My Math Forum diophantine reciprocals: amount of solutions

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 March 18th, 2014, 01:22 PM #1 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 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 ?
 March 19th, 2014, 12:45 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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).
 March 19th, 2014, 05:58 AM #3 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 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? Thanks in advance
 March 19th, 2014, 01:48 PM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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!
 March 23rd, 2014, 01:24 AM #5 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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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