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 gelatine1 March 18th, 2014 01:22 PM

diophantine reciprocals: amount of solutions

Hello,

I was reading about solving an equation of the form $\frac {1}{x}+\frac {1}{y}=\frac {1}{z}$ on this website: http://www.cut-the-knot.org/arithmet...shtml#solution
I understood the derivation of the solutions but now I am wondering how to quickly determine the amount of distinct solutions of the equation with a given z.

the solutions are $x=km(m+n) \;\; y=kn(m+n) \; \; z=kmn$
Is there someone who could help me out with this ?

 mathbalarka March 19th, 2014 12:45 AM

Re: diophantine reciprocals: amount of solutions

It's a good exercise for you to prove that there are $\sigma_0(z^2)$ solutions in total. Now subtract the ones you get modulo the symmetric equivalence (x, y) ~ (y, x).

 gelatine1 March 19th, 2014 05:58 AM

Re: diophantine reciprocals: amount of solutions

Thanks for your reply. I have no clue how to even start the proof of that. I am not quite familiar with those. But I think it is interesting so maybe you could give me a hint on how to start the proof?

 mathbalarka March 19th, 2014 01:48 PM

Re: diophantine reciprocals: amount of solutions

OK, here's a method :

$\frac1{x} + \frac1{y}= \frac1{z}$

Now let $(x, y)= k$ so that this is

$\frac1{k} \left ( \frac1{a} + \frac1{b} )= \frac1{z}$

Which is, by rearranging,

$z( a + b)= kab$

We note that $(a, b)= 1$, thus $a$ divides $z$, i.e., $z= a \cdot n$

$n ( a + b)= kb$

Similarly, $b$ divides $n$. Hence the number of solutions can precisely be found by counting

$\sum_{a | z} \sum_{b | z/a} 1$

Excercise : Do the counting!

 mathbalarka March 23rd, 2014 01:24 AM

Re: diophantine reciprocals: amount of solutions

A sneaky guy in MSE did some nonobvious trick there which gives essentially a 1-line proof :

$\frac1{n+d}+\frac1{n+\frac{n^2}{d}}= \frac1{n}$

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