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.

#NumberTheory

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Comments

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 - 5 years, 4 months ago

汶良林 - 5 years, 4 months ago

for example:

x = 3, y = 4, z = 5

a = 20, b = 15, c = 40

汶良林 - 5 years, 4 months ago

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

Pi Han Goh - 5 years, 5 months ago

Thank you very much!

Ananth Jayadev - 5 years, 5 months ago

Cool!

Kamalpreet Singh - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$