1

Given

${1 \over x} + {1 \over y} = {1 \over n}$

where $x,y,n$ are positive integers.

Find all possible values of $n$ such that there are only $5$ ordered pairs of $(x,y)$ which satisfies the equation.

I tried taking LCM and trying to factorize it, but I am unable to do so. Please help.

PS- Do not use calculus in your reasoning or solution as I am preparing for an Olympiad which does not allow calculus. You can still upload an answer involving calculus for other's to understand.

G. Chiusole
  • 5,486
Jayant Jha
  • 552
  • Ohhk I am sorry. – Jayant Jha Aug 12 '19 at 07:55
  • Yes some ideas can be taken from there and the factorisation is given as well. Thanks. But it would be great if someone would provide a solution to this question. – Jayant Jha Aug 12 '19 at 07:58
  • The answer to the original of this "duplicate" is a "hint". I think OP meant to ask for the clarification of that hint since they were unable to solve the problem given the "hint". The same answer was again given here, and OP asking for a clarification. Based on this, I would not close this as a duplicate, but rather it be given a more detailed answer. – Vepir Aug 12 '19 at 08:49

3 Answers3

3

If $${1 \over x} + {1 \over y} = {1 \over n}$$

Then $$(x-n)(y-n)=n^2\, \qquad \forall xy \neq \, 0$$

  • So what could be the values of $n$ such that there are only there are only 5 ordered pairs of $(x,y)$ ? – Jayant Jha Aug 12 '19 at 08:06
  • Every solution to $ab=n^2$ would lead to a unique solution to the original problem $x=a+n, y=b+n$. So you would just need to identify the $n$ such that $d(n^2)=5$. –  Aug 12 '19 at 09:30
3

Let $d=(x,y)$ and take $a,b$ such that $x=ad$ and $y=bd$ (so that $(a,b)=1$). Then $$\frac{1}{da}+\frac{1}{db}=\frac{1}{n}$$ or $$n(a+b)=dab$$ $(a+b,a)=(a+b,b)=1$, so the solutions to this problem in terms of $n$ are generated from precisely the ways to choose $a$ and $b$ such that (a,b)=1 and $ab\mid n$.

If $n$ is prime, there are three solutions: (a,b) = (n, 1), (1,n), or (1,1). If n=$p^2$ for some prime p, there are five solutions: ($p^2$,1), (1,$p^2$), (p,1), (1,p),(1,1). The interested reader could go on to calculate that $n=p^k$ has $2k+1$ solutions and $n$ having two distinct prime factors has at least nine solutions.

Therefore, the problem has five solutions if and only if $n$ is the square of a prime.

1

Some months ago I was thinking about the same problem and I have written a correct 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

This function can be rewritten (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 $ξ^2$) as $\lambda(\xi)=\frac{d(\xi^2)+1}2$, where $d(n)$ is the number of divisors of $n$.

Matteo
  • 6,581