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I was given this problem but I have no idea how to prove it. Any ideas?

Let $n$ be a positive integer. The number of positive integral solutions to $\frac{1}{n}$ = $\frac{1}{x}$ + $\frac{1}{y}$ is $\tau(n^2)$ where $\tau(n)$ is the number of divisors of $n$.

Henry Swanson
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Chris
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2 Answers2

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We set out to show that the equation $\frac{1}{n} = \frac{1}{x} + \frac{1}{y}$ (for nonzero $n$, $x$, and $y$) is equivalent to $(x-n)(y-n)=n^2$:

Multiply by $nxy$ to get $xy=ny+nx$. Add $n^2$ to both sides and subtract $nx+ny$ from both sides to get $xy-nx-ny+n^2=n^2$. Then the LHS can be factored, yielding $(x-n)(y-n)=n^2$ as claimed. Now you just need to reason about what positive integer values of $x$ and $y$ are possible.

Dustan Levenstein
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  • im struggling to do that. How do you do it? – Chris Jul 15 '14 at 20:18
  • @Chris see my edit. Note also that I changed $(n-x)(n-y)$ to $(x-n)(y-n)$, which makes a little more sense contextually. – Dustan Levenstein Jul 15 '14 at 20:24
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    Another way of doing this - and a natural way of arguing from scratch, is to note that both fractions on the right-hand side are positive, so each is less than $\frac 1n$ and $x,y\gt n$. So set $a=x-n, b=y-n$. Then $\frac 1n=\frac 1{a+n}+\frac 1{b+n}$ and clearing fractions gives $(a+n)(b+n)=n(b+n)+n(a+n)$ which reduces to $ab=n^2$ and $a$ can be chosen to be any positive factor of $n$. – Mark Bennet Jul 15 '14 at 21:29
  • @MarkBennet nice! I probably did something similar to that when I first stumbled on this solution years ago. – Dustan Levenstein Jul 15 '14 at 21:34
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This is because the equation.

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

If the square lay on multipliers. $n^2=ks$

Then the solution can be written.

$$x=n+s$$

$$y=n+k$$

Although in the General case can be any character. Though it is necessary to mention another solution.

For the equation: $$\frac{1}{X}+\frac{1}{Y}=\frac{1}{A}$$

You can write a simple solution if the number on the decomposition factors as follows: $$A=(k-t)(k+t)$$

then: $$X=2k(k+t)$$ $$Y=2k(k-t)$$ or: $$X=2t(k-t)$$ $$Y=-2t(k+t)$$

Although these formulas give one and the same solution.

individ
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