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Let $S(n)$ denote the number of ordered pairs $(x,y)$ satisfying $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n>1$ and $x,y,nāˆˆN$

1) Find the value of $S(6)$.

2) Show that if $n$ is prime then $S(n)=3$ always. I couldn't get a correct method to proceed.

Martin Sleziak
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Aditya Kumar
  • 1,491

2 Answers2

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Note that for $x,y,n\in\mathbb N$, we have $$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}\iff ny+nx=xy\iff (x-n)(y-n)=n^2.$$

mathlove
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Since $x,y$ are positive integers, we have $x>n$ and $y>n$. Write $x=n+a$ and $y=n+b$. Substitute this and you get $n^2=a\cdot b$. Retracing the steps, we find that $n^2$ is written as a product of two numbers say, $p$ and $q$, then $$\frac{1}{n+p}+\frac{1}{n+q}=\frac{1}{n}$$ Thus the number of solutions of our original equation for a given $n$ is the same as the divisors of $n^2$. Let $n={p_1}^\alpha_1{p_2}^\alpha_2\cdots {p_k}^\alpha_k.$ Therefore you want the solutions of the equation $$(2\alpha_1+1)(2\alpha_2+1)\cdots (2\alpha_k+1)=6.$$(From the formula of the number of divisors). I hope you can take it from here.