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Variation on Pythagorean Theorem

I already know about the Pythagorean Theorem, where \(a^2 + b^2 = c^2\). Now I'm wondering if there can be a solution where \(\frac {1}{a^2} + \frac {1} {b^2} = \frac {1}{c^2}\). 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
11 months, 1 week ago

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We can put it into the form of

\[a^2+b^2 = (\frac{ab}{c})^2\]

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

\[(a, b, c) = (15x, 20x, 12x), (175x, 600x, 168x)\] Kay Xspre · 10 months, 1 week ago

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汶良 林 · 10 months, 1 week ago

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@汶良 林 for example:

x = 3, y = 4, z = 5

a = 20, b = 15, c = 40 汶良 林 · 10 months, 1 week ago

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\(a = 20, b =15, c= 12 \) Pi Han Goh · 11 months, 1 week ago

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@Pi Han Goh Thank you very much! Ananth Jayadev · 11 months, 1 week ago

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Cool! Kamalpreet Singh · 10 months, 1 week ago

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