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?

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.

2 years, 11 months 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

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

- 2 years, 10 months ago

- 2 years, 10 months ago

for example:

x = 3, y = 4, z = 5

a = 20, b = 15, c = 40

- 2 years, 10 months ago

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

- 2 years, 11 months ago

Thank you very much!

- 2 years, 11 months ago

Cool!

- 2 years, 10 months ago