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I've tried too much but unable to find its integral solutions. I've tried to solve it by assuming $\frac 1x$, $\frac 1y$ and $\frac 1z$ as sides of a right angled triangle and applying pythagoras theorem.

sgrmshrsm7
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  • So you enforce $x,y,z \in \mathbb{Z}$ ? Otherwise $x = 1/3,\ y = 1/4,\ z = 1/5$ works as well as all the Pythagorean triples. – Zubzub Mar 03 '17 at 10:51
  • Let $a=1/x$, $b=1/y$ and $c=1/z$. Integral solutions to $a^2+b^2=c^2$ are known as Pythagorean Triples; see https://en.wikipedia.org/wiki/Pythagorean_triple – mlc Mar 03 '17 at 10:55
  • @mle You are not completely wrong. The equation actually is related to pythagorean triples (See my answer) – Peter Mar 03 '17 at 11:25
  • http://math.stackexchange.com/questions/1634944/integer-solutions-for-frac1x2-frac1y2-frac1z2 – individ Mar 03 '17 at 12:43

1 Answers1

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We can assume $x,y,z$ to be positive integers. Then, the given equation is equivlane to $$x^2z^2+y^2z^2=x^2y^2$$

The solutions with $1\le x,y,z\le 100$ are :

? for(x=1,100,for(y=1,100,for(z=1,100,if(1/x^2+1/y^2==1/z^2,print(x," ",y," ",z)
))))
15 20 12
20 15 12
30 40 24
40 30 24
45 60 36
60 45 36
60 80 48
75 100 60
80 60 48
100 75 60

If you have a pythagorean triple $(a,b,c)$, then, $(bc,ac,ab)$ solves the given equation, so there are infinite many solutions.

Peter
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