Variation on Pythagorean Theorem

I already know about the Pythagorean Theorem, where a2+b2=c2a^2 + b^2 = c^2. Now I'm wondering if there can be a solution where 1a2+1b2=1c2\frac {1}{a^2} + \frac {1} {b^2} = \frac {1}{c^2}. Is this possible?

Note: If this question has already been asked, please show me a link where this question was asked so I can look at that.

If this is impossible tell me how it is impossible to get a solution (or solutions) for this equation.

If this is possible tell me how it is possible to get a solution (or solutions) for this equation.

Note by Ananth Jayadev
3 years, 9 months ago

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We can put it into the form of

a2+b2=(abc)2a^2+b^2 = (\frac{ab}{c})^2

of which one answer is c=aba2+b2c = \frac{ab}{\sqrt{a^2+b^2}}. Some answer of (a, b, c) for natural number xx is

(a,b,c)=(15x,20x,12x),(175x,600x,168x)(a, b, c) = (15x, 20x, 12x), (175x, 600x, 168x)

Kay Xspre - 3 years, 8 months ago

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a=20,b=15,c=12a = 20, b =15, c= 12

Pi Han Goh - 3 years, 9 months ago

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Thank you very much!

Ananth Jayadev - 3 years, 9 months ago

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汶良 林 - 3 years, 8 months ago

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for example:

x = 3, y = 4, z = 5

a = 20, b = 15, c = 40

汶良 林 - 3 years, 8 months ago

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Cool!

Kamalpreet Singh - 3 years, 8 months ago

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