_2^{200}=\left(p_1 p_2^{100}\right)^2 $ is a perfect square.

In general, if $N^2$ has odd numbers of divisors, then $N$ must be a perfect square.